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Section 4.4
Curve Sketching
V63.0121.041, Calculus I
New York University
November 17, 2010
Announcements
Quiz 4 this week in recitation on 3.3, 3.4, 3.5, 3.7
There is class on November 24
. . . . . .
. . . . . .
Announcements
Quiz 4 this week in
recitation on 3.3, 3.4, 3.5,
3.7
There is class on
November 24
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 2 / 55
. . . . . .
Objectives
given a function, graph it
completely, indicating
zeroes (if easy)
asymptotes if applicable
critical points
local/global max/min
inflection points
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 3 / 55
. . . . . .
Why?
Graphing functions is like
dissection
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 4 / 55
. . . . . .
Why?
Graphing functions is like
dissection … or diagramming
sentences
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 4 / 55
. . . . . .
Why?
Graphing functions is like
dissection … or diagramming
sentences
You can really know a lot about
a function when you know all of
its anatomy.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 4 / 55
. . . . . .
The Increasing/Decreasing Test
Theorem (The Increasing/Decreasing Test)
If f′
> 0 on (a, b), then f is increasing on (a, b). If f′
< 0 on (a, b), then f
is decreasing on (a, b).
Example
Here f(x) = x3
+ x2
, and f′
(x) = 3x2
+ 2x.
..
f(x)
.
f′
(x)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 5 / 55
. . . . . .
Testing for Concavity
Theorem (Concavity Test)
If f′′
(x) > 0 for all x in (a, b), then the graph of f is concave upward on
(a, b) If f′′
(x) < 0 for all x in (a, b), then the graph of f is concave
downward on (a, b).
Example
Here f(x) = x3
+ x2
, f′
(x) = 3x2
+ 2x, and f′′
(x) = 6x + 2.
..
f(x)
.
f′
(x)
.
f′′
(x)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 6 / 55
. . . . . .
Graphing Checklist
To graph a function f, follow this plan:
0. Find when f is positive, negative, zero,
not defined.
1. Find f′
and form its sign chart. Conclude
information about increasing/decreasing
and local max/min.
2. Find f′′
and form its sign chart. Conclude
concave up/concave down and inflection.
3. Put together a big chart to assemble
monotonicity and concavity data
4. Graph!
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 7 / 55
. . . . . .
Outline
Simple examples
A cubic function
A quartic function
More Examples
Points of nondifferentiability
Horizontal asymptotes
Vertical asymptotes
Trigonometric and polynomial together
Logarithmic
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 8 / 55
. . . . . .
Graphing a cubic
Example
Graph f(x) = 2x3
− 3x2
− 12x.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 9 / 55
. . . . . .
Graphing a cubic
Example
Graph f(x) = 2x3
− 3x2
− 12x.
(Step 0) First, let’s find the zeros. We can at least factor out one power
of x:
f(x) = x(2x2
− 3x − 12)
so f(0) = 0. The other factor is a quadratic, so we the other two roots
are
x =
3 ±
√
32
− 4(2)(−12)
4
=
3 ±
√
105
4
It’s OK to skip this step for now since the roots are so complicated.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 9 / 55
. . . . . .
Step 1: Monotonicity
f(x) = 2x3
− 3x2
− 12x
=⇒ f′
(x) = 6x2
− 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
Step 1: Monotonicity
f(x) = 2x3
− 3x2
− 12x
=⇒ f′
(x) = 6x2
− 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
.. x − 2..
2
.
x + 1
..
−1
.
f′
(x)
.
f(x)
..
2
..
−1
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
Step 1: Monotonicity
f(x) = 2x3
− 3x2
− 12x
=⇒ f′
(x) = 6x2
− 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
.. x − 2..
2
.− . −. +.
x + 1
..
−1
.
f′
(x)
.
f(x)
..
2
..
−1
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
Step 1: Monotonicity
f(x) = 2x3
− 3x2
− 12x
=⇒ f′
(x) = 6x2
− 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
.. x − 2..
2
.− . −. +.
x + 1
..
−1
.
+
.
+
.
−
.
f′
(x)
.
f(x)
..
2
..
−1
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
Step 1: Monotonicity
f(x) = 2x3
− 3x2
− 12x
=⇒ f′
(x) = 6x2
− 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
.. x − 2..
2
.− . −. +.
x + 1
..
−1
.
+
.
+
.
−
.
f′
(x)
.
f(x)
..
2
..
−1
.
+
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
Step 1: Monotonicity
f(x) = 2x3
− 3x2
− 12x
=⇒ f′
(x) = 6x2
− 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
.. x − 2..
2
.− . −. +.
x + 1
..
−1
.
+
.
+
.
−
.
f′
(x)
.
f(x)
..
2
..
−1
.
+
.
−
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
Step 1: Monotonicity
f(x) = 2x3
− 3x2
− 12x
=⇒ f′
(x) = 6x2
− 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
.. x − 2..
2
.− . −. +.
x + 1
..
−1
.
+
.
+
.
−
.
f′
(x)
.
f(x)
..
2
..
−1
.
+
.
−
.
+
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
Step 1: Monotonicity
f(x) = 2x3
− 3x2
− 12x
=⇒ f′
(x) = 6x2
− 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
.. x − 2..
2
.− . −. +.
x + 1
..
−1
.
+
.
+
.
−
.
f′
(x)
.
f(x)
..
2
..
−1
.
+
.
−
.
+
.
↗
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
Step 1: Monotonicity
f(x) = 2x3
− 3x2
− 12x
=⇒ f′
(x) = 6x2
− 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
.. x − 2..
2
.− . −. +.
x + 1
..
−1
.
+
.
+
.
−
.
f′
(x)
.
f(x)
..
2
..
−1
.
+
.
−
.
+
.
↗
.
↘
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
Step 1: Monotonicity
f(x) = 2x3
− 3x2
− 12x
=⇒ f′
(x) = 6x2
− 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
.. x − 2..
2
.− . −. +.
x + 1
..
−1
.
+
.
+
.
−
.
f′
(x)
.
f(x)
..
2
..
−1
.
+
.
−
.
+
.
↗
.
↘
.
↗
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
Step 1: Monotonicity
f(x) = 2x3
− 3x2
− 12x
=⇒ f′
(x) = 6x2
− 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
.. x − 2..
2
.− . −. +.
x + 1
..
−1
.
+
.
+
.
−
.
f′
(x)
.
f(x)
..
2
..
−1
.
+
.
−
.
+
.
↗
.
↘
.
↗
.
max
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
Step 1: Monotonicity
f(x) = 2x3
− 3x2
− 12x
=⇒ f′
(x) = 6x2
− 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
.. x − 2..
2
.− . −. +.
x + 1
..
−1
.
+
.
+
.
−
.
f′
(x)
.
f(x)
..
2
..
−1
.
+
.
−
.
+
.
↗
.
↘
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 6x2
− 6x − 12
=⇒ f′′
(x) = 12x − 6 = 6(2x − 1)
Another sign chart: .
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 11 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 6x2
− 6x − 12
=⇒ f′′
(x) = 12x − 6 = 6(2x − 1)
Another sign chart: ..
f′′
(x)
.
f(x)
..
1/2
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 11 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 6x2
− 6x − 12
=⇒ f′′
(x) = 12x − 6 = 6(2x − 1)
Another sign chart: ..
f′′
(x)
.
f(x)
..
1/2
.
−−
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 11 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 6x2
− 6x − 12
=⇒ f′′
(x) = 12x − 6 = 6(2x − 1)
Another sign chart: ..
f′′
(x)
.
f(x)
..
1/2
.
−−
.
++
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 11 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 6x2
− 6x − 12
=⇒ f′′
(x) = 12x − 6 = 6(2x − 1)
Another sign chart: ..
f′′
(x)
.
f(x)
..
1/2
.
−−
.
++
.
⌢
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 11 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 6x2
− 6x − 12
=⇒ f′′
(x) = 12x − 6 = 6(2x − 1)
Another sign chart: ..
f′′
(x)
.
f(x)
..
1/2
.
−−
.
++
.
⌢
.
⌣
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 11 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 6x2
− 6x − 12
=⇒ f′′
(x) = 12x − 6 = 6(2x − 1)
Another sign chart: ..
f′′
(x)
.
f(x)
..
1/2
.
−−
.
++
.
⌢
.
⌣
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 11 / 55
. . . . . .
Step 3: One sign chart to rule them all
Remember, f(x) = 2x3
− 3x2
− 12x.
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 12 / 55
. . . . . .
Step 3: One sign chart to rule them all
Remember, f(x) = 2x3
− 3x2
− 12x.
..
f′
(x)
.
monotonicity
..
−1
..
2
. +.
↗
.− .
↘
. −.
↘
.+ .
↗
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 12 / 55
. . . . . .
Step 3: One sign chart to rule them all
Remember, f(x) = 2x3
− 3x2
− 12x.
..
f′
(x)
.
monotonicity
..
−1
..
2
. +.
↗
.− .
↘
. −.
↘
.+ .
↗
.
f′′
(x)
.
concavity
..
1/2
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 12 / 55
. . . . . .
Step 3: One sign chart to rule them all
Remember, f(x) = 2x3
− 3x2
− 12x.
..
f′
(x)
.
monotonicity
..
−1
..
2
. +.
↗
.− .
↘
. −.
↘
.+ .
↗
.
f′′
(x)
.
concavity
..
1/2
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
f(x)
.
shape of f
..
−1
.
7
.
max
..
2
.
−20
.
min
..
1/2
.
−61/2
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 12 / 55
. . . . . .
Combinations of monotonicity and concavity
..
I
.
II
.
III
.
IV
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 13 / 55
. . . . . .
Combinations of monotonicity and concavity
..
I
.
II
.
III
.
IV
.
decreasing,
concave
down
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 13 / 55
. . . . . .
Combinations of monotonicity and concavity
..
I
.
II
.
III
.
IV
.
decreasing,
concave
down
.
increasing,
concave
down
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 13 / 55
. . . . . .
Combinations of monotonicity and concavity
..
I
.
II
.
III
.
IV
.
decreasing,
concave
down
.
increasing,
concave
down
.
decreasing,
concave up
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 13 / 55
. . . . . .
Combinations of monotonicity and concavity
..
I
.
II
.
III
.
IV
.
decreasing,
concave
down
.
increasing,
concave
down
.
decreasing,
concave up
.
increasing,
concave up
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 13 / 55
. . . . . .
Step 3: One sign chart to rule them all
Remember, f(x) = 2x3
− 3x2
− 12x.
..
f′
(x)
.
monotonicity
..
−1
..
2
. +.
↗
.− .
↘
. −.
↘
.+ .
↗
.
f′′
(x)
.
concavity
..
1/2
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
f(x)
.
shape of f
..
−1
.
7
.
max
..
2
.
−20
.
min
..
1/2
.
−61/2
.
IP
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 14 / 55
. . . . . .
Step 3: One sign chart to rule them all
Remember, f(x) = 2x3
− 3x2
− 12x.
..
f′
(x)
.
monotonicity
..
−1
..
2
. +.
↗
.− .
↘
. −.
↘
.+ .
↗
.
f′′
(x)
.
concavity
..
1/2
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
f(x)
.
shape of f
..
−1
.
7
.
max
..
2
.
−20
.
min
..
1/2
.
−61/2
.
IP
..
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 14 / 55
. . . . . .
Step 3: One sign chart to rule them all
Remember, f(x) = 2x3
− 3x2
− 12x.
..
f′
(x)
.
monotonicity
..
−1
..
2
. +.
↗
.− .
↘
. −.
↘
.+ .
↗
.
f′′
(x)
.
concavity
..
1/2
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
f(x)
.
shape of f
..
−1
.
7
.
max
..
2
.
−20
.
min
..
1/2
.
−61/2
.
IP
...
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 14 / 55
. . . . . .
Step 3: One sign chart to rule them all
Remember, f(x) = 2x3
− 3x2
− 12x.
..
f′
(x)
.
monotonicity
..
−1
..
2
. +.
↗
.− .
↘
. −.
↘
.+ .
↗
.
f′′
(x)
.
concavity
..
1/2
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
f(x)
.
shape of f
..
−1
.
7
.
max
..
2
.
−20
.
min
..
1/2
.
−61/2
.
IP
....
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 14 / 55
. . . . . .
Step 4: Graph
..
f(x) = 2x3
− 3x2
− 12x
. x.
f(x)
.
f(x)
.
shape of f
..
−1
.
7
.
max
..
2
.
−20
.
min
..
1/2
.
−61/2
.
IP
......
(
3−
√
105
4 , 0
)
..
(−1, 7)
..
(0, 0)
..
(1/2, −61/2)
..
(2, −20)
..
(
3+
√
105
4 , 0
)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 15 / 55
. . . . . .
Step 4: Graph
..
f(x) = 2x3
− 3x2
− 12x
. x.
f(x)
.
f(x)
.
shape of f
..
−1
.
7
.
max
..
2
.
−20
.
min
..
1/2
.
−61/2
.
IP
......
(
3−
√
105
4 , 0
)
..
(−1, 7)
..
(0, 0)
..
(1/2, −61/2)
..
(2, −20)
..
(
3+
√
105
4 , 0
)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 15 / 55
. . . . . .
Step 4: Graph
..
f(x) = 2x3
− 3x2
− 12x
. x.
f(x)
.
f(x)
.
shape of f
..
−1
.
7
.
max
..
2
.
−20
.
min
..
1/2
.
−61/2
.
IP
......
(
3−
√
105
4 , 0
)
..
(−1, 7)
..
(0, 0)
..
(1/2, −61/2)
..
(2, −20)
..
(
3+
√
105
4 , 0
)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 15 / 55
. . . . . .
Step 4: Graph
..
f(x) = 2x3
− 3x2
− 12x
. x.
f(x)
.
f(x)
.
shape of f
..
−1
.
7
.
max
..
2
.
−20
.
min
..
1/2
.
−61/2
.
IP
......
(
3−
√
105
4 , 0
)
..
(−1, 7)
..
(0, 0)
..
(1/2, −61/2)
..
(2, −20)
..
(
3+
√
105
4 , 0
)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 15 / 55
. . . . . .
Step 4: Graph
..
f(x) = 2x3
− 3x2
− 12x
. x.
f(x)
.
f(x)
.
shape of f
..
−1
.
7
.
max
..
2
.
−20
.
min
..
1/2
.
−61/2
.
IP
......
(
3−
√
105
4 , 0
)
..
(−1, 7)
..
(0, 0)
..
(1/2, −61/2)
..
(2, −20)
..
(
3+
√
105
4 , 0
)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 15 / 55
. . . . . .
Graphing a quartic
Example
Graph f(x) = x4
− 4x3
+ 10
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 16 / 55
. . . . . .
Graphing a quartic
Example
Graph f(x) = x4
− 4x3
+ 10
(Step 0) We know f(0) = 10 and lim
x→±∞
f(x) = +∞. Not too many other
points on the graph are evident.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 16 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4
− 4x3
+ 10
=⇒ f′
(x) = 4x3
− 12x2
= 4x2
(x − 3)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4
− 4x3
+ 10
=⇒ f′
(x) = 4x3
− 12x2
= 4x2
(x − 3)
We make its sign chart.
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4
− 4x3
+ 10
=⇒ f′
(x) = 4x3
− 12x2
= 4x2
(x − 3)
We make its sign chart.
.. 4x2..
0
.0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4
− 4x3
+ 10
=⇒ f′
(x) = 4x3
− 12x2
= 4x2
(x − 3)
We make its sign chart.
.. 4x2..
0
.0.+
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4
− 4x3
+ 10
=⇒ f′
(x) = 4x3
− 12x2
= 4x2
(x − 3)
We make its sign chart.
.. 4x2..
0
.0.+ . +
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4
− 4x3
+ 10
=⇒ f′
(x) = 4x3
− 12x2
= 4x2
(x − 3)
We make its sign chart.
.. 4x2..
0
.0.+ . +. +
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4
− 4x3
+ 10
=⇒ f′
(x) = 4x3
− 12x2
= 4x2
(x − 3)
We make its sign chart.
.. 4x2..
0
.0.+ . +. +.
(x − 3)
..
3
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4
− 4x3
+ 10
=⇒ f′
(x) = 4x3
− 12x2
= 4x2
(x − 3)
We make its sign chart.
.. 4x2..
0
.0.+ . +. +.
(x − 3)
..
3
.
0
.
−
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4
− 4x3
+ 10
=⇒ f′
(x) = 4x3
− 12x2
= 4x2
(x − 3)
We make its sign chart.
.. 4x2..
0
.0.+ . +. +.
(x − 3)
..
3
.
0
.
−
.
−
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4
− 4x3
+ 10
=⇒ f′
(x) = 4x3
− 12x2
= 4x2
(x − 3)
We make its sign chart.
.. 4x2..
0
.0.+ . +. +.
(x − 3)
..
3
.
0
.
−
.
−
.
+
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4
− 4x3
+ 10
=⇒ f′
(x) = 4x3
− 12x2
= 4x2
(x − 3)
We make its sign chart.
.. 4x2..
0
.0.+ . +. +.
(x − 3)
..
3
.
0
.
−
.
−
.
+
.
f′
(x)
.
f(x)
..
3
.
0
..
0
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4
− 4x3
+ 10
=⇒ f′
(x) = 4x3
− 12x2
= 4x2
(x − 3)
We make its sign chart.
.. 4x2..
0
.0.+ . +. +.
(x − 3)
..
3
.
0
.
−
.
−
.
+
.
f′
(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4
− 4x3
+ 10
=⇒ f′
(x) = 4x3
− 12x2
= 4x2
(x − 3)
We make its sign chart.
.. 4x2..
0
.0.+ . +. +.
(x − 3)
..
3
.
0
.
−
.
−
.
+
.
f′
(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4
− 4x3
+ 10
=⇒ f′
(x) = 4x3
− 12x2
= 4x2
(x − 3)
We make its sign chart.
.. 4x2..
0
.0.+ . +. +.
(x − 3)
..
3
.
0
.
−
.
−
.
+
.
f′
(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4
− 4x3
+ 10
=⇒ f′
(x) = 4x3
− 12x2
= 4x2
(x − 3)
We make its sign chart.
.. 4x2..
0
.0.+ . +. +.
(x − 3)
..
3
.
0
.
−
.
−
.
+
.
f′
(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
.
↘
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4
− 4x3
+ 10
=⇒ f′
(x) = 4x3
− 12x2
= 4x2
(x − 3)
We make its sign chart.
.. 4x2..
0
.0.+ . +. +.
(x − 3)
..
3
.
0
.
−
.
−
.
+
.
f′
(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
.
↘
.
↘
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4
− 4x3
+ 10
=⇒ f′
(x) = 4x3
− 12x2
= 4x2
(x − 3)
We make its sign chart.
.. 4x2..
0
.0.+ . +. +.
(x − 3)
..
3
.
0
.
−
.
−
.
+
.
f′
(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
.
↘
.
↘
.
↗
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4
− 4x3
+ 10
=⇒ f′
(x) = 4x3
− 12x2
= 4x2
(x − 3)
We make its sign chart.
.. 4x2..
0
.0.+ . +. +.
(x − 3)
..
3
.
0
.
−
.
−
.
+
.
f′
(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
.
↘
.
↘
.
↗
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 4x3
− 12x2
=⇒ f′′
(x) = 12x2
− 24x = 12x(x − 2)
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 4x3
− 12x2
=⇒ f′′
(x) = 12x2
− 24x = 12x(x − 2)
Here is its sign chart:
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 4x3
− 12x2
=⇒ f′′
(x) = 12x2
− 24x = 12x(x − 2)
Here is its sign chart:
.. 12x..
0
.0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 4x3
− 12x2
=⇒ f′′
(x) = 12x2
− 24x = 12x(x − 2)
Here is its sign chart:
.. 12x..
0
.0.−
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 4x3
− 12x2
=⇒ f′′
(x) = 12x2
− 24x = 12x(x − 2)
Here is its sign chart:
.. 12x..
0
.0.− . +
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 4x3
− 12x2
=⇒ f′′
(x) = 12x2
− 24x = 12x(x − 2)
Here is its sign chart:
.. 12x..
0
.0.− . +. +
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 4x3
− 12x2
=⇒ f′′
(x) = 12x2
− 24x = 12x(x − 2)
Here is its sign chart:
.. 12x..
0
.0.− . +. +.
x − 2
..
2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 4x3
− 12x2
=⇒ f′′
(x) = 12x2
− 24x = 12x(x − 2)
Here is its sign chart:
.. 12x..
0
.0.− . +. +.
x − 2
..
2
.
0
.
−
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 4x3
− 12x2
=⇒ f′′
(x) = 12x2
− 24x = 12x(x − 2)
Here is its sign chart:
.. 12x..
0
.0.− . +. +.
x − 2
..
2
.
0
.
−
.
−
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 4x3
− 12x2
=⇒ f′′
(x) = 12x2
− 24x = 12x(x − 2)
Here is its sign chart:
.. 12x..
0
.0.− . +. +.
x − 2
..
2
.
0
.
−
.
−
.
+
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 4x3
− 12x2
=⇒ f′′
(x) = 12x2
− 24x = 12x(x − 2)
Here is its sign chart:
.. 12x..
0
.0.− . +. +.
x − 2
..
2
.
0
.
−
.
−
.
+
.
f′′
(x)
.
f(x)
..
0
.
0
..
2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 4x3
− 12x2
=⇒ f′′
(x) = 12x2
− 24x = 12x(x − 2)
Here is its sign chart:
.. 12x..
0
.0.− . +. +.
x − 2
..
2
.
0
.
−
.
−
.
+
.
f′′
(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 4x3
− 12x2
=⇒ f′′
(x) = 12x2
− 24x = 12x(x − 2)
Here is its sign chart:
.. 12x..
0
.0.− . +. +.
x − 2
..
2
.
0
.
−
.
−
.
+
.
f′′
(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 4x3
− 12x2
=⇒ f′′
(x) = 12x2
− 24x = 12x(x − 2)
Here is its sign chart:
.. 12x..
0
.0.− . +. +.
x − 2
..
2
.
0
.
−
.
−
.
+
.
f′′
(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 4x3
− 12x2
=⇒ f′′
(x) = 12x2
− 24x = 12x(x − 2)
Here is its sign chart:
.. 12x..
0
.0.− . +. +.
x − 2
..
2
.
0
.
−
.
−
.
+
.
f′′
(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 4x3
− 12x2
=⇒ f′′
(x) = 12x2
− 24x = 12x(x − 2)
Here is its sign chart:
.. 12x..
0
.0.− . +. +.
x − 2
..
2
.
0
.
−
.
−
.
+
.
f′′
(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 4x3
− 12x2
=⇒ f′′
(x) = 12x2
− 24x = 12x(x − 2)
Here is its sign chart:
.. 12x..
0
.0.− . +. +.
x − 2
..
2
.
0
.
−
.
−
.
+
.
f′′
(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
.
⌣
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 4x3
− 12x2
=⇒ f′′
(x) = 12x2
− 24x = 12x(x − 2)
Here is its sign chart:
.. 12x..
0
.0.− . +. +.
x − 2
..
2
.
0
.
−
.
−
.
+
.
f′′
(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
.
⌣
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′
(x) = 4x3
− 12x2
=⇒ f′′
(x) = 12x2
− 24x = 12x(x − 2)
Here is its sign chart:
.. 12x..
0
.0.− . +. +.
x − 2
..
2
.
0
.
−
.
−
.
+
.
f′′
(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
.
⌣
.
IP
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 3: Grand Unified Sign Chart
Remember, f(x) = x4
− 4x3
+ 10.
..
f′
(x)
.
monotonicity
..
3
.
0
..
0
.
0
.
−
.
↘
.
−
.
↘
.
−
.
↘
.
+
.
↗
.
f′′
(x)
.
concavity
..
0
.
0
..
2
.
0
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
f(x)
.
shape
..
0
.
10
.
IP
..
2
.
−6
.
IP
..
3
.
−17
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 19 / 55
. . . . . .
Step 3: Grand Unified Sign Chart
Remember, f(x) = x4
− 4x3
+ 10.
..
f′
(x)
.
monotonicity
..
3
.
0
..
0
.
0
.
−
.
↘
.
−
.
↘
.
−
.
↘
.
+
.
↗
.
f′′
(x)
.
concavity
..
0
.
0
..
2
.
0
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
f(x)
.
shape
..
0
.
10
.
IP
..
2
.
−6
.
IP
..
3
.
−17
.
min
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 19 / 55
. . . . . .
Step 3: Grand Unified Sign Chart
Remember, f(x) = x4
− 4x3
+ 10.
..
f′
(x)
.
monotonicity
..
3
.
0
..
0
.
0
.
−
.
↘
.
−
.
↘
.
−
.
↘
.
+
.
↗
.
f′′
(x)
.
concavity
..
0
.
0
..
2
.
0
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
f(x)
.
shape
..
0
.
10
.
IP
..
2
.
−6
.
IP
..
3
.
−17
.
min
..
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 19 / 55
. . . . . .
Step 3: Grand Unified Sign Chart
Remember, f(x) = x4
− 4x3
+ 10.
..
f′
(x)
.
monotonicity
..
3
.
0
..
0
.
0
.
−
.
↘
.
−
.
↘
.
−
.
↘
.
+
.
↗
.
f′′
(x)
.
concavity
..
0
.
0
..
2
.
0
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
f(x)
.
shape
..
0
.
10
.
IP
..
2
.
−6
.
IP
..
3
.
−17
.
min
...
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 19 / 55
. . . . . .
Step 3: Grand Unified Sign Chart
Remember, f(x) = x4
− 4x3
+ 10.
..
f′
(x)
.
monotonicity
..
3
.
0
..
0
.
0
.
−
.
↘
.
−
.
↘
.
−
.
↘
.
+
.
↗
.
f′′
(x)
.
concavity
..
0
.
0
..
2
.
0
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
f(x)
.
shape
..
0
.
10
.
IP
..
2
.
−6
.
IP
..
3
.
−17
.
min
....
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 19 / 55
. . . . . .
Step 4: Graph
..
f(x) = x4
− 4x3
+ 10
. x.
y
.
f(x)
.
shape
..
0
.
10
.
IP
..
2
.
−6
.
IP
..
3
.
−17
.
min
......
(0, 10)
..
(2, −6)
..
(3, −17)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 20 / 55
. . . . . .
Step 4: Graph
..
f(x) = x4
− 4x3
+ 10
. x.
y
.
f(x)
.
shape
..
0
.
10
.
IP
..
2
.
−6
.
IP
..
3
.
−17
.
min
......
(0, 10)
..
(2, −6)
..
(3, −17)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 20 / 55
. . . . . .
Step 4: Graph
..
f(x) = x4
− 4x3
+ 10
. x.
y
.
f(x)
.
shape
..
0
.
10
.
IP
..
2
.
−6
.
IP
..
3
.
−17
.
min
......
(0, 10)
..
(2, −6)
..
(3, −17)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 20 / 55
. . . . . .
Step 4: Graph
..
f(x) = x4
− 4x3
+ 10
. x.
y
.
f(x)
.
shape
..
0
.
10
.
IP
..
2
.
−6
.
IP
..
3
.
−17
.
min
......
(0, 10)
..
(2, −6)
..
(3, −17)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 20 / 55
. . . . . .
Step 4: Graph
..
f(x) = x4
− 4x3
+ 10
. x.
y
.
f(x)
.
shape
..
0
.
10
.
IP
..
2
.
−6
.
IP
..
3
.
−17
.
min
......
(0, 10)
..
(2, −6)
..
(3, −17)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 20 / 55
. . . . . .
Outline
Simple examples
A cubic function
A quartic function
More Examples
Points of nondifferentiability
Horizontal asymptotes
Vertical asymptotes
Trigonometric and polynomial together
Logarithmic
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 21 / 55
. . . . . .
Graphing a function with a cusp
Example
Graph f(x) = x +
√
|x|
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 22 / 55
. . . . . .
Graphing a function with a cusp
Example
Graph f(x) = x +
√
|x|
This function looks strange because of the absolute value. But
whenever we become nervous, we can just take cases.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 22 / 55
. . . . . .
Step 0: Finding Zeroes
f(x) = x +
√
|x|
First, look at f by itself. We can tell that f(0) = 0 and that f(x) > 0 if
x is positive.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 23 / 55
. . . . . .
Step 0: Finding Zeroes
f(x) = x +
√
|x|
First, look at f by itself. We can tell that f(0) = 0 and that f(x) > 0 if
x is positive.
Are there negative numbers which are zeroes for f?
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 23 / 55
. . . . . .
Step 0: Finding Zeroes
f(x) = x +
√
|x|
First, look at f by itself. We can tell that f(0) = 0 and that f(x) > 0 if
x is positive.
Are there negative numbers which are zeroes for f?
x +
√
−x = 0
√
−x = −x
−x = x2
x2
+ x = 0
The only solutions are x = 0 and x = −1.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 23 / 55
. . . . . .
Step 0: Asymptotic behavior
f(x) = x +
√
|x|
lim
x→∞
f(x) = ∞, because both terms tend to ∞.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 24 / 55
. . . . . .
Step 0: Asymptotic behavior
f(x) = x +
√
|x|
lim
x→∞
f(x) = ∞, because both terms tend to ∞.
lim
x→−∞
f(x) is indeterminate of the form −∞ + ∞. It’s the same as
lim
y→+∞
(−y +
√
y)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 24 / 55
. . . . . .
Step 0: Asymptotic behavior
f(x) = x +
√
|x|
lim
x→∞
f(x) = ∞, because both terms tend to ∞.
lim
x→−∞
f(x) is indeterminate of the form −∞ + ∞. It’s the same as
lim
y→+∞
(−y +
√
y)
lim
y→+∞
(−y +
√
y) = lim
y→∞
(
√
y − y) ·
√
y + y
√
y + y
= lim
y→∞
y − y2
√
y + y
= −∞
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 24 / 55
. . . . . .
Step 1: The derivative
Remember, f(x) = x +
√
|x|.
To find f′
, first assume x > 0. Then
f′
(x) =
d
dx
(
x +
√
x
)
= 1 +
1
2
√
x
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 25 / 55
. . . . . .
Step 1: The derivative
Remember, f(x) = x +
√
|x|.
To find f′
, first assume x > 0. Then
f′
(x) =
d
dx
(
x +
√
x
)
= 1 +
1
2
√
x
Notice
f′
(x) > 0 when x > 0 (so no critical points here)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 25 / 55
. . . . . .
Step 1: The derivative
Remember, f(x) = x +
√
|x|.
To find f′
, first assume x > 0. Then
f′
(x) =
d
dx
(
x +
√
x
)
= 1 +
1
2
√
x
Notice
f′
(x) > 0 when x > 0 (so no critical points here)
lim
x→0+
f′
(x) = ∞ (so 0 is a critical point)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 25 / 55
. . . . . .
Step 1: The derivative
Remember, f(x) = x +
√
|x|.
To find f′
, first assume x > 0. Then
f′
(x) =
d
dx
(
x +
√
x
)
= 1 +
1
2
√
x
Notice
f′
(x) > 0 when x > 0 (so no critical points here)
lim
x→0+
f′
(x) = ∞ (so 0 is a critical point)
lim
x→∞
f′
(x) = 1 (so the graph is asymptotic to a line of slope 1)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 25 / 55
. . . . . .
Step 1: The derivative
Remember, f(x) = x +
√
|x|.
If x is negative, we have
f′
(x) =
d
dx
(
x +
√
−x
)
= 1 −
1
2
√
−x
Notice
lim
x→0−
f′
(x) = −∞ (other side of the critical point)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 26 / 55
. . . . . .
Step 1: The derivative
Remember, f(x) = x +
√
|x|.
If x is negative, we have
f′
(x) =
d
dx
(
x +
√
−x
)
= 1 −
1
2
√
−x
Notice
lim
x→0−
f′
(x) = −∞ (other side of the critical point)
lim
x→−∞
f′
(x) = 1 (asymptotic to a line of slope 1)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 26 / 55
. . . . . .
Step 1: The derivative
Remember, f(x) = x +
√
|x|.
If x is negative, we have
f′
(x) =
d
dx
(
x +
√
−x
)
= 1 −
1
2
√
−x
Notice
lim
x→0−
f′
(x) = −∞ (other side of the critical point)
lim
x→−∞
f′
(x) = 1 (asymptotic to a line of slope 1)
f′
(x) = 0 when
1 −
1
2
√
−x
= 0 =⇒
√
−x =
1
2
=⇒ −x =
1
4
=⇒ x = −
1
4
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 26 / 55
. . . . . .
Step 1: Monotonicity
f′
(x) =



1 +
1
2
√
x
if x > 0
1 −
1
2
√
−x
if x < 0
We can’t make a multi-factor sign chart because of the absolute value,
but we can test points in between critical points.
..
f′
(x)
.
f(x)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
. . . . . .
Step 1: Monotonicity
f′
(x) =



1 +
1
2
√
x
if x > 0
1 −
1
2
√
−x
if x < 0
We can’t make a multi-factor sign chart because of the absolute value,
but we can test points in between critical points.
..
f′
(x)
.
f(x)
..
−1
4
.0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
. . . . . .
Step 1: Monotonicity
f′
(x) =



1 +
1
2
√
x
if x > 0
1 −
1
2
√
−x
if x < 0
We can’t make a multi-factor sign chart because of the absolute value,
but we can test points in between critical points.
..
f′
(x)
.
f(x)
..
−1
4
.0 ..
0
.∞
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
. . . . . .
Step 1: Monotonicity
f′
(x) =



1 +
1
2
√
x
if x > 0
1 −
1
2
√
−x
if x < 0
We can’t make a multi-factor sign chart because of the absolute value,
but we can test points in between critical points.
..
f′
(x)
.
f(x)
..
−1
4
.0 ..
0
.∞.+
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
. . . . . .
Step 1: Monotonicity
f′
(x) =



1 +
1
2
√
x
if x > 0
1 −
1
2
√
−x
if x < 0
We can’t make a multi-factor sign chart because of the absolute value,
but we can test points in between critical points.
..
f′
(x)
.
f(x)
..
−1
4
.0 ..
0
.∞.+ .−
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
. . . . . .
Step 1: Monotonicity
f′
(x) =



1 +
1
2
√
x
if x > 0
1 −
1
2
√
−x
if x < 0
We can’t make a multi-factor sign chart because of the absolute value,
but we can test points in between critical points.
..
f′
(x)
.
f(x)
..
−1
4
.0 ..
0
.∞.+ .− . +
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
. . . . . .
Step 1: Monotonicity
f′
(x) =



1 +
1
2
√
x
if x > 0
1 −
1
2
√
−x
if x < 0
We can’t make a multi-factor sign chart because of the absolute value,
but we can test points in between critical points.
..
f′
(x)
.
f(x)
..
−1
4
.0 ..
0
.∞.+ .− . +.
↗
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
. . . . . .
Step 1: Monotonicity
f′
(x) =



1 +
1
2
√
x
if x > 0
1 −
1
2
√
−x
if x < 0
We can’t make a multi-factor sign chart because of the absolute value,
but we can test points in between critical points.
..
f′
(x)
.
f(x)
..
−1
4
.0 ..
0
.∞.+ .− . +.
↗
.
↘
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
. . . . . .
Step 1: Monotonicity
f′
(x) =



1 +
1
2
√
x
if x > 0
1 −
1
2
√
−x
if x < 0
We can’t make a multi-factor sign chart because of the absolute value,
but we can test points in between critical points.
..
f′
(x)
.
f(x)
..
−1
4
.0 ..
0
.∞.+ .− . +.
↗
.
↘
.
↗
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
. . . . . .
Step 1: Monotonicity
f′
(x) =



1 +
1
2
√
x
if x > 0
1 −
1
2
√
−x
if x < 0
We can’t make a multi-factor sign chart because of the absolute value,
but we can test points in between critical points.
..
f′
(x)
.
f(x)
..
−1
4
.0 ..
0
.∞.+ .− . +.
↗
.
↘
.
↗
.
max
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
. . . . . .
Step 1: Monotonicity
f′
(x) =



1 +
1
2
√
x
if x > 0
1 −
1
2
√
−x
if x < 0
We can’t make a multi-factor sign chart because of the absolute value,
but we can test points in between critical points.
..
f′
(x)
.
f(x)
..
−1
4
.0 ..
0
.∞.+ .− . +.
↗
.
↘
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
. . . . . .
Step 2: Concavity
If x > 0, then
f′′
(x) =
d
dx
(
1 +
1
2
x−1/2
)
= −
1
4
x−3/2
This is negative whenever x > 0.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 28 / 55
. . . . . .
Step 2: Concavity
If x > 0, then
f′′
(x) =
d
dx
(
1 +
1
2
x−1/2
)
= −
1
4
x−3/2
This is negative whenever x > 0.
If x < 0, then
f′′
(x) =
d
dx
(
1 −
1
2
(−x)−1/2
)
= −
1
4
(−x)−3/2
which is also always negative for negative x.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 28 / 55
. . . . . .
Step 2: Concavity
If x > 0, then
f′′
(x) =
d
dx
(
1 +
1
2
x−1/2
)
= −
1
4
x−3/2
This is negative whenever x > 0.
If x < 0, then
f′′
(x) =
d
dx
(
1 −
1
2
(−x)−1/2
)
= −
1
4
(−x)−3/2
which is also always negative for negative x.
In other words, f′′
(x) = −
1
4
|x|−3/2
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 28 / 55
. . . . . .
Step 2: Concavity
If x > 0, then
f′′
(x) =
d
dx
(
1 +
1
2
x−1/2
)
= −
1
4
x−3/2
This is negative whenever x > 0.
If x < 0, then
f′′
(x) =
d
dx
(
1 −
1
2
(−x)−1/2
)
= −
1
4
(−x)−3/2
which is also always negative for negative x.
In other words, f′′
(x) = −
1
4
|x|−3/2
.
Here is the sign chart:
..
f′′
(x)
.
f(x)
..
0
.−∞.−− .
⌢
... −−.
⌢
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 28 / 55
. . . . . .
Step 3: Synthesis
Now we can put these things together.
f(x) = x +
√
|x|
..
f′
(x)
.
monotonicity
..
−1
4
.0 ..
0
.∞.+1 .
↗
.+ .
↗
.− .
↘
. +.
↗
. +1.
↗
.
f′′
(x)
.
concavity
..
0
.
−∞
.
−−
.
⌢
.
−−
.
⌢
.
−−
.
⌢
.
−∞
.
⌢
.
−∞
.
⌢
.
f(x)
.
shape
..
−1
.
0
.
zero
..
−1
4
.
1
4
.
max
..
0
.
0
.
min
.
−∞
.
+∞
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 29 / 55
. . . . . .
Step 3: Synthesis
Now we can put these things together.
f(x) = x +
√
|x|
..
f′
(x)
.
monotonicity
..
−1
4
.0 ..
0
.∞.+1 .
↗
.+ .
↗
.− .
↘
. +.
↗
. +1.
↗
.
f′′
(x)
.
concavity
..
0
.
−∞
.
−−
.
⌢
.
−−
.
⌢
.
−−
.
⌢
.
−∞
.
⌢
.
−∞
.
⌢
.
f(x)
.
shape
..
−1
.
0
.
zero
..
−1
4
.
1
4
.
max
..
0
.
0
.
min
.
−∞
.
+∞
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 29 / 55
. . . . . .
Step 3: Synthesis
Now we can put these things together.
f(x) = x +
√
|x|
..
f′
(x)
.
monotonicity
..
−1
4
.0 ..
0
.∞.+1 .
↗
.+ .
↗
.− .
↘
. +.
↗
. +1.
↗
.
f′′
(x)
.
concavity
..
0
.
−∞
.
−−
.
⌢
.
−−
.
⌢
.
−−
.
⌢
.
−∞
.
⌢
.
−∞
.
⌢
.
f(x)
.
shape
..
−1
.
0
.
zero
..
−1
4
.
1
4
.
max
..
0
.
0
.
min
.
−∞
.
+∞
..
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 29 / 55
. . . . . .
Step 3: Synthesis
Now we can put these things together.
f(x) = x +
√
|x|
..
f′
(x)
.
monotonicity
..
−1
4
.0 ..
0
.∞.+1 .
↗
.+ .
↗
.− .
↘
. +.
↗
. +1.
↗
.
f′′
(x)
.
concavity
..
0
.
−∞
.
−−
.
⌢
.
−−
.
⌢
.
−−
.
⌢
.
−∞
.
⌢
.
−∞
.
⌢
.
f(x)
.
shape
..
−1
.
0
.
zero
..
−1
4
.
1
4
.
max
..
0
.
0
.
min
.
−∞
.
+∞
...
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 29 / 55
. . . . . .
Step 3: Synthesis
Now we can put these things together.
f(x) = x +
√
|x|
..
f′
(x)
.
monotonicity
..
−1
4
.0 ..
0
.∞.+1 .
↗
.+ .
↗
.− .
↘
. +.
↗
. +1.
↗
.
f′′
(x)
.
concavity
..
0
.
−∞
.
−−
.
⌢
.
−−
.
⌢
.
−−
.
⌢
.
−∞
.
⌢
.
−∞
.
⌢
.
f(x)
.
shape
..
−1
.
0
.
zero
..
−1
4
.
1
4
.
max
..
0
.
0
.
min
.
−∞
.
+∞
....
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 29 / 55
. . . . . .
Graph
f(x) = x +
√
|x|
..
f(x)
.
shape
..
−1
.
0
.
zero
.
−∞
.
+∞
..
−1
4
.
1
4
.
max
.
−∞
.
+∞
..
0
.
0
.
min
.
−∞
.
+∞
..... x.
f(x)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 30 / 55
. . . . . .
Graph
f(x) = x +
√
|x|
..
f(x)
.
shape
..
−1
.
0
.
zero
.
−∞
.
+∞
..
−1
4
.
1
4
.
max
.
−∞
.
+∞
..
0
.
0
.
min
.
−∞
.
+∞
..... x.
f(x)
..
(−1, 0)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 30 / 55
. . . . . .
Graph
f(x) = x +
√
|x|
..
f(x)
.
shape
..
−1
.
0
.
zero
.
−∞
.
+∞
..
−1
4
.
1
4
.
max
.
−∞
.
+∞
..
0
.
0
.
min
.
−∞
.
+∞
..... x.
f(x)
..
(−1, 0)
..
(−1
4 , 1
4 )
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 30 / 55
. . . . . .
Graph
f(x) = x +
√
|x|
..
f(x)
.
shape
..
−1
.
0
.
zero
.
−∞
.
+∞
..
−1
4
.
1
4
.
max
.
−∞
.
+∞
..
0
.
0
.
min
.
−∞
.
+∞
..... x.
f(x)
..
(−1, 0)
..
(−1
4 , 1
4 )
..
(0, 0)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 30 / 55
. . . . . .
Graph
f(x) = x +
√
|x|
..
f(x)
.
shape
..
−1
.
0
.
zero
.
−∞
.
+∞
..
−1
4
.
1
4
.
max
.
−∞
.
+∞
..
0
.
0
.
min
.
−∞
.
+∞
..... x.
f(x)
..
(−1, 0)
..
(−1
4 , 1
4 )
..
(0, 0)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 30 / 55
. . . . . .
Graph
f(x) = x +
√
|x|
..
f(x)
.
shape
..
−1
.
0
.
zero
.
−∞
.
+∞
..
−1
4
.
1
4
.
max
.
−∞
.
+∞
..
0
.
0
.
min
.
−∞
.
+∞
..... x.
f(x)
..
(−1, 0)
..
(−1
4 , 1
4 )
..
(0, 0)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 30 / 55
. . . . . .
Graph
f(x) = x +
√
|x|
..
f(x)
.
shape
..
−1
.
0
.
zero
.
−∞
.
+∞
..
−1
4
.
1
4
.
max
.
−∞
.
+∞
..
0
.
0
.
min
.
−∞
.
+∞
..... x.
f(x)
..
(−1, 0)
..
(−1
4 , 1
4 )
..
(0, 0)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 30 / 55
. . . . . .
Example with Horizontal Asymptotes
Example
Graph f(x) = xe−x2
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 31 / 55
. . . . . .
Example with Horizontal Asymptotes
Example
Graph f(x) = xe−x2
Before taking derivatives, we notice that f is odd, that f(0) = 0, and
lim
x→∞
f(x) = 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 31 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.
1 +
√
2x
..
−
√
1/2
.
0
.
f′
(x)
.
f(x)
..
−
√
1/2
.
0
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.+ .
1 +
√
2x
..
−
√
1/2
.
0
.
f′
(x)
.
f(x)
..
−
√
1/2
.
0
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.+ .+.
1 +
√
2x
..
−
√
1/2
.
0
.
f′
(x)
.
f(x)
..
−
√
1/2
.
0
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.+ .+. −.
1 +
√
2x
..
−
√
1/2
.
0
.
f′
(x)
.
f(x)
..
−
√
1/2
.
0
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.+ .+. −.
1 +
√
2x
..
−
√
1/2
.
0
.
−
.
f′
(x)
.
f(x)
..
−
√
1/2
.
0
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.+ .+. −.
1 +
√
2x
..
−
√
1/2
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
−
√
1/2
.
0
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.+ .+. −.
1 +
√
2x
..
−
√
1/2
.
0
.
−
.
+
.
+
.
f′
(x)
.
f(x)
..
−
√
1/2
.
0
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.+ .+. −.
1 +
√
2x
..
−
√
1/2
.
0
.
−
.
+
.
+
.
f′
(x)
.
f(x)
.
−
..
−
√
1/2
.
0
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.+ .+. −.
1 +
√
2x
..
−
√
1/2
.
0
.
−
.
+
.
+
.
f′
(x)
.
f(x)
.
−
.
+
..
−
√
1/2
.
0
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.+ .+. −.
1 +
√
2x
..
−
√
1/2
.
0
.
−
.
+
.
+
.
f′
(x)
.
f(x)
.
−
.
↘
.
+
.
−
..
−
√
1/2
.
0
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.+ .+. −.
1 +
√
2x
..
−
√
1/2
.
0
.
−
.
+
.
+
.
f′
(x)
.
f(x)
.
−
.
↘
.
+
.
↗
.
−
..
−
√
1/2
.
0
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.+ .+. −.
1 +
√
2x
..
−
√
1/2
.
0
.
−
.
+
.
+
.
f′
(x)
.
f(x)
.
−
.
↘
.
+
.
↗
.
−
.
↘
..
−
√
1/2
.
0
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.+ .+. −.
1 +
√
2x
..
−
√
1/2
.
0
.
−
.
+
.
+
.
f′
(x)
.
f(x)
.
−
.
↘
.
+
.
↗
.
−
.
↘
..
−
√
1/2
.
0
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.+ .+. −.
1 +
√
2x
..
−
√
1/2
.
0
.
−
.
+
.
+
.
f′
(x)
.
f(x)
.
−
.
↘
.
+
.
↗
.
−
.
↘
..
−
√
1/2
.
0
.
min
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.+ .+. −.
1 +
√
2x
..
−
√
1/2
.
0
.
−
.
+
.
+
.
f′
(x)
.
f(x)
.
−
.
↘
.
+
.
↗
.
−
.
↘
..
−
√
1/2
.
0
.
min
..
√
1/2
.
0
.
max
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.
√
2x −
√
3
..
√
3/2
.
0
.
√
2x +
√
3
..
−
√
3/2
.
0
.
f′′
(x)
.
f(x)
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .
√
2x −
√
3
..
√
3/2
.
0
.
√
2x +
√
3
..
−
√
3/2
.
0
.
f′′
(x)
.
f(x)
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .− .
√
2x −
√
3
..
√
3/2
.
0
.
√
2x +
√
3
..
−
√
3/2
.
0
.
f′′
(x)
.
f(x)
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .− . +.
√
2x −
√
3
..
√
3/2
.
0
.
√
2x +
√
3
..
−
√
3/2
.
0
.
f′′
(x)
.
f(x)
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .− . +. +.
√
2x −
√
3
..
√
3/2
.
0
.
√
2x +
√
3
..
−
√
3/2
.
0
.
f′′
(x)
.
f(x)
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .− . +. +.
√
2x −
√
3
..
√
3/2
.
0
.
−
.
√
2x +
√
3
..
−
√
3/2
.
0
.
f′′
(x)
.
f(x)
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .− . +. +.
√
2x −
√
3
..
√
3/2
.
0
.
−
.
−
.
√
2x +
√
3
..
−
√
3/2
.
0
.
f′′
(x)
.
f(x)
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .− . +. +.
√
2x −
√
3
..
√
3/2
.
0
.
−
.
−
.
−
.
√
2x +
√
3
..
−
√
3/2
.
0
.
f′′
(x)
.
f(x)
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .− . +. +.
√
2x −
√
3
..
√
3/2
.
0
.
−
.
−
.
−
.
+
.
√
2x +
√
3
..
−
√
3/2
.
0
.
f′′
(x)
.
f(x)
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .− . +. +.
√
2x −
√
3
..
√
3/2
.
0
.
−
.
−
.
−
.
+
.
√
2x +
√
3
..
−
√
3/2
.
0
.
−
.
f′′
(x)
.
f(x)
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .− . +. +.
√
2x −
√
3
..
√
3/2
.
0
.
−
.
−
.
−
.
+
.
√
2x +
√
3
..
−
√
3/2
.
0
.
−
.
+
.
f′′
(x)
.
f(x)
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .− . +. +.
√
2x −
√
3
..
√
3/2
.
0
.
−
.
−
.
−
.
+
.
√
2x +
√
3
..
−
√
3/2
.
0
.
−
.
+
.
+
.
f′′
(x)
.
f(x)
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .− . +. +.
√
2x −
√
3
..
√
3/2
.
0
.
−
.
−
.
−
.
+
.
√
2x +
√
3
..
−
√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′
(x)
.
f(x)
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .− . +. +.
√
2x −
√
3
..
√
3/2
.
0
.
−
.
−
.
−
.
+
.
√
2x +
√
3
..
−
√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′
(x)
.
f(x)
.
−−
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .− . +. +.
√
2x −
√
3
..
√
3/2
.
0
.
−
.
−
.
−
.
+
.
√
2x +
√
3
..
−
√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′
(x)
.
f(x)
.
−−
.
++
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .− . +. +.
√
2x −
√
3
..
√
3/2
.
0
.
−
.
−
.
−
.
+
.
√
2x +
√
3
..
−
√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′
(x)
.
f(x)
.
−−
.
++
.
−−
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .− . +. +.
√
2x −
√
3
..
√
3/2
.
0
.
−
.
−
.
−
.
+
.
√
2x +
√
3
..
−
√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′
(x)
.
f(x)
.
−−
.
⌢
.
++
.
−−
.
++
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .− . +. +.
√
2x −
√
3
..
√
3/2
.
0
.
−
.
−
.
−
.
+
.
√
2x +
√
3
..
−
√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′
(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
++
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .− . +. +.
√
2x −
√
3
..
√
3/2
.
0
.
−
.
−
.
−
.
+
.
√
2x +
√
3
..
−
√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′
(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .− . +. +.
√
2x −
√
3
..
√
3/2
.
0
.
−
.
−
.
−
.
+
.
√
2x +
√
3
..
−
√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′
(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .− . +. +.
√
2x −
√
3
..
√
3/2
.
0
.
−
.
−
.
−
.
+
.
√
2x +
√
3
..
−
√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′
(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .− . +. +.
√
2x −
√
3
..
√
3/2
.
0
.
−
.
−
.
−
.
+
.
√
2x +
√
3
..
−
√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′
(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−
√
3/2
.
0
.
IP
..
0
.
0
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .− . +. +.
√
2x −
√
3
..
√
3/2
.
0
.
−
.
−
.
−
.
+
.
√
2x +
√
3
..
−
√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′
(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−
√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√
3/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = (1 − 2x2
)e−x2
, we know
f′′
(x) = (−4x)e−x2
+ (1 − 2x2
)e−x2
(−2x) =
(
4x3
− 6x
)
e−x2
= 2x(2x2
− 3)e−x2
.. 2x..
0
.0.− .− . +. +.
√
2x −
√
3
..
√
3/2
.
0
.
−
.
−
.
−
.
+
.
√
2x +
√
3
..
−
√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′
(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−
√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√
3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 3: Synthesis
f(x) = xe−x2
..
f′
(x)
.
monotonicity
..
−
√
1/2
.0 .. √
1/2
. 0.− .
↘
.− .
↘
.+ .
↗
. +.
↗
. −.
↘
. −.
↘
.
f′′
(x)
.
concavity
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−
√
1/2
.
− 1√
2e
.
min
..
√
1/2
.
1√
2e
.
max
..
−
√
3/2
.
−
√
3
2e3
.
IP
..
0
.
0
.
IP
..
√
3/2
.
√
3
2e3
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 55
. . . . . .
Step 3: Synthesis
f(x) = xe−x2
..
f′
(x)
.
monotonicity
..
−
√
1/2
.0 .. √
1/2
. 0.− .
↘
.− .
↘
.+ .
↗
. +.
↗
. −.
↘
. −.
↘
.
f′′
(x)
.
concavity
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−
√
1/2
.
− 1√
2e
.
min
..
√
1/2
.
1√
2e
.
max
..
−
√
3/2
.
−
√
3
2e3
.
IP
..
0
.
0
.
IP
..
√
3/2
.
√
3
2e3
.
IP
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 55
. . . . . .
Step 3: Synthesis
f(x) = xe−x2
..
f′
(x)
.
monotonicity
..
−
√
1/2
.0 .. √
1/2
. 0.− .
↘
.− .
↘
.+ .
↗
. +.
↗
. −.
↘
. −.
↘
.
f′′
(x)
.
concavity
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−
√
1/2
.
− 1√
2e
.
min
..
√
1/2
.
1√
2e
.
max
..
−
√
3/2
.
−
√
3
2e3
.
IP
..
0
.
0
.
IP
..
√
3/2
.
√
3
2e3
.
IP
..
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 55
. . . . . .
Step 3: Synthesis
f(x) = xe−x2
..
f′
(x)
.
monotonicity
..
−
√
1/2
.0 .. √
1/2
. 0.− .
↘
.− .
↘
.+ .
↗
. +.
↗
. −.
↘
. −.
↘
.
f′′
(x)
.
concavity
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−
√
1/2
.
− 1√
2e
.
min
..
√
1/2
.
1√
2e
.
max
..
−
√
3/2
.
−
√
3
2e3
.
IP
..
0
.
0
.
IP
..
√
3/2
.
√
3
2e3
.
IP
...
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 55
. . . . . .
Step 3: Synthesis
f(x) = xe−x2
..
f′
(x)
.
monotonicity
..
−
√
1/2
.0 .. √
1/2
. 0.− .
↘
.− .
↘
.+ .
↗
. +.
↗
. −.
↘
. −.
↘
.
f′′
(x)
.
concavity
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−
√
1/2
.
− 1√
2e
.
min
..
√
1/2
.
1√
2e
.
max
..
−
√
3/2
.
−
√
3
2e3
.
IP
..
0
.
0
.
IP
..
√
3/2
.
√
3
2e3
.
IP
....
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 55
. . . . . .
Step 3: Synthesis
f(x) = xe−x2
..
f′
(x)
.
monotonicity
..
−
√
1/2
.0 .. √
1/2
. 0.− .
↘
.− .
↘
.+ .
↗
. +.
↗
. −.
↘
. −.
↘
.
f′′
(x)
.
concavity
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−
√
1/2
.
− 1√
2e
.
min
..
√
1/2
.
1√
2e
.
max
..
−
√
3/2
.
−
√
3
2e3
.
IP
..
0
.
0
.
IP
..
√
3/2
.
√
3
2e3
.
IP
.....
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 55
. . . . . .
Step 3: Synthesis
f(x) = xe−x2
..
f′
(x)
.
monotonicity
..
−
√
1/2
.0 .. √
1/2
. 0.− .
↘
.− .
↘
.+ .
↗
. +.
↗
. −.
↘
. −.
↘
.
f′′
(x)
.
concavity
..
−
√
3/2
.
0
..
0
.
0
..
√
3/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−
√
1/2
.
− 1√
2e
.
min
..
√
1/2
.
1√
2e
.
max
..
−
√
3/2
.
−
√
3
2e3
.
IP
..
0
.
0
.
IP
..
√
3/2
.
√
3
2e3
.
IP
......
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 55
. . . . . .
Step 4: Graph
..
x
.
f(x)
.
f(x) = xe−x2
..
(
−
√
1/2, − 1√
2e
)
..
(√
1/2, 1√
2e
)
..
(
−
√
3/2, −
√
3
2e3
)
..
(0, 0)
..
(
√
3/2,
√
3
2e3
)
.
f(x)
.
shape
..
−
√
1/2
.
− 1√
2e
.
min
.. √
1/2
.
1√
2e
.
max
..
−
√
3/2
.
−
√
3
2e3
.
IP
..
0
.0.
IP
.. √
3/2
.
√
3
2e3
.
IP
......
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
. . . . . .
Step 4: Graph
..
x
.
f(x)
.
f(x) = xe−x2
..
(
−
√
1/2, − 1√
2e
)
..
(√
1/2, 1√
2e
)
..
(
−
√
3/2, −
√
3
2e3
)
..
(0, 0)
..
(
√
3/2,
√
3
2e3
)
.
f(x)
.
shape
..
−
√
1/2
.
− 1√
2e
.
min
.. √
1/2
.
1√
2e
.
max
..
−
√
3/2
.
−
√
3
2e3
.
IP
..
0
.0.
IP
.. √
3/2
.
√
3
2e3
.
IP
......
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
. . . . . .
Step 4: Graph
..
x
.
f(x)
.
f(x) = xe−x2
..
(
−
√
1/2, − 1√
2e
)
..
(√
1/2, 1√
2e
)
..
(
−
√
3/2, −
√
3
2e3
)
..
(0, 0)
..
(
√
3/2,
√
3
2e3
)
.
f(x)
.
shape
..
−
√
1/2
.
− 1√
2e
.
min
.. √
1/2
.
1√
2e
.
max
..
−
√
3/2
.
−
√
3
2e3
.
IP
..
0
.0.
IP
.. √
3/2
.
√
3
2e3
.
IP
......
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
. . . . . .
Step 4: Graph
..
x
.
f(x)
.
f(x) = xe−x2
..
(
−
√
1/2, − 1√
2e
)
..
(√
1/2, 1√
2e
)
..
(
−
√
3/2, −
√
3
2e3
)
..
(0, 0)
..
(
√
3/2,
√
3
2e3
)
.
f(x)
.
shape
..
−
√
1/2
.
− 1√
2e
.
min
.. √
1/2
.
1√
2e
.
max
..
−
√
3/2
.
−
√
3
2e3
.
IP
..
0
.0.
IP
.. √
3/2
.
√
3
2e3
.
IP
......
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
. . . . . .
Step 4: Graph
..
x
.
f(x)
.
f(x) = xe−x2
..
(
−
√
1/2, − 1√
2e
)
..
(√
1/2, 1√
2e
)
..
(
−
√
3/2, −
√
3
2e3
)
..
(0, 0)
..
(
√
3/2,
√
3
2e3
)
.
f(x)
.
shape
..
−
√
1/2
.
− 1√
2e
.
min
.. √
1/2
.
1√
2e
.
max
..
−
√
3/2
.
−
√
3
2e3
.
IP
..
0
.0.
IP
.. √
3/2
.
√
3
2e3
.
IP
......
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
. . . . . .
Step 4: Graph
..
x
.
f(x)
.
f(x) = xe−x2
..
(
−
√
1/2, − 1√
2e
)
..
(√
1/2, 1√
2e
)
..
(
−
√
3/2, −
√
3
2e3
)
..
(0, 0)
..
(
√
3/2,
√
3
2e3
)
.
f(x)
.
shape
..
−
√
1/2
.
− 1√
2e
.
min
.. √
1/2
.
1√
2e
.
max
..
−
√
3/2
.
−
√
3
2e3
.
IP
..
0
.0.
IP
.. √
3/2
.
√
3
2e3
.
IP
......
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
. . . . . .
Step 4: Graph
..
x
.
f(x)
.
f(x) = xe−x2
..
(
−
√
1/2, − 1√
2e
)
..
(√
1/2, 1√
2e
)
..
(
−
√
3/2, −
√
3
2e3
)
..
(0, 0)
..
(
√
3/2,
√
3
2e3
)
.
f(x)
.
shape
..
−
√
1/2
.
− 1√
2e
.
min
.. √
1/2
.
1√
2e
.
max
..
−
√
3/2
.
−
√
3
2e3
.
IP
..
0
.0.
IP
.. √
3/2
.
√
3
2e3
.
IP
......
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
. . . . . .
Example with Vertical Asymptotes
Example
Graph f(x) =
1
x
+
1
x2
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 36 / 55
. . . . . .
Step 0
Find when f is positive, negative, zero, not defined.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 37 / 55
. . . . . .
Step 0
Find when f is positive, negative, zero, not defined. We need to factor f:
f(x) =
1
x
+
1
x2
=
x + 1
x2
.
This means f is 0 at −1 and has trouble at 0. In fact,
lim
x→0
x + 1
x2
= ∞,
so x = 0 is a vertical asymptote of the graph.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 37 / 55
. . . . . .
Step 0
Find when f is positive, negative, zero, not defined. We need to factor f:
f(x) =
1
x
+
1
x2
=
x + 1
x2
.
This means f is 0 at −1 and has trouble at 0. In fact,
lim
x→0
x + 1
x2
= ∞,
so x = 0 is a vertical asymptote of the graph. We can make a sign
chart as follows:
.. x + 1..0 .
−1
.− . +.
x2
..
0
.
0
.
+
.
+
.
f(x)
..
∞
.
0
..
0
.
−1
.
−
.
+
.
+
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 37 / 55
. . . . . .
Step 0, continued
For horizontal asymptotes, notice that
lim
x→∞
x + 1
x2
= 0,
so y = 0 is a horizontal asymptote of the graph. The same is true at
−∞.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 38 / 55
. . . . . .
Step 1: Monotonicity
We have
f′
(x) = −
1
x2
−
2
x3
= −
x + 2
x3
.
The critical points are x = −2 and x = 0. We have the following sign
chart:
.. −(x + 2)..0 .
−2
.+ . −.
x3
..
0
.
0
.
−
.
+
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 55
. . . . . .
Step 1: Monotonicity
We have
f′
(x) = −
1
x2
−
2
x3
= −
x + 2
x3
.
The critical points are x = −2 and x = 0. We have the following sign
chart:
.. −(x + 2)..0 .
−2
.+ . −.
x3
..
0
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 55
. . . . . .
Step 1: Monotonicity
We have
f′
(x) = −
1
x2
−
2
x3
= −
x + 2
x3
.
The critical points are x = −2 and x = 0. We have the following sign
chart:
.. −(x + 2)..0 .
−2
.+ . −.
x3
..
0
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 55
. . . . . .
Step 1: Monotonicity
We have
f′
(x) = −
1
x2
−
2
x3
= −
x + 2
x3
.
The critical points are x = −2 and x = 0. We have the following sign
chart:
.. −(x + 2)..0 .
−2
.+ . −.
x3
..
0
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 55
. . . . . .
Step 1: Monotonicity
We have
f′
(x) = −
1
x2
−
2
x3
= −
x + 2
x3
.
The critical points are x = −2 and x = 0. We have the following sign
chart:
.. −(x + 2)..0 .
−2
.+ . −.
x3
..
0
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 55
. . . . . .
Step 1: Monotonicity
We have
f′
(x) = −
1
x2
−
2
x3
= −
x + 2
x3
.
The critical points are x = −2 and x = 0. We have the following sign
chart:
.. −(x + 2)..0 .
−2
.+ . −.
x3
..
0
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 55
. . . . . .
Step 1: Monotonicity
We have
f′
(x) = −
1
x2
−
2
x3
= −
x + 2
x3
.
The critical points are x = −2 and x = 0. We have the following sign
chart:
.. −(x + 2)..0 .
−2
.+ . −.
x3
..
0
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
min
.
VA
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 55
. . . . . .
Step 2: Concavity
We have
f′′
(x) =
2
x3
+
6
x4
=
2(x + 3)
x4
.
The critical points of f′
are −3 and 0. Sign chart:
.. (x + 3)..0 .
−3
.− . +.
x4
..
0
.
0
.
+
.
+
.
f′′
(x)
.
f(x)
..
∞
.
0
..
0
.
−3
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
. . . . . .
Step 2: Concavity
We have
f′′
(x) =
2
x3
+
6
x4
=
2(x + 3)
x4
.
The critical points of f′
are −3 and 0. Sign chart:
.. (x + 3)..0 .
−3
.− . +.
x4
..
0
.
0
.
+
.
+
.
f′′
(x)
.
f(x)
..
∞
.
0
..
0
.
−3
.
−−
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
. . . . . .
Step 2: Concavity
We have
f′′
(x) =
2
x3
+
6
x4
=
2(x + 3)
x4
.
The critical points of f′
are −3 and 0. Sign chart:
.. (x + 3)..0 .
−3
.− . +.
x4
..
0
.
0
.
+
.
+
.
f′′
(x)
.
f(x)
..
∞
.
0
..
0
.
−3
.
−−
.
++
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
. . . . . .
Step 2: Concavity
We have
f′′
(x) =
2
x3
+
6
x4
=
2(x + 3)
x4
.
The critical points of f′
are −3 and 0. Sign chart:
.. (x + 3)..0 .
−3
.− . +.
x4
..
0
.
0
.
+
.
+
.
f′′
(x)
.
f(x)
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
. . . . . .
Step 2: Concavity
We have
f′′
(x) =
2
x3
+
6
x4
=
2(x + 3)
x4
.
The critical points of f′
are −3 and 0. Sign chart:
.. (x + 3)..0 .
−3
.− . +.
x4
..
0
.
0
.
+
.
+
.
f′′
(x)
.
f(x)
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
. . . . . .
Step 2: Concavity
We have
f′′
(x) =
2
x3
+
6
x4
=
2(x + 3)
x4
.
The critical points of f′
are −3 and 0. Sign chart:
.. (x + 3)..0 .
−3
.− . +.
x4
..
0
.
0
.
+
.
+
.
f′′
(x)
.
f(x)
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
. . . . . .
Step 2: Concavity
We have
f′′
(x) =
2
x3
+
6
x4
=
2(x + 3)
x4
.
The critical points of f′
are −3 and 0. Sign chart:
.. (x + 3)..0 .
−3
.− . +.
x4
..
0
.
0
.
+
.
+
.
f′′
(x)
.
f(x)
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
. . . . . .
Step 2: Concavity
We have
f′′
(x) =
2
x3
+
6
x4
=
2(x + 3)
x4
.
The critical points of f′
are −3 and 0. Sign chart:
.. (x + 3)..0 .
−3
.− . +.
x4
..
0
.
0
.
+
.
+
.
f′′
(x)
.
f(x)
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
. . . . . .
Step 2: Concavity
We have
f′′
(x) =
2
x3
+
6
x4
=
2(x + 3)
x4
.
The critical points of f′
are −3 and 0. Sign chart:
.. (x + 3)..0 .
−3
.− . +.
x4
..
0
.
0
.
+
.
+
.
f′′
(x)
.
f(x)
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
IP
.
VA
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
f
.
shape of f
..
∞
.
0
..
0
.
−1
..
−2
.
−1/4
..
−3
.
−2/9
.
−∞
.
0
.
∞
.
0
.
−
.
+
.
+
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
f
.
shape of f
..
∞
.
0
..
0
.
−1
..
−2
.
−1/4
..
−3
.
−2/9
.
−∞
.
0
.
∞
.
0
.
−
.
+
.
+
.
HA
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
f
.
shape of f
..
∞
.
0
..
0
.
−1
..
−2
.
−1/4
..
−3
.
−2/9
.
−∞
.
0
.
∞
.
0
.
−
.
+
.
+
.
HA
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
f
.
shape of f
..
∞
.
0
..
0
.
−1
..
−2
.
−1/4
..
−3
.
−2/9
.
−∞
.
0
.
∞
.
0
.
−
.
+
.
+
.
HA
..
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
f
.
shape of f
..
∞
.
0
..
0
.
−1
..
−2
.
−1/4
..
−3
.
−2/9
.
−∞
.
0
.
∞
.
0
.
−
.
+
.
+
.
HA
..
IP
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
f
.
shape of f
..
∞
.
0
..
0
.
−1
..
−2
.
−1/4
..
−3
.
−2/9
.
−∞
.
0
.
∞
.
0
.
−
.
+
.
+
.
HA
..
IP
..
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
f
.
shape of f
..
∞
.
0
..
0
.
−1
..
−2
.
−1/4
..
−3
.
−2/9
.
−∞
.
0
.
∞
.
0
.
−
.
+
.
+
.
HA
..
IP
..
min
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
f
.
shape of f
..
∞
.
0
..
0
.
−1
..
−2
.
−1/4
..
−3
.
−2/9
.
−∞
.
0
.
∞
.
0
.
−
.
+
.
+
.
HA
..
IP
..
min
..
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
f
.
shape of f
..
∞
.
0
..
0
.
−1
..
−2
.
−1/4
..
−3
.
−2/9
.
−∞
.
0
.
∞
.
0
.
−
.
+
.
+
.
HA
..
IP
..
min
..
0
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
f
.
shape of f
..
∞
.
0
..
0
.
−1
..
−2
.
−1/4
..
−3
.
−2/9
.
−∞
.
0
.
∞
.
0
.
−
.
+
.
+
.
HA
..
IP
..
min
..
0
..
VA
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
f
.
shape of f
..
∞
.
0
..
0
.
−1
..
−2
.
−1/4
..
−3
.
−2/9
.
−∞
.
0
.
∞
.
0
.
−
.
+
.
+
.
HA
..
IP
..
min
..
0
..
VA
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
f
.
shape of f
..
∞
.
0
..
0
.
−1
..
−2
.
−1/4
..
−3
.
−2/9
.
−∞
.
0
.
∞
.
0
.
−
.
+
.
+
.
HA
..
IP
..
min
..
0
..
VA
..
HA
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
. . . . . .
Step 4: Graph
.. x.
y
..
(−3, −2/9)
..
(−2, −1/4)
.
f
.
shape of f
..
∞
.
0
..
0
.
−1
..
−2
.
−1/4
..
−3
.
−2/9
.
−∞
.
0
.
∞
.
0
.
−
.
+
.
+
.
HA
..
IP
..
min
..
0
..
VA
..
HA
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 42 / 55
. . . . . .
Trigonometric and polynomial together
Problem
Graph f(x) = cos x − x
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 43 / 55
. . . . . .
Step 0: intercepts and asymptotes
f(0) = 1 and f(−π/2) = −π/2. So by the Intermediate Value
Theorem there is a zero in between. We don’t know it’s precise
value, though.
Since −1 ≤ cos x ≤ 1 for all x, we have
−1 − x ≤ cos x − x ≤ 1 − x
for all x. This means that lim
x→∞
f(x) = −∞ and lim
x→−∞
f(x) = ∞.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 44 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = cos x − x, then f′
(x) = − sin x − 1 = (−1)(sin x + 1).
f′
(x) = 0 if x = 3π/2 + 2πk, where k is any integer
f′
(x) is periodic with period 2π
Since −1 ≤ sin x ≤ 1 for all x, we have
0 ≤ sin x + 1 ≤ 2 =⇒ −2 ≤ (−1)(sin x + 1) ≤ 0
for all x. This means f′
(x) is negative at all other points.
..
f′
(x)
.
f(x)
..
−π/2
.0 ..
3π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 45 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = cos x − x, then f′
(x) = − sin x − 1 = (−1)(sin x + 1).
f′
(x) = 0 if x = 3π/2 + 2πk, where k is any integer
f′
(x) is periodic with period 2π
Since −1 ≤ sin x ≤ 1 for all x, we have
0 ≤ sin x + 1 ≤ 2 =⇒ −2 ≤ (−1)(sin x + 1) ≤ 0
for all x. This means f′
(x) is negative at all other points.
..
f′
(x)
.
f(x)
..
−π/2
.0 ..
3π/2
. 0..
7π/2
. 0. −
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 45 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = cos x − x, then f′
(x) = − sin x − 1 = (−1)(sin x + 1).
f′
(x) = 0 if x = 3π/2 + 2πk, where k is any integer
f′
(x) is periodic with period 2π
Since −1 ≤ sin x ≤ 1 for all x, we have
0 ≤ sin x + 1 ≤ 2 =⇒ −2 ≤ (−1)(sin x + 1) ≤ 0
for all x. This means f′
(x) is negative at all other points.
..
f′
(x)
.
f(x)
..
−π/2
.0 ..
3π/2
. 0..
7π/2
. 0. −. −
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 45 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = cos x − x, then f′
(x) = − sin x − 1 = (−1)(sin x + 1).
f′
(x) = 0 if x = 3π/2 + 2πk, where k is any integer
f′
(x) is periodic with period 2π
Since −1 ≤ sin x ≤ 1 for all x, we have
0 ≤ sin x + 1 ≤ 2 =⇒ −2 ≤ (−1)(sin x + 1) ≤ 0
for all x. This means f′
(x) is negative at all other points.
..
f′
(x)
.
f(x)
..
−π/2
.0 ..
3π/2
. 0..
7π/2
. 0. −.
↘
. −
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 45 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = cos x − x, then f′
(x) = − sin x − 1 = (−1)(sin x + 1).
f′
(x) = 0 if x = 3π/2 + 2πk, where k is any integer
f′
(x) is periodic with period 2π
Since −1 ≤ sin x ≤ 1 for all x, we have
0 ≤ sin x + 1 ≤ 2 =⇒ −2 ≤ (−1)(sin x + 1) ≤ 0
for all x. This means f′
(x) is negative at all other points.
..
f′
(x)
.
f(x)
..
−π/2
.0 ..
3π/2
. 0..
7π/2
. 0. −.
↘
. −.
↘
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 45 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
..
−π/2
.0 ..
π/2
. 0..
3π/2
. 0..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−..
−π/2
.0 ..
π/2
. 0..
3π/2
. 0..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−. ++..
−π/2
.0 ..
π/2
. 0..
3π/2
. 0..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−. ++. −−..
−π/2
.0 ..
π/2
. 0..
3π/2
. 0..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−. ++. −−. ++..
−π/2
.0 ..
π/2
. 0..
3π/2
. 0..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−.
⌢
. ++. −−. ++..
−π/2
.0 ..
π/2
. 0..
3π/2
. 0..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−.
⌢
. ++.
⌣
. −−. ++..
−π/2
.0 ..
π/2
. 0..
3π/2
. 0..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−.
⌢
. ++.
⌣
. −−.
⌢
. ++..
−π/2
.0 ..
π/2
. 0..
3π/2
. 0..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−.
⌢
. ++.
⌣
. −−.
⌢
. ++.
⌣
..
−π/2
.0 ..
π/2
. 0..
3π/2
. 0..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−.
⌢
. ++.
⌣
. −−.
⌢
. ++.
⌣
..
−π/2
.0 .
IP
..
π/2
. 0..
3π/2
. 0..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−.
⌢
. ++.
⌣
. −−.
⌢
. ++.
⌣
..
−π/2
.0 .
IP
..
π/2
. 0.
IP
..
3π/2
. 0..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−.
⌢
. ++.
⌣
. −−.
⌢
. ++.
⌣
..
−π/2
.0 .
IP
..
π/2
. 0.
IP
..
3π/2
. 0.
IP
..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−.
⌢
. ++.
⌣
. −−.
⌢
. ++.
⌣
..
−π/2
.0 .
IP
..
π/2
. 0.
IP
..
3π/2
. 0.
IP
..
5π/2
. 0.
IP
..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
. . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−.
⌢
. ++.
⌣
. −−.
⌢
. ++.
⌣
..
−π/2
.0 .
IP
..
π/2
. 0.
IP
..
3π/2
. 0.
IP
..
5π/2
. 0.
IP
..
7π/2
. 0.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
. . . . . .
Step 3: Synthesis
..
f′
(x)
.
mono
..
−π/2
.0 ..
3π/2
. 0..
7π/2
. 0. −.
↘
. −.
↘
.
f′′
(x)
.
conc
..
−π/2
.
0
..
π/2
.
0
..
3π/2
.
0
..
5π/2
.
0
..
7π/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−π/2
.
π/2
.
IP
..
π/2
.
−π/2
.
IP
..
3π/2
.
−3π/2
.
IP
..
5π/2
.
−5π/2
.
IP
..
7π/2
.
−7π/2
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 47 / 55
. . . . . .
Step 3: Synthesis
..
f′
(x)
.
mono
..
−π/2
.0 ..
3π/2
. 0..
7π/2
. 0. −.
↘
. −.
↘
.
f′′
(x)
.
conc
..
−π/2
.
0
..
π/2
.
0
..
3π/2
.
0
..
5π/2
.
0
..
7π/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−π/2
.
π/2
.
IP
..
π/2
.
−π/2
.
IP
..
3π/2
.
−3π/2
.
IP
..
5π/2
.
−5π/2
.
IP
..
7π/2
.
−7π/2
.
IP
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 47 / 55
. . . . . .
Step 3: Synthesis
..
f′
(x)
.
mono
..
−π/2
.0 ..
3π/2
. 0..
7π/2
. 0. −.
↘
. −.
↘
.
f′′
(x)
.
conc
..
−π/2
.
0
..
π/2
.
0
..
3π/2
.
0
..
5π/2
.
0
..
7π/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−π/2
.
π/2
.
IP
..
π/2
.
−π/2
.
IP
..
3π/2
.
−3π/2
.
IP
..
5π/2
.
−5π/2
.
IP
..
7π/2
.
−7π/2
.
IP
..
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 47 / 55
. . . . . .
Step 3: Synthesis
..
f′
(x)
.
mono
..
−π/2
.0 ..
3π/2
. 0..
7π/2
. 0. −.
↘
. −.
↘
.
f′′
(x)
.
conc
..
−π/2
.
0
..
π/2
.
0
..
3π/2
.
0
..
5π/2
.
0
..
7π/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−π/2
.
π/2
.
IP
..
π/2
.
−π/2
.
IP
..
3π/2
.
−3π/2
.
IP
..
5π/2
.
−5π/2
.
IP
..
7π/2
.
−7π/2
.
IP
...
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 47 / 55
. . . . . .
Step 3: Synthesis
..
f′
(x)
.
mono
..
−π/2
.0 ..
3π/2
. 0..
7π/2
. 0. −.
↘
. −.
↘
.
f′′
(x)
.
conc
..
−π/2
.
0
..
π/2
.
0
..
3π/2
.
0
..
5π/2
.
0
..
7π/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−π/2
.
π/2
.
IP
..
π/2
.
−π/2
.
IP
..
3π/2
.
−3π/2
.
IP
..
5π/2
.
−5π/2
.
IP
..
7π/2
.
−7π/2
.
IP
....
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 47 / 55
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)
Lesson 21: Curve Sketching (Section 041 slides)

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Lesson 21: Curve Sketching (Section 041 slides)

  • 1. .. Section 4.4 Curve Sketching V63.0121.041, Calculus I New York University November 17, 2010 Announcements Quiz 4 this week in recitation on 3.3, 3.4, 3.5, 3.7 There is class on November 24 . . . . . .
  • 2. . . . . . . Announcements Quiz 4 this week in recitation on 3.3, 3.4, 3.5, 3.7 There is class on November 24 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 2 / 55
  • 3. . . . . . . Objectives given a function, graph it completely, indicating zeroes (if easy) asymptotes if applicable critical points local/global max/min inflection points V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 3 / 55
  • 4. . . . . . . Why? Graphing functions is like dissection V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 4 / 55
  • 5. . . . . . . Why? Graphing functions is like dissection … or diagramming sentences V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 4 / 55
  • 6. . . . . . . Why? Graphing functions is like dissection … or diagramming sentences You can really know a lot about a function when you know all of its anatomy. V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 4 / 55
  • 7. . . . . . . The Increasing/Decreasing Test Theorem (The Increasing/Decreasing Test) If f′ > 0 on (a, b), then f is increasing on (a, b). If f′ < 0 on (a, b), then f is decreasing on (a, b). Example Here f(x) = x3 + x2 , and f′ (x) = 3x2 + 2x. .. f(x) . f′ (x) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 5 / 55
  • 8. . . . . . . Testing for Concavity Theorem (Concavity Test) If f′′ (x) > 0 for all x in (a, b), then the graph of f is concave upward on (a, b) If f′′ (x) < 0 for all x in (a, b), then the graph of f is concave downward on (a, b). Example Here f(x) = x3 + x2 , f′ (x) = 3x2 + 2x, and f′′ (x) = 6x + 2. .. f(x) . f′ (x) . f′′ (x) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 6 / 55
  • 9. . . . . . . Graphing Checklist To graph a function f, follow this plan: 0. Find when f is positive, negative, zero, not defined. 1. Find f′ and form its sign chart. Conclude information about increasing/decreasing and local max/min. 2. Find f′′ and form its sign chart. Conclude concave up/concave down and inflection. 3. Put together a big chart to assemble monotonicity and concavity data 4. Graph! V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 7 / 55
  • 10. . . . . . . Outline Simple examples A cubic function A quartic function More Examples Points of nondifferentiability Horizontal asymptotes Vertical asymptotes Trigonometric and polynomial together Logarithmic V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 8 / 55
  • 11. . . . . . . Graphing a cubic Example Graph f(x) = 2x3 − 3x2 − 12x. V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 9 / 55
  • 12. . . . . . . Graphing a cubic Example Graph f(x) = 2x3 − 3x2 − 12x. (Step 0) First, let’s find the zeros. We can at least factor out one power of x: f(x) = x(2x2 − 3x − 12) so f(0) = 0. The other factor is a quadratic, so we the other two roots are x = 3 ± √ 32 − 4(2)(−12) 4 = 3 ± √ 105 4 It’s OK to skip this step for now since the roots are so complicated. V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 9 / 55
  • 13. . . . . . . Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: . V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
  • 14. . . . . . . Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: .. x − 2.. 2 . x + 1 .. −1 . f′ (x) . f(x) .. 2 .. −1 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
  • 15. . . . . . . Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: .. x − 2.. 2 .− . −. +. x + 1 .. −1 . f′ (x) . f(x) .. 2 .. −1 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
  • 16. . . . . . . Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: .. x − 2.. 2 .− . −. +. x + 1 .. −1 . + . + . − . f′ (x) . f(x) .. 2 .. −1 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
  • 17. . . . . . . Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: .. x − 2.. 2 .− . −. +. x + 1 .. −1 . + . + . − . f′ (x) . f(x) .. 2 .. −1 . + V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
  • 18. . . . . . . Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: .. x − 2.. 2 .− . −. +. x + 1 .. −1 . + . + . − . f′ (x) . f(x) .. 2 .. −1 . + . − V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
  • 19. . . . . . . Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: .. x − 2.. 2 .− . −. +. x + 1 .. −1 . + . + . − . f′ (x) . f(x) .. 2 .. −1 . + . − . + V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
  • 20. . . . . . . Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: .. x − 2.. 2 .− . −. +. x + 1 .. −1 . + . + . − . f′ (x) . f(x) .. 2 .. −1 . + . − . + . ↗ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
  • 21. . . . . . . Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: .. x − 2.. 2 .− . −. +. x + 1 .. −1 . + . + . − . f′ (x) . f(x) .. 2 .. −1 . + . − . + . ↗ . ↘ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
  • 22. . . . . . . Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: .. x − 2.. 2 .− . −. +. x + 1 .. −1 . + . + . − . f′ (x) . f(x) .. 2 .. −1 . + . − . + . ↗ . ↘ . ↗ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
  • 23. . . . . . . Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: .. x − 2.. 2 .− . −. +. x + 1 .. −1 . + . + . − . f′ (x) . f(x) .. 2 .. −1 . + . − . + . ↗ . ↘ . ↗ . max V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
  • 24. . . . . . . Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: .. x − 2.. 2 .− . −. +. x + 1 .. −1 . + . + . − . f′ (x) . f(x) .. 2 .. −1 . + . − . + . ↗ . ↘ . ↗ . max . min V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
  • 25. . . . . . . Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 11 / 55
  • 26. . . . . . . Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: .. f′′ (x) . f(x) .. 1/2 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 11 / 55
  • 27. . . . . . . Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: .. f′′ (x) . f(x) .. 1/2 . −− V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 11 / 55
  • 28. . . . . . . Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: .. f′′ (x) . f(x) .. 1/2 . −− . ++ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 11 / 55
  • 29. . . . . . . Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: .. f′′ (x) . f(x) .. 1/2 . −− . ++ . ⌢ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 11 / 55
  • 30. . . . . . . Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: .. f′′ (x) . f(x) .. 1/2 . −− . ++ . ⌢ . ⌣ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 11 / 55
  • 31. . . . . . . Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: .. f′′ (x) . f(x) .. 1/2 . −− . ++ . ⌢ . ⌣ . IP V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 11 / 55
  • 32. . . . . . . Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. . V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 12 / 55
  • 33. . . . . . . Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. .. f′ (x) . monotonicity .. −1 .. 2 . +. ↗ .− . ↘ . −. ↘ .+ . ↗ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 12 / 55
  • 34. . . . . . . Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. .. f′ (x) . monotonicity .. −1 .. 2 . +. ↗ .− . ↘ . −. ↘ .+ . ↗ . f′′ (x) . concavity .. 1/2 . −− . ⌢ . −− . ⌢ . ++ . ⌣ . ++ . ⌣ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 12 / 55
  • 35. . . . . . . Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. .. f′ (x) . monotonicity .. −1 .. 2 . +. ↗ .− . ↘ . −. ↘ .+ . ↗ . f′′ (x) . concavity .. 1/2 . −− . ⌢ . −− . ⌢ . ++ . ⌣ . ++ . ⌣ . f(x) . shape of f .. −1 . 7 . max .. 2 . −20 . min .. 1/2 . −61/2 . IP V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 12 / 55
  • 36. . . . . . . Combinations of monotonicity and concavity .. I . II . III . IV V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 13 / 55
  • 37. . . . . . . Combinations of monotonicity and concavity .. I . II . III . IV . decreasing, concave down V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 13 / 55
  • 38. . . . . . . Combinations of monotonicity and concavity .. I . II . III . IV . decreasing, concave down . increasing, concave down V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 13 / 55
  • 39. . . . . . . Combinations of monotonicity and concavity .. I . II . III . IV . decreasing, concave down . increasing, concave down . decreasing, concave up V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 13 / 55
  • 40. . . . . . . Combinations of monotonicity and concavity .. I . II . III . IV . decreasing, concave down . increasing, concave down . decreasing, concave up . increasing, concave up V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 13 / 55
  • 41. . . . . . . Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. .. f′ (x) . monotonicity .. −1 .. 2 . +. ↗ .− . ↘ . −. ↘ .+ . ↗ . f′′ (x) . concavity .. 1/2 . −− . ⌢ . −− . ⌢ . ++ . ⌣ . ++ . ⌣ . f(x) . shape of f .. −1 . 7 . max .. 2 . −20 . min .. 1/2 . −61/2 . IP . V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 14 / 55
  • 42. . . . . . . Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. .. f′ (x) . monotonicity .. −1 .. 2 . +. ↗ .− . ↘ . −. ↘ .+ . ↗ . f′′ (x) . concavity .. 1/2 . −− . ⌢ . −− . ⌢ . ++ . ⌣ . ++ . ⌣ . f(x) . shape of f .. −1 . 7 . max .. 2 . −20 . min .. 1/2 . −61/2 . IP .. V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 14 / 55
  • 43. . . . . . . Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. .. f′ (x) . monotonicity .. −1 .. 2 . +. ↗ .− . ↘ . −. ↘ .+ . ↗ . f′′ (x) . concavity .. 1/2 . −− . ⌢ . −− . ⌢ . ++ . ⌣ . ++ . ⌣ . f(x) . shape of f .. −1 . 7 . max .. 2 . −20 . min .. 1/2 . −61/2 . IP ... V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 14 / 55
  • 44. . . . . . . Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. .. f′ (x) . monotonicity .. −1 .. 2 . +. ↗ .− . ↘ . −. ↘ .+ . ↗ . f′′ (x) . concavity .. 1/2 . −− . ⌢ . −− . ⌢ . ++ . ⌣ . ++ . ⌣ . f(x) . shape of f .. −1 . 7 . max .. 2 . −20 . min .. 1/2 . −61/2 . IP .... V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 14 / 55
  • 45. . . . . . . Step 4: Graph .. f(x) = 2x3 − 3x2 − 12x . x. f(x) . f(x) . shape of f .. −1 . 7 . max .. 2 . −20 . min .. 1/2 . −61/2 . IP ...... ( 3− √ 105 4 , 0 ) .. (−1, 7) .. (0, 0) .. (1/2, −61/2) .. (2, −20) .. ( 3+ √ 105 4 , 0 ) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 15 / 55
  • 46. . . . . . . Step 4: Graph .. f(x) = 2x3 − 3x2 − 12x . x. f(x) . f(x) . shape of f .. −1 . 7 . max .. 2 . −20 . min .. 1/2 . −61/2 . IP ...... ( 3− √ 105 4 , 0 ) .. (−1, 7) .. (0, 0) .. (1/2, −61/2) .. (2, −20) .. ( 3+ √ 105 4 , 0 ) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 15 / 55
  • 47. . . . . . . Step 4: Graph .. f(x) = 2x3 − 3x2 − 12x . x. f(x) . f(x) . shape of f .. −1 . 7 . max .. 2 . −20 . min .. 1/2 . −61/2 . IP ...... ( 3− √ 105 4 , 0 ) .. (−1, 7) .. (0, 0) .. (1/2, −61/2) .. (2, −20) .. ( 3+ √ 105 4 , 0 ) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 15 / 55
  • 48. . . . . . . Step 4: Graph .. f(x) = 2x3 − 3x2 − 12x . x. f(x) . f(x) . shape of f .. −1 . 7 . max .. 2 . −20 . min .. 1/2 . −61/2 . IP ...... ( 3− √ 105 4 , 0 ) .. (−1, 7) .. (0, 0) .. (1/2, −61/2) .. (2, −20) .. ( 3+ √ 105 4 , 0 ) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 15 / 55
  • 49. . . . . . . Step 4: Graph .. f(x) = 2x3 − 3x2 − 12x . x. f(x) . f(x) . shape of f .. −1 . 7 . max .. 2 . −20 . min .. 1/2 . −61/2 . IP ...... ( 3− √ 105 4 , 0 ) .. (−1, 7) .. (0, 0) .. (1/2, −61/2) .. (2, −20) .. ( 3+ √ 105 4 , 0 ) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 15 / 55
  • 50. . . . . . . Graphing a quartic Example Graph f(x) = x4 − 4x3 + 10 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 16 / 55
  • 51. . . . . . . Graphing a quartic Example Graph f(x) = x4 − 4x3 + 10 (Step 0) We know f(0) = 10 and lim x→±∞ f(x) = +∞. Not too many other points on the graph are evident. V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 16 / 55
  • 52. . . . . . . Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
  • 53. . . . . . . Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
  • 54. . . . . . . Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. .. 4x2.. 0 .0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
  • 55. . . . . . . Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. .. 4x2.. 0 .0.+ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
  • 56. . . . . . . Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. .. 4x2.. 0 .0.+ . + V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
  • 57. . . . . . . Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. .. 4x2.. 0 .0.+ . +. + V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
  • 58. . . . . . . Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. .. 4x2.. 0 .0.+ . +. +. (x − 3) .. 3 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
  • 59. . . . . . . Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. .. 4x2.. 0 .0.+ . +. +. (x − 3) .. 3 . 0 . − V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
  • 60. . . . . . . Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. .. 4x2.. 0 .0.+ . +. +. (x − 3) .. 3 . 0 . − . − V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
  • 61. . . . . . . Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. .. 4x2.. 0 .0.+ . +. +. (x − 3) .. 3 . 0 . − . − . + V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
  • 62. . . . . . . Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. .. 4x2.. 0 .0.+ . +. +. (x − 3) .. 3 . 0 . − . − . + . f′ (x) . f(x) .. 3 . 0 .. 0 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
  • 63. . . . . . . Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. .. 4x2.. 0 .0.+ . +. +. (x − 3) .. 3 . 0 . − . − . + . f′ (x) . f(x) .. 3 . 0 .. 0 . 0 . − V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
  • 64. . . . . . . Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. .. 4x2.. 0 .0.+ . +. +. (x − 3) .. 3 . 0 . − . − . + . f′ (x) . f(x) .. 3 . 0 .. 0 . 0 . − . − V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
  • 65. . . . . . . Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. .. 4x2.. 0 .0.+ . +. +. (x − 3) .. 3 . 0 . − . − . + . f′ (x) . f(x) .. 3 . 0 .. 0 . 0 . − . − . + V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
  • 66. . . . . . . Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. .. 4x2.. 0 .0.+ . +. +. (x − 3) .. 3 . 0 . − . − . + . f′ (x) . f(x) .. 3 . 0 .. 0 . 0 . − . − . + . ↘ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
  • 67. . . . . . . Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. .. 4x2.. 0 .0.+ . +. +. (x − 3) .. 3 . 0 . − . − . + . f′ (x) . f(x) .. 3 . 0 .. 0 . 0 . − . − . + . ↘ . ↘ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
  • 68. . . . . . . Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. .. 4x2.. 0 .0.+ . +. +. (x − 3) .. 3 . 0 . − . − . + . f′ (x) . f(x) .. 3 . 0 .. 0 . 0 . − . − . + . ↘ . ↘ . ↗ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
  • 69. . . . . . . Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. .. 4x2.. 0 .0.+ . +. +. (x − 3) .. 3 . 0 . − . − . + . f′ (x) . f(x) .. 3 . 0 .. 0 . 0 . − . − . + . ↘ . ↘ . ↗ . min V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
  • 70. . . . . . . Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) . V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
  • 71. . . . . . . Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: . V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
  • 72. . . . . . . Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: .. 12x.. 0 .0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
  • 73. . . . . . . Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: .. 12x.. 0 .0.− V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
  • 74. . . . . . . Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: .. 12x.. 0 .0.− . + V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
  • 75. . . . . . . Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: .. 12x.. 0 .0.− . +. + V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
  • 76. . . . . . . Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: .. 12x.. 0 .0.− . +. +. x − 2 .. 2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
  • 77. . . . . . . Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: .. 12x.. 0 .0.− . +. +. x − 2 .. 2 . 0 . − V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
  • 78. . . . . . . Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: .. 12x.. 0 .0.− . +. +. x − 2 .. 2 . 0 . − . − V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
  • 79. . . . . . . Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: .. 12x.. 0 .0.− . +. +. x − 2 .. 2 . 0 . − . − . + V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
  • 80. . . . . . . Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: .. 12x.. 0 .0.− . +. +. x − 2 .. 2 . 0 . − . − . + . f′′ (x) . f(x) .. 0 . 0 .. 2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
  • 81. . . . . . . Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: .. 12x.. 0 .0.− . +. +. x − 2 .. 2 . 0 . − . − . + . f′′ (x) . f(x) .. 0 . 0 .. 2 . 0 . ++ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
  • 82. . . . . . . Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: .. 12x.. 0 .0.− . +. +. x − 2 .. 2 . 0 . − . − . + . f′′ (x) . f(x) .. 0 . 0 .. 2 . 0 . ++ . −− V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
  • 83. . . . . . . Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: .. 12x.. 0 .0.− . +. +. x − 2 .. 2 . 0 . − . − . + . f′′ (x) . f(x) .. 0 . 0 .. 2 . 0 . ++ . −− . ++ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
  • 84. . . . . . . Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: .. 12x.. 0 .0.− . +. +. x − 2 .. 2 . 0 . − . − . + . f′′ (x) . f(x) .. 0 . 0 .. 2 . 0 . ++ . −− . ++ . ⌣ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
  • 85. . . . . . . Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: .. 12x.. 0 .0.− . +. +. x − 2 .. 2 . 0 . − . − . + . f′′ (x) . f(x) .. 0 . 0 .. 2 . 0 . ++ . −− . ++ . ⌣ . ⌢ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
  • 86. . . . . . . Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: .. 12x.. 0 .0.− . +. +. x − 2 .. 2 . 0 . − . − . + . f′′ (x) . f(x) .. 0 . 0 .. 2 . 0 . ++ . −− . ++ . ⌣ . ⌢ . ⌣ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
  • 87. . . . . . . Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: .. 12x.. 0 .0.− . +. +. x − 2 .. 2 . 0 . − . − . + . f′′ (x) . f(x) .. 0 . 0 .. 2 . 0 . ++ . −− . ++ . ⌣ . ⌢ . ⌣ . IP V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
  • 88. . . . . . . Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: .. 12x.. 0 .0.− . +. +. x − 2 .. 2 . 0 . − . − . + . f′′ (x) . f(x) .. 0 . 0 .. 2 . 0 . ++ . −− . ++ . ⌣ . ⌢ . ⌣ . IP . IP V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
  • 89. . . . . . . Step 3: Grand Unified Sign Chart Remember, f(x) = x4 − 4x3 + 10. .. f′ (x) . monotonicity .. 3 . 0 .. 0 . 0 . − . ↘ . − . ↘ . − . ↘ . + . ↗ . f′′ (x) . concavity .. 0 . 0 .. 2 . 0 . ++ . ⌣ . −− . ⌢ . ++ . ⌣ . ++ . ⌣ . f(x) . shape .. 0 . 10 . IP .. 2 . −6 . IP .. 3 . −17 . min V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 19 / 55
  • 90. . . . . . . Step 3: Grand Unified Sign Chart Remember, f(x) = x4 − 4x3 + 10. .. f′ (x) . monotonicity .. 3 . 0 .. 0 . 0 . − . ↘ . − . ↘ . − . ↘ . + . ↗ . f′′ (x) . concavity .. 0 . 0 .. 2 . 0 . ++ . ⌣ . −− . ⌢ . ++ . ⌣ . ++ . ⌣ . f(x) . shape .. 0 . 10 . IP .. 2 . −6 . IP .. 3 . −17 . min . V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 19 / 55
  • 91. . . . . . . Step 3: Grand Unified Sign Chart Remember, f(x) = x4 − 4x3 + 10. .. f′ (x) . monotonicity .. 3 . 0 .. 0 . 0 . − . ↘ . − . ↘ . − . ↘ . + . ↗ . f′′ (x) . concavity .. 0 . 0 .. 2 . 0 . ++ . ⌣ . −− . ⌢ . ++ . ⌣ . ++ . ⌣ . f(x) . shape .. 0 . 10 . IP .. 2 . −6 . IP .. 3 . −17 . min .. V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 19 / 55
  • 92. . . . . . . Step 3: Grand Unified Sign Chart Remember, f(x) = x4 − 4x3 + 10. .. f′ (x) . monotonicity .. 3 . 0 .. 0 . 0 . − . ↘ . − . ↘ . − . ↘ . + . ↗ . f′′ (x) . concavity .. 0 . 0 .. 2 . 0 . ++ . ⌣ . −− . ⌢ . ++ . ⌣ . ++ . ⌣ . f(x) . shape .. 0 . 10 . IP .. 2 . −6 . IP .. 3 . −17 . min ... V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 19 / 55
  • 93. . . . . . . Step 3: Grand Unified Sign Chart Remember, f(x) = x4 − 4x3 + 10. .. f′ (x) . monotonicity .. 3 . 0 .. 0 . 0 . − . ↘ . − . ↘ . − . ↘ . + . ↗ . f′′ (x) . concavity .. 0 . 0 .. 2 . 0 . ++ . ⌣ . −− . ⌢ . ++ . ⌣ . ++ . ⌣ . f(x) . shape .. 0 . 10 . IP .. 2 . −6 . IP .. 3 . −17 . min .... V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 19 / 55
  • 94. . . . . . . Step 4: Graph .. f(x) = x4 − 4x3 + 10 . x. y . f(x) . shape .. 0 . 10 . IP .. 2 . −6 . IP .. 3 . −17 . min ...... (0, 10) .. (2, −6) .. (3, −17) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 20 / 55
  • 95. . . . . . . Step 4: Graph .. f(x) = x4 − 4x3 + 10 . x. y . f(x) . shape .. 0 . 10 . IP .. 2 . −6 . IP .. 3 . −17 . min ...... (0, 10) .. (2, −6) .. (3, −17) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 20 / 55
  • 96. . . . . . . Step 4: Graph .. f(x) = x4 − 4x3 + 10 . x. y . f(x) . shape .. 0 . 10 . IP .. 2 . −6 . IP .. 3 . −17 . min ...... (0, 10) .. (2, −6) .. (3, −17) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 20 / 55
  • 97. . . . . . . Step 4: Graph .. f(x) = x4 − 4x3 + 10 . x. y . f(x) . shape .. 0 . 10 . IP .. 2 . −6 . IP .. 3 . −17 . min ...... (0, 10) .. (2, −6) .. (3, −17) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 20 / 55
  • 98. . . . . . . Step 4: Graph .. f(x) = x4 − 4x3 + 10 . x. y . f(x) . shape .. 0 . 10 . IP .. 2 . −6 . IP .. 3 . −17 . min ...... (0, 10) .. (2, −6) .. (3, −17) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 20 / 55
  • 99. . . . . . . Outline Simple examples A cubic function A quartic function More Examples Points of nondifferentiability Horizontal asymptotes Vertical asymptotes Trigonometric and polynomial together Logarithmic V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 21 / 55
  • 100. . . . . . . Graphing a function with a cusp Example Graph f(x) = x + √ |x| V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 22 / 55
  • 101. . . . . . . Graphing a function with a cusp Example Graph f(x) = x + √ |x| This function looks strange because of the absolute value. But whenever we become nervous, we can just take cases. V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 22 / 55
  • 102. . . . . . . Step 0: Finding Zeroes f(x) = x + √ |x| First, look at f by itself. We can tell that f(0) = 0 and that f(x) > 0 if x is positive. V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 23 / 55
  • 103. . . . . . . Step 0: Finding Zeroes f(x) = x + √ |x| First, look at f by itself. We can tell that f(0) = 0 and that f(x) > 0 if x is positive. Are there negative numbers which are zeroes for f? V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 23 / 55
  • 104. . . . . . . Step 0: Finding Zeroes f(x) = x + √ |x| First, look at f by itself. We can tell that f(0) = 0 and that f(x) > 0 if x is positive. Are there negative numbers which are zeroes for f? x + √ −x = 0 √ −x = −x −x = x2 x2 + x = 0 The only solutions are x = 0 and x = −1. V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 23 / 55
  • 105. . . . . . . Step 0: Asymptotic behavior f(x) = x + √ |x| lim x→∞ f(x) = ∞, because both terms tend to ∞. V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 24 / 55
  • 106. . . . . . . Step 0: Asymptotic behavior f(x) = x + √ |x| lim x→∞ f(x) = ∞, because both terms tend to ∞. lim x→−∞ f(x) is indeterminate of the form −∞ + ∞. It’s the same as lim y→+∞ (−y + √ y) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 24 / 55
  • 107. . . . . . . Step 0: Asymptotic behavior f(x) = x + √ |x| lim x→∞ f(x) = ∞, because both terms tend to ∞. lim x→−∞ f(x) is indeterminate of the form −∞ + ∞. It’s the same as lim y→+∞ (−y + √ y) lim y→+∞ (−y + √ y) = lim y→∞ ( √ y − y) · √ y + y √ y + y = lim y→∞ y − y2 √ y + y = −∞ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 24 / 55
  • 108. . . . . . . Step 1: The derivative Remember, f(x) = x + √ |x|. To find f′ , first assume x > 0. Then f′ (x) = d dx ( x + √ x ) = 1 + 1 2 √ x V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 25 / 55
  • 109. . . . . . . Step 1: The derivative Remember, f(x) = x + √ |x|. To find f′ , first assume x > 0. Then f′ (x) = d dx ( x + √ x ) = 1 + 1 2 √ x Notice f′ (x) > 0 when x > 0 (so no critical points here) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 25 / 55
  • 110. . . . . . . Step 1: The derivative Remember, f(x) = x + √ |x|. To find f′ , first assume x > 0. Then f′ (x) = d dx ( x + √ x ) = 1 + 1 2 √ x Notice f′ (x) > 0 when x > 0 (so no critical points here) lim x→0+ f′ (x) = ∞ (so 0 is a critical point) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 25 / 55
  • 111. . . . . . . Step 1: The derivative Remember, f(x) = x + √ |x|. To find f′ , first assume x > 0. Then f′ (x) = d dx ( x + √ x ) = 1 + 1 2 √ x Notice f′ (x) > 0 when x > 0 (so no critical points here) lim x→0+ f′ (x) = ∞ (so 0 is a critical point) lim x→∞ f′ (x) = 1 (so the graph is asymptotic to a line of slope 1) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 25 / 55
  • 112. . . . . . . Step 1: The derivative Remember, f(x) = x + √ |x|. If x is negative, we have f′ (x) = d dx ( x + √ −x ) = 1 − 1 2 √ −x Notice lim x→0− f′ (x) = −∞ (other side of the critical point) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 26 / 55
  • 113. . . . . . . Step 1: The derivative Remember, f(x) = x + √ |x|. If x is negative, we have f′ (x) = d dx ( x + √ −x ) = 1 − 1 2 √ −x Notice lim x→0− f′ (x) = −∞ (other side of the critical point) lim x→−∞ f′ (x) = 1 (asymptotic to a line of slope 1) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 26 / 55
  • 114. . . . . . . Step 1: The derivative Remember, f(x) = x + √ |x|. If x is negative, we have f′ (x) = d dx ( x + √ −x ) = 1 − 1 2 √ −x Notice lim x→0− f′ (x) = −∞ (other side of the critical point) lim x→−∞ f′ (x) = 1 (asymptotic to a line of slope 1) f′ (x) = 0 when 1 − 1 2 √ −x = 0 =⇒ √ −x = 1 2 =⇒ −x = 1 4 =⇒ x = − 1 4 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 26 / 55
  • 115. . . . . . . Step 1: Monotonicity f′ (x) =    1 + 1 2 √ x if x > 0 1 − 1 2 √ −x if x < 0 We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. .. f′ (x) . f(x) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
  • 116. . . . . . . Step 1: Monotonicity f′ (x) =    1 + 1 2 √ x if x > 0 1 − 1 2 √ −x if x < 0 We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. .. f′ (x) . f(x) .. −1 4 .0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
  • 117. . . . . . . Step 1: Monotonicity f′ (x) =    1 + 1 2 √ x if x > 0 1 − 1 2 √ −x if x < 0 We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. .. f′ (x) . f(x) .. −1 4 .0 .. 0 .∞ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
  • 118. . . . . . . Step 1: Monotonicity f′ (x) =    1 + 1 2 √ x if x > 0 1 − 1 2 √ −x if x < 0 We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. .. f′ (x) . f(x) .. −1 4 .0 .. 0 .∞.+ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
  • 119. . . . . . . Step 1: Monotonicity f′ (x) =    1 + 1 2 √ x if x > 0 1 − 1 2 √ −x if x < 0 We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. .. f′ (x) . f(x) .. −1 4 .0 .. 0 .∞.+ .− V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
  • 120. . . . . . . Step 1: Monotonicity f′ (x) =    1 + 1 2 √ x if x > 0 1 − 1 2 √ −x if x < 0 We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. .. f′ (x) . f(x) .. −1 4 .0 .. 0 .∞.+ .− . + V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
  • 121. . . . . . . Step 1: Monotonicity f′ (x) =    1 + 1 2 √ x if x > 0 1 − 1 2 √ −x if x < 0 We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. .. f′ (x) . f(x) .. −1 4 .0 .. 0 .∞.+ .− . +. ↗ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
  • 122. . . . . . . Step 1: Monotonicity f′ (x) =    1 + 1 2 √ x if x > 0 1 − 1 2 √ −x if x < 0 We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. .. f′ (x) . f(x) .. −1 4 .0 .. 0 .∞.+ .− . +. ↗ . ↘ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
  • 123. . . . . . . Step 1: Monotonicity f′ (x) =    1 + 1 2 √ x if x > 0 1 − 1 2 √ −x if x < 0 We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. .. f′ (x) . f(x) .. −1 4 .0 .. 0 .∞.+ .− . +. ↗ . ↘ . ↗ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
  • 124. . . . . . . Step 1: Monotonicity f′ (x) =    1 + 1 2 √ x if x > 0 1 − 1 2 √ −x if x < 0 We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. .. f′ (x) . f(x) .. −1 4 .0 .. 0 .∞.+ .− . +. ↗ . ↘ . ↗ . max V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
  • 125. . . . . . . Step 1: Monotonicity f′ (x) =    1 + 1 2 √ x if x > 0 1 − 1 2 √ −x if x < 0 We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. .. f′ (x) . f(x) .. −1 4 .0 .. 0 .∞.+ .− . +. ↗ . ↘ . ↗ . max . min V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
  • 126. . . . . . . Step 2: Concavity If x > 0, then f′′ (x) = d dx ( 1 + 1 2 x−1/2 ) = − 1 4 x−3/2 This is negative whenever x > 0. V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 28 / 55
  • 127. . . . . . . Step 2: Concavity If x > 0, then f′′ (x) = d dx ( 1 + 1 2 x−1/2 ) = − 1 4 x−3/2 This is negative whenever x > 0. If x < 0, then f′′ (x) = d dx ( 1 − 1 2 (−x)−1/2 ) = − 1 4 (−x)−3/2 which is also always negative for negative x. V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 28 / 55
  • 128. . . . . . . Step 2: Concavity If x > 0, then f′′ (x) = d dx ( 1 + 1 2 x−1/2 ) = − 1 4 x−3/2 This is negative whenever x > 0. If x < 0, then f′′ (x) = d dx ( 1 − 1 2 (−x)−1/2 ) = − 1 4 (−x)−3/2 which is also always negative for negative x. In other words, f′′ (x) = − 1 4 |x|−3/2 . V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 28 / 55
  • 129. . . . . . . Step 2: Concavity If x > 0, then f′′ (x) = d dx ( 1 + 1 2 x−1/2 ) = − 1 4 x−3/2 This is negative whenever x > 0. If x < 0, then f′′ (x) = d dx ( 1 − 1 2 (−x)−1/2 ) = − 1 4 (−x)−3/2 which is also always negative for negative x. In other words, f′′ (x) = − 1 4 |x|−3/2 . Here is the sign chart: .. f′′ (x) . f(x) .. 0 .−∞.−− . ⌢ ... −−. ⌢ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 28 / 55
  • 130. . . . . . . Step 3: Synthesis Now we can put these things together. f(x) = x + √ |x| .. f′ (x) . monotonicity .. −1 4 .0 .. 0 .∞.+1 . ↗ .+ . ↗ .− . ↘ . +. ↗ . +1. ↗ . f′′ (x) . concavity .. 0 . −∞ . −− . ⌢ . −− . ⌢ . −− . ⌢ . −∞ . ⌢ . −∞ . ⌢ . f(x) . shape .. −1 . 0 . zero .. −1 4 . 1 4 . max .. 0 . 0 . min . −∞ . +∞ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 29 / 55
  • 131. . . . . . . Step 3: Synthesis Now we can put these things together. f(x) = x + √ |x| .. f′ (x) . monotonicity .. −1 4 .0 .. 0 .∞.+1 . ↗ .+ . ↗ .− . ↘ . +. ↗ . +1. ↗ . f′′ (x) . concavity .. 0 . −∞ . −− . ⌢ . −− . ⌢ . −− . ⌢ . −∞ . ⌢ . −∞ . ⌢ . f(x) . shape .. −1 . 0 . zero .. −1 4 . 1 4 . max .. 0 . 0 . min . −∞ . +∞ . V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 29 / 55
  • 132. . . . . . . Step 3: Synthesis Now we can put these things together. f(x) = x + √ |x| .. f′ (x) . monotonicity .. −1 4 .0 .. 0 .∞.+1 . ↗ .+ . ↗ .− . ↘ . +. ↗ . +1. ↗ . f′′ (x) . concavity .. 0 . −∞ . −− . ⌢ . −− . ⌢ . −− . ⌢ . −∞ . ⌢ . −∞ . ⌢ . f(x) . shape .. −1 . 0 . zero .. −1 4 . 1 4 . max .. 0 . 0 . min . −∞ . +∞ .. V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 29 / 55
  • 133. . . . . . . Step 3: Synthesis Now we can put these things together. f(x) = x + √ |x| .. f′ (x) . monotonicity .. −1 4 .0 .. 0 .∞.+1 . ↗ .+ . ↗ .− . ↘ . +. ↗ . +1. ↗ . f′′ (x) . concavity .. 0 . −∞ . −− . ⌢ . −− . ⌢ . −− . ⌢ . −∞ . ⌢ . −∞ . ⌢ . f(x) . shape .. −1 . 0 . zero .. −1 4 . 1 4 . max .. 0 . 0 . min . −∞ . +∞ ... V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 29 / 55
  • 134. . . . . . . Step 3: Synthesis Now we can put these things together. f(x) = x + √ |x| .. f′ (x) . monotonicity .. −1 4 .0 .. 0 .∞.+1 . ↗ .+ . ↗ .− . ↘ . +. ↗ . +1. ↗ . f′′ (x) . concavity .. 0 . −∞ . −− . ⌢ . −− . ⌢ . −− . ⌢ . −∞ . ⌢ . −∞ . ⌢ . f(x) . shape .. −1 . 0 . zero .. −1 4 . 1 4 . max .. 0 . 0 . min . −∞ . +∞ .... V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 29 / 55
  • 135. . . . . . . Graph f(x) = x + √ |x| .. f(x) . shape .. −1 . 0 . zero . −∞ . +∞ .. −1 4 . 1 4 . max . −∞ . +∞ .. 0 . 0 . min . −∞ . +∞ ..... x. f(x) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 30 / 55
  • 136. . . . . . . Graph f(x) = x + √ |x| .. f(x) . shape .. −1 . 0 . zero . −∞ . +∞ .. −1 4 . 1 4 . max . −∞ . +∞ .. 0 . 0 . min . −∞ . +∞ ..... x. f(x) .. (−1, 0) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 30 / 55
  • 137. . . . . . . Graph f(x) = x + √ |x| .. f(x) . shape .. −1 . 0 . zero . −∞ . +∞ .. −1 4 . 1 4 . max . −∞ . +∞ .. 0 . 0 . min . −∞ . +∞ ..... x. f(x) .. (−1, 0) .. (−1 4 , 1 4 ) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 30 / 55
  • 138. . . . . . . Graph f(x) = x + √ |x| .. f(x) . shape .. −1 . 0 . zero . −∞ . +∞ .. −1 4 . 1 4 . max . −∞ . +∞ .. 0 . 0 . min . −∞ . +∞ ..... x. f(x) .. (−1, 0) .. (−1 4 , 1 4 ) .. (0, 0) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 30 / 55
  • 139. . . . . . . Graph f(x) = x + √ |x| .. f(x) . shape .. −1 . 0 . zero . −∞ . +∞ .. −1 4 . 1 4 . max . −∞ . +∞ .. 0 . 0 . min . −∞ . +∞ ..... x. f(x) .. (−1, 0) .. (−1 4 , 1 4 ) .. (0, 0) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 30 / 55
  • 140. . . . . . . Graph f(x) = x + √ |x| .. f(x) . shape .. −1 . 0 . zero . −∞ . +∞ .. −1 4 . 1 4 . max . −∞ . +∞ .. 0 . 0 . min . −∞ . +∞ ..... x. f(x) .. (−1, 0) .. (−1 4 , 1 4 ) .. (0, 0) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 30 / 55
  • 141. . . . . . . Graph f(x) = x + √ |x| .. f(x) . shape .. −1 . 0 . zero . −∞ . +∞ .. −1 4 . 1 4 . max . −∞ . +∞ .. 0 . 0 . min . −∞ . +∞ ..... x. f(x) .. (−1, 0) .. (−1 4 , 1 4 ) .. (0, 0) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 30 / 55
  • 142. . . . . . . Example with Horizontal Asymptotes Example Graph f(x) = xe−x2 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 31 / 55
  • 143. . . . . . . Example with Horizontal Asymptotes Example Graph f(x) = xe−x2 Before taking derivatives, we notice that f is odd, that f(0) = 0, and lim x→∞ f(x) = 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 31 / 55
  • 144. . . . . . . Step 1: Monotonicity If f(x) = xe−x2 , then f′ (x) = 1 · e−x2 + xe−x2 (−2x) = ( 1 − 2x2 ) e−x2 = ( 1 − √ 2x ) ( 1 + √ 2x ) e−x2 The factor e−x2 is always positive so it doesn’t figure into the sign of f′ (x). So our sign chart looks like this: .. 1 − √ 2x.. √ 1/2 . 0. 1 + √ 2x .. − √ 1/2 . 0 . f′ (x) . f(x) .. − √ 1/2 . 0 .. √ 1/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
  • 145. . . . . . . Step 1: Monotonicity If f(x) = xe−x2 , then f′ (x) = 1 · e−x2 + xe−x2 (−2x) = ( 1 − 2x2 ) e−x2 = ( 1 − √ 2x ) ( 1 + √ 2x ) e−x2 The factor e−x2 is always positive so it doesn’t figure into the sign of f′ (x). So our sign chart looks like this: .. 1 − √ 2x.. √ 1/2 . 0.+ . 1 + √ 2x .. − √ 1/2 . 0 . f′ (x) . f(x) .. − √ 1/2 . 0 .. √ 1/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
  • 146. . . . . . . Step 1: Monotonicity If f(x) = xe−x2 , then f′ (x) = 1 · e−x2 + xe−x2 (−2x) = ( 1 − 2x2 ) e−x2 = ( 1 − √ 2x ) ( 1 + √ 2x ) e−x2 The factor e−x2 is always positive so it doesn’t figure into the sign of f′ (x). So our sign chart looks like this: .. 1 − √ 2x.. √ 1/2 . 0.+ .+. 1 + √ 2x .. − √ 1/2 . 0 . f′ (x) . f(x) .. − √ 1/2 . 0 .. √ 1/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
  • 147. . . . . . . Step 1: Monotonicity If f(x) = xe−x2 , then f′ (x) = 1 · e−x2 + xe−x2 (−2x) = ( 1 − 2x2 ) e−x2 = ( 1 − √ 2x ) ( 1 + √ 2x ) e−x2 The factor e−x2 is always positive so it doesn’t figure into the sign of f′ (x). So our sign chart looks like this: .. 1 − √ 2x.. √ 1/2 . 0.+ .+. −. 1 + √ 2x .. − √ 1/2 . 0 . f′ (x) . f(x) .. − √ 1/2 . 0 .. √ 1/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
  • 148. . . . . . . Step 1: Monotonicity If f(x) = xe−x2 , then f′ (x) = 1 · e−x2 + xe−x2 (−2x) = ( 1 − 2x2 ) e−x2 = ( 1 − √ 2x ) ( 1 + √ 2x ) e−x2 The factor e−x2 is always positive so it doesn’t figure into the sign of f′ (x). So our sign chart looks like this: .. 1 − √ 2x.. √ 1/2 . 0.+ .+. −. 1 + √ 2x .. − √ 1/2 . 0 . − . f′ (x) . f(x) .. − √ 1/2 . 0 .. √ 1/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
  • 149. . . . . . . Step 1: Monotonicity If f(x) = xe−x2 , then f′ (x) = 1 · e−x2 + xe−x2 (−2x) = ( 1 − 2x2 ) e−x2 = ( 1 − √ 2x ) ( 1 + √ 2x ) e−x2 The factor e−x2 is always positive so it doesn’t figure into the sign of f′ (x). So our sign chart looks like this: .. 1 − √ 2x.. √ 1/2 . 0.+ .+. −. 1 + √ 2x .. − √ 1/2 . 0 . − . + . f′ (x) . f(x) .. − √ 1/2 . 0 .. √ 1/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
  • 150. . . . . . . Step 1: Monotonicity If f(x) = xe−x2 , then f′ (x) = 1 · e−x2 + xe−x2 (−2x) = ( 1 − 2x2 ) e−x2 = ( 1 − √ 2x ) ( 1 + √ 2x ) e−x2 The factor e−x2 is always positive so it doesn’t figure into the sign of f′ (x). So our sign chart looks like this: .. 1 − √ 2x.. √ 1/2 . 0.+ .+. −. 1 + √ 2x .. − √ 1/2 . 0 . − . + . + . f′ (x) . f(x) .. − √ 1/2 . 0 .. √ 1/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
  • 151. . . . . . . Step 1: Monotonicity If f(x) = xe−x2 , then f′ (x) = 1 · e−x2 + xe−x2 (−2x) = ( 1 − 2x2 ) e−x2 = ( 1 − √ 2x ) ( 1 + √ 2x ) e−x2 The factor e−x2 is always positive so it doesn’t figure into the sign of f′ (x). So our sign chart looks like this: .. 1 − √ 2x.. √ 1/2 . 0.+ .+. −. 1 + √ 2x .. − √ 1/2 . 0 . − . + . + . f′ (x) . f(x) . − .. − √ 1/2 . 0 .. √ 1/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
  • 152. . . . . . . Step 1: Monotonicity If f(x) = xe−x2 , then f′ (x) = 1 · e−x2 + xe−x2 (−2x) = ( 1 − 2x2 ) e−x2 = ( 1 − √ 2x ) ( 1 + √ 2x ) e−x2 The factor e−x2 is always positive so it doesn’t figure into the sign of f′ (x). So our sign chart looks like this: .. 1 − √ 2x.. √ 1/2 . 0.+ .+. −. 1 + √ 2x .. − √ 1/2 . 0 . − . + . + . f′ (x) . f(x) . − . + .. − √ 1/2 . 0 .. √ 1/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
  • 153. . . . . . . Step 1: Monotonicity If f(x) = xe−x2 , then f′ (x) = 1 · e−x2 + xe−x2 (−2x) = ( 1 − 2x2 ) e−x2 = ( 1 − √ 2x ) ( 1 + √ 2x ) e−x2 The factor e−x2 is always positive so it doesn’t figure into the sign of f′ (x). So our sign chart looks like this: .. 1 − √ 2x.. √ 1/2 . 0.+ .+. −. 1 + √ 2x .. − √ 1/2 . 0 . − . + . + . f′ (x) . f(x) . − . ↘ . + . − .. − √ 1/2 . 0 .. √ 1/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
  • 154. . . . . . . Step 1: Monotonicity If f(x) = xe−x2 , then f′ (x) = 1 · e−x2 + xe−x2 (−2x) = ( 1 − 2x2 ) e−x2 = ( 1 − √ 2x ) ( 1 + √ 2x ) e−x2 The factor e−x2 is always positive so it doesn’t figure into the sign of f′ (x). So our sign chart looks like this: .. 1 − √ 2x.. √ 1/2 . 0.+ .+. −. 1 + √ 2x .. − √ 1/2 . 0 . − . + . + . f′ (x) . f(x) . − . ↘ . + . ↗ . − .. − √ 1/2 . 0 .. √ 1/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
  • 155. . . . . . . Step 1: Monotonicity If f(x) = xe−x2 , then f′ (x) = 1 · e−x2 + xe−x2 (−2x) = ( 1 − 2x2 ) e−x2 = ( 1 − √ 2x ) ( 1 + √ 2x ) e−x2 The factor e−x2 is always positive so it doesn’t figure into the sign of f′ (x). So our sign chart looks like this: .. 1 − √ 2x.. √ 1/2 . 0.+ .+. −. 1 + √ 2x .. − √ 1/2 . 0 . − . + . + . f′ (x) . f(x) . − . ↘ . + . ↗ . − . ↘ .. − √ 1/2 . 0 .. √ 1/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
  • 156. . . . . . . Step 1: Monotonicity If f(x) = xe−x2 , then f′ (x) = 1 · e−x2 + xe−x2 (−2x) = ( 1 − 2x2 ) e−x2 = ( 1 − √ 2x ) ( 1 + √ 2x ) e−x2 The factor e−x2 is always positive so it doesn’t figure into the sign of f′ (x). So our sign chart looks like this: .. 1 − √ 2x.. √ 1/2 . 0.+ .+. −. 1 + √ 2x .. − √ 1/2 . 0 . − . + . + . f′ (x) . f(x) . − . ↘ . + . ↗ . − . ↘ .. − √ 1/2 . 0 .. √ 1/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
  • 157. . . . . . . Step 1: Monotonicity If f(x) = xe−x2 , then f′ (x) = 1 · e−x2 + xe−x2 (−2x) = ( 1 − 2x2 ) e−x2 = ( 1 − √ 2x ) ( 1 + √ 2x ) e−x2 The factor e−x2 is always positive so it doesn’t figure into the sign of f′ (x). So our sign chart looks like this: .. 1 − √ 2x.. √ 1/2 . 0.+ .+. −. 1 + √ 2x .. − √ 1/2 . 0 . − . + . + . f′ (x) . f(x) . − . ↘ . + . ↗ . − . ↘ .. − √ 1/2 . 0 . min .. √ 1/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
  • 158. . . . . . . Step 1: Monotonicity If f(x) = xe−x2 , then f′ (x) = 1 · e−x2 + xe−x2 (−2x) = ( 1 − 2x2 ) e−x2 = ( 1 − √ 2x ) ( 1 + √ 2x ) e−x2 The factor e−x2 is always positive so it doesn’t figure into the sign of f′ (x). So our sign chart looks like this: .. 1 − √ 2x.. √ 1/2 . 0.+ .+. −. 1 + √ 2x .. − √ 1/2 . 0 . − . + . + . f′ (x) . f(x) . − . ↘ . + . ↗ . − . ↘ .. − √ 1/2 . 0 . min .. √ 1/2 . 0 . max V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
  • 159. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0. √ 2x − √ 3 .. √ 3/2 . 0 . √ 2x + √ 3 .. − √ 3/2 . 0 . f′′ (x) . f(x) .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 160. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− . √ 2x − √ 3 .. √ 3/2 . 0 . √ 2x + √ 3 .. − √ 3/2 . 0 . f′′ (x) . f(x) .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 161. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− .− . √ 2x − √ 3 .. √ 3/2 . 0 . √ 2x + √ 3 .. − √ 3/2 . 0 . f′′ (x) . f(x) .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 162. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− .− . +. √ 2x − √ 3 .. √ 3/2 . 0 . √ 2x + √ 3 .. − √ 3/2 . 0 . f′′ (x) . f(x) .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 163. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− .− . +. +. √ 2x − √ 3 .. √ 3/2 . 0 . √ 2x + √ 3 .. − √ 3/2 . 0 . f′′ (x) . f(x) .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 164. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− .− . +. +. √ 2x − √ 3 .. √ 3/2 . 0 . − . √ 2x + √ 3 .. − √ 3/2 . 0 . f′′ (x) . f(x) .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 165. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− .− . +. +. √ 2x − √ 3 .. √ 3/2 . 0 . − . − . √ 2x + √ 3 .. − √ 3/2 . 0 . f′′ (x) . f(x) .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 166. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− .− . +. +. √ 2x − √ 3 .. √ 3/2 . 0 . − . − . − . √ 2x + √ 3 .. − √ 3/2 . 0 . f′′ (x) . f(x) .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 167. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− .− . +. +. √ 2x − √ 3 .. √ 3/2 . 0 . − . − . − . + . √ 2x + √ 3 .. − √ 3/2 . 0 . f′′ (x) . f(x) .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 168. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− .− . +. +. √ 2x − √ 3 .. √ 3/2 . 0 . − . − . − . + . √ 2x + √ 3 .. − √ 3/2 . 0 . − . f′′ (x) . f(x) .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 169. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− .− . +. +. √ 2x − √ 3 .. √ 3/2 . 0 . − . − . − . + . √ 2x + √ 3 .. − √ 3/2 . 0 . − . + . f′′ (x) . f(x) .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 170. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− .− . +. +. √ 2x − √ 3 .. √ 3/2 . 0 . − . − . − . + . √ 2x + √ 3 .. − √ 3/2 . 0 . − . + . + . f′′ (x) . f(x) .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 171. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− .− . +. +. √ 2x − √ 3 .. √ 3/2 . 0 . − . − . − . + . √ 2x + √ 3 .. − √ 3/2 . 0 . − . + . + . + . f′′ (x) . f(x) .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 172. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− .− . +. +. √ 2x − √ 3 .. √ 3/2 . 0 . − . − . − . + . √ 2x + √ 3 .. − √ 3/2 . 0 . − . + . + . + . f′′ (x) . f(x) . −− .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 173. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− .− . +. +. √ 2x − √ 3 .. √ 3/2 . 0 . − . − . − . + . √ 2x + √ 3 .. − √ 3/2 . 0 . − . + . + . + . f′′ (x) . f(x) . −− . ++ .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 174. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− .− . +. +. √ 2x − √ 3 .. √ 3/2 . 0 . − . − . − . + . √ 2x + √ 3 .. − √ 3/2 . 0 . − . + . + . + . f′′ (x) . f(x) . −− . ++ . −− .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 175. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− .− . +. +. √ 2x − √ 3 .. √ 3/2 . 0 . − . − . − . + . √ 2x + √ 3 .. − √ 3/2 . 0 . − . + . + . + . f′′ (x) . f(x) . −− . ⌢ . ++ . −− . ++ .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 176. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− .− . +. +. √ 2x − √ 3 .. √ 3/2 . 0 . − . − . − . + . √ 2x + √ 3 .. − √ 3/2 . 0 . − . + . + . + . f′′ (x) . f(x) . −− . ⌢ . ++ . ⌣ . −− . ++ .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 177. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− .− . +. +. √ 2x − √ 3 .. √ 3/2 . 0 . − . − . − . + . √ 2x + √ 3 .. − √ 3/2 . 0 . − . + . + . + . f′′ (x) . f(x) . −− . ⌢ . ++ . ⌣ . −− . ⌢ . ++ .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 178. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− .− . +. +. √ 2x − √ 3 .. √ 3/2 . 0 . − . − . − . + . √ 2x + √ 3 .. − √ 3/2 . 0 . − . + . + . + . f′′ (x) . f(x) . −− . ⌢ . ++ . ⌣ . −− . ⌢ . ++ . ⌣ .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 179. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− .− . +. +. √ 2x − √ 3 .. √ 3/2 . 0 . − . − . − . + . √ 2x + √ 3 .. − √ 3/2 . 0 . − . + . + . + . f′′ (x) . f(x) . −− . ⌢ . ++ . ⌣ . −− . ⌢ . ++ . ⌣ .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 180. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− .− . +. +. √ 2x − √ 3 .. √ 3/2 . 0 . − . − . − . + . √ 2x + √ 3 .. − √ 3/2 . 0 . − . + . + . + . f′′ (x) . f(x) . −− . ⌢ . ++ . ⌣ . −− . ⌢ . ++ . ⌣ .. − √ 3/2 . 0 . IP .. 0 . 0 .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 181. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− .− . +. +. √ 2x − √ 3 .. √ 3/2 . 0 . − . − . − . + . √ 2x + √ 3 .. − √ 3/2 . 0 . − . + . + . + . f′′ (x) . f(x) . −− . ⌢ . ++ . ⌣ . −− . ⌢ . ++ . ⌣ .. − √ 3/2 . 0 . IP .. 0 . 0 . IP .. √ 3/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 182. . . . . . . Step 2: Concavity If f′ (x) = (1 − 2x2 )e−x2 , we know f′′ (x) = (−4x)e−x2 + (1 − 2x2 )e−x2 (−2x) = ( 4x3 − 6x ) e−x2 = 2x(2x2 − 3)e−x2 .. 2x.. 0 .0.− .− . +. +. √ 2x − √ 3 .. √ 3/2 . 0 . − . − . − . + . √ 2x + √ 3 .. − √ 3/2 . 0 . − . + . + . + . f′′ (x) . f(x) . −− . ⌢ . ++ . ⌣ . −− . ⌢ . ++ . ⌣ .. − √ 3/2 . 0 . IP .. 0 . 0 . IP .. √ 3/2 . 0 . IP V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
  • 183. . . . . . . Step 3: Synthesis f(x) = xe−x2 .. f′ (x) . monotonicity .. − √ 1/2 .0 .. √ 1/2 . 0.− . ↘ .− . ↘ .+ . ↗ . +. ↗ . −. ↘ . −. ↘ . f′′ (x) . concavity .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 . −− . ⌢ . ++ . ⌣ . ++ . ⌣ . −− . ⌢ . −− . ⌢ . ++ . ⌣ . f(x) . shape .. − √ 1/2 . − 1√ 2e . min .. √ 1/2 . 1√ 2e . max .. − √ 3/2 . − √ 3 2e3 . IP .. 0 . 0 . IP .. √ 3/2 . √ 3 2e3 . IP V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 55
  • 184. . . . . . . Step 3: Synthesis f(x) = xe−x2 .. f′ (x) . monotonicity .. − √ 1/2 .0 .. √ 1/2 . 0.− . ↘ .− . ↘ .+ . ↗ . +. ↗ . −. ↘ . −. ↘ . f′′ (x) . concavity .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 . −− . ⌢ . ++ . ⌣ . ++ . ⌣ . −− . ⌢ . −− . ⌢ . ++ . ⌣ . f(x) . shape .. − √ 1/2 . − 1√ 2e . min .. √ 1/2 . 1√ 2e . max .. − √ 3/2 . − √ 3 2e3 . IP .. 0 . 0 . IP .. √ 3/2 . √ 3 2e3 . IP . V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 55
  • 185. . . . . . . Step 3: Synthesis f(x) = xe−x2 .. f′ (x) . monotonicity .. − √ 1/2 .0 .. √ 1/2 . 0.− . ↘ .− . ↘ .+ . ↗ . +. ↗ . −. ↘ . −. ↘ . f′′ (x) . concavity .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 . −− . ⌢ . ++ . ⌣ . ++ . ⌣ . −− . ⌢ . −− . ⌢ . ++ . ⌣ . f(x) . shape .. − √ 1/2 . − 1√ 2e . min .. √ 1/2 . 1√ 2e . max .. − √ 3/2 . − √ 3 2e3 . IP .. 0 . 0 . IP .. √ 3/2 . √ 3 2e3 . IP .. V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 55
  • 186. . . . . . . Step 3: Synthesis f(x) = xe−x2 .. f′ (x) . monotonicity .. − √ 1/2 .0 .. √ 1/2 . 0.− . ↘ .− . ↘ .+ . ↗ . +. ↗ . −. ↘ . −. ↘ . f′′ (x) . concavity .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 . −− . ⌢ . ++ . ⌣ . ++ . ⌣ . −− . ⌢ . −− . ⌢ . ++ . ⌣ . f(x) . shape .. − √ 1/2 . − 1√ 2e . min .. √ 1/2 . 1√ 2e . max .. − √ 3/2 . − √ 3 2e3 . IP .. 0 . 0 . IP .. √ 3/2 . √ 3 2e3 . IP ... V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 55
  • 187. . . . . . . Step 3: Synthesis f(x) = xe−x2 .. f′ (x) . monotonicity .. − √ 1/2 .0 .. √ 1/2 . 0.− . ↘ .− . ↘ .+ . ↗ . +. ↗ . −. ↘ . −. ↘ . f′′ (x) . concavity .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 . −− . ⌢ . ++ . ⌣ . ++ . ⌣ . −− . ⌢ . −− . ⌢ . ++ . ⌣ . f(x) . shape .. − √ 1/2 . − 1√ 2e . min .. √ 1/2 . 1√ 2e . max .. − √ 3/2 . − √ 3 2e3 . IP .. 0 . 0 . IP .. √ 3/2 . √ 3 2e3 . IP .... V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 55
  • 188. . . . . . . Step 3: Synthesis f(x) = xe−x2 .. f′ (x) . monotonicity .. − √ 1/2 .0 .. √ 1/2 . 0.− . ↘ .− . ↘ .+ . ↗ . +. ↗ . −. ↘ . −. ↘ . f′′ (x) . concavity .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 . −− . ⌢ . ++ . ⌣ . ++ . ⌣ . −− . ⌢ . −− . ⌢ . ++ . ⌣ . f(x) . shape .. − √ 1/2 . − 1√ 2e . min .. √ 1/2 . 1√ 2e . max .. − √ 3/2 . − √ 3 2e3 . IP .. 0 . 0 . IP .. √ 3/2 . √ 3 2e3 . IP ..... V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 55
  • 189. . . . . . . Step 3: Synthesis f(x) = xe−x2 .. f′ (x) . monotonicity .. − √ 1/2 .0 .. √ 1/2 . 0.− . ↘ .− . ↘ .+ . ↗ . +. ↗ . −. ↘ . −. ↘ . f′′ (x) . concavity .. − √ 3/2 . 0 .. 0 . 0 .. √ 3/2 . 0 . −− . ⌢ . ++ . ⌣ . ++ . ⌣ . −− . ⌢ . −− . ⌢ . ++ . ⌣ . f(x) . shape .. − √ 1/2 . − 1√ 2e . min .. √ 1/2 . 1√ 2e . max .. − √ 3/2 . − √ 3 2e3 . IP .. 0 . 0 . IP .. √ 3/2 . √ 3 2e3 . IP ...... V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 55
  • 190. . . . . . . Step 4: Graph .. x . f(x) . f(x) = xe−x2 .. ( − √ 1/2, − 1√ 2e ) .. (√ 1/2, 1√ 2e ) .. ( − √ 3/2, − √ 3 2e3 ) .. (0, 0) .. ( √ 3/2, √ 3 2e3 ) . f(x) . shape .. − √ 1/2 . − 1√ 2e . min .. √ 1/2 . 1√ 2e . max .. − √ 3/2 . − √ 3 2e3 . IP .. 0 .0. IP .. √ 3/2 . √ 3 2e3 . IP ...... V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
  • 191. . . . . . . Step 4: Graph .. x . f(x) . f(x) = xe−x2 .. ( − √ 1/2, − 1√ 2e ) .. (√ 1/2, 1√ 2e ) .. ( − √ 3/2, − √ 3 2e3 ) .. (0, 0) .. ( √ 3/2, √ 3 2e3 ) . f(x) . shape .. − √ 1/2 . − 1√ 2e . min .. √ 1/2 . 1√ 2e . max .. − √ 3/2 . − √ 3 2e3 . IP .. 0 .0. IP .. √ 3/2 . √ 3 2e3 . IP ...... V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
  • 192. . . . . . . Step 4: Graph .. x . f(x) . f(x) = xe−x2 .. ( − √ 1/2, − 1√ 2e ) .. (√ 1/2, 1√ 2e ) .. ( − √ 3/2, − √ 3 2e3 ) .. (0, 0) .. ( √ 3/2, √ 3 2e3 ) . f(x) . shape .. − √ 1/2 . − 1√ 2e . min .. √ 1/2 . 1√ 2e . max .. − √ 3/2 . − √ 3 2e3 . IP .. 0 .0. IP .. √ 3/2 . √ 3 2e3 . IP ...... V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
  • 193. . . . . . . Step 4: Graph .. x . f(x) . f(x) = xe−x2 .. ( − √ 1/2, − 1√ 2e ) .. (√ 1/2, 1√ 2e ) .. ( − √ 3/2, − √ 3 2e3 ) .. (0, 0) .. ( √ 3/2, √ 3 2e3 ) . f(x) . shape .. − √ 1/2 . − 1√ 2e . min .. √ 1/2 . 1√ 2e . max .. − √ 3/2 . − √ 3 2e3 . IP .. 0 .0. IP .. √ 3/2 . √ 3 2e3 . IP ...... V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
  • 194. . . . . . . Step 4: Graph .. x . f(x) . f(x) = xe−x2 .. ( − √ 1/2, − 1√ 2e ) .. (√ 1/2, 1√ 2e ) .. ( − √ 3/2, − √ 3 2e3 ) .. (0, 0) .. ( √ 3/2, √ 3 2e3 ) . f(x) . shape .. − √ 1/2 . − 1√ 2e . min .. √ 1/2 . 1√ 2e . max .. − √ 3/2 . − √ 3 2e3 . IP .. 0 .0. IP .. √ 3/2 . √ 3 2e3 . IP ...... V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
  • 195. . . . . . . Step 4: Graph .. x . f(x) . f(x) = xe−x2 .. ( − √ 1/2, − 1√ 2e ) .. (√ 1/2, 1√ 2e ) .. ( − √ 3/2, − √ 3 2e3 ) .. (0, 0) .. ( √ 3/2, √ 3 2e3 ) . f(x) . shape .. − √ 1/2 . − 1√ 2e . min .. √ 1/2 . 1√ 2e . max .. − √ 3/2 . − √ 3 2e3 . IP .. 0 .0. IP .. √ 3/2 . √ 3 2e3 . IP ...... V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
  • 196. . . . . . . Step 4: Graph .. x . f(x) . f(x) = xe−x2 .. ( − √ 1/2, − 1√ 2e ) .. (√ 1/2, 1√ 2e ) .. ( − √ 3/2, − √ 3 2e3 ) .. (0, 0) .. ( √ 3/2, √ 3 2e3 ) . f(x) . shape .. − √ 1/2 . − 1√ 2e . min .. √ 1/2 . 1√ 2e . max .. − √ 3/2 . − √ 3 2e3 . IP .. 0 .0. IP .. √ 3/2 . √ 3 2e3 . IP ...... V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
  • 197. . . . . . . Example with Vertical Asymptotes Example Graph f(x) = 1 x + 1 x2 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 36 / 55
  • 198. . . . . . . Step 0 Find when f is positive, negative, zero, not defined. V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 37 / 55
  • 199. . . . . . . Step 0 Find when f is positive, negative, zero, not defined. We need to factor f: f(x) = 1 x + 1 x2 = x + 1 x2 . This means f is 0 at −1 and has trouble at 0. In fact, lim x→0 x + 1 x2 = ∞, so x = 0 is a vertical asymptote of the graph. V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 37 / 55
  • 200. . . . . . . Step 0 Find when f is positive, negative, zero, not defined. We need to factor f: f(x) = 1 x + 1 x2 = x + 1 x2 . This means f is 0 at −1 and has trouble at 0. In fact, lim x→0 x + 1 x2 = ∞, so x = 0 is a vertical asymptote of the graph. We can make a sign chart as follows: .. x + 1..0 . −1 .− . +. x2 .. 0 . 0 . + . + . f(x) .. ∞ . 0 .. 0 . −1 . − . + . + V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 37 / 55
  • 201. . . . . . . Step 0, continued For horizontal asymptotes, notice that lim x→∞ x + 1 x2 = 0, so y = 0 is a horizontal asymptote of the graph. The same is true at −∞. V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 38 / 55
  • 202. . . . . . . Step 1: Monotonicity We have f′ (x) = − 1 x2 − 2 x3 = − x + 2 x3 . The critical points are x = −2 and x = 0. We have the following sign chart: .. −(x + 2)..0 . −2 .+ . −. x3 .. 0 . 0 . − . + V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 55
  • 203. . . . . . . Step 1: Monotonicity We have f′ (x) = − 1 x2 − 2 x3 = − x + 2 x3 . The critical points are x = −2 and x = 0. We have the following sign chart: .. −(x + 2)..0 . −2 .+ . −. x3 .. 0 . 0 . − . + . f′ (x) . f(x) .. ∞ . 0 .. 0 . −2 . − . + . − V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 55
  • 204. . . . . . . Step 1: Monotonicity We have f′ (x) = − 1 x2 − 2 x3 = − x + 2 x3 . The critical points are x = −2 and x = 0. We have the following sign chart: .. −(x + 2)..0 . −2 .+ . −. x3 .. 0 . 0 . − . + . f′ (x) . f(x) .. ∞ . 0 .. 0 . −2 . − . + . − . ↘ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 55
  • 205. . . . . . . Step 1: Monotonicity We have f′ (x) = − 1 x2 − 2 x3 = − x + 2 x3 . The critical points are x = −2 and x = 0. We have the following sign chart: .. −(x + 2)..0 . −2 .+ . −. x3 .. 0 . 0 . − . + . f′ (x) . f(x) .. ∞ . 0 .. 0 . −2 . − . + . − . ↘ . ↗ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 55
  • 206. . . . . . . Step 1: Monotonicity We have f′ (x) = − 1 x2 − 2 x3 = − x + 2 x3 . The critical points are x = −2 and x = 0. We have the following sign chart: .. −(x + 2)..0 . −2 .+ . −. x3 .. 0 . 0 . − . + . f′ (x) . f(x) .. ∞ . 0 .. 0 . −2 . − . + . − . ↘ . ↗ . ↘ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 55
  • 207. . . . . . . Step 1: Monotonicity We have f′ (x) = − 1 x2 − 2 x3 = − x + 2 x3 . The critical points are x = −2 and x = 0. We have the following sign chart: .. −(x + 2)..0 . −2 .+ . −. x3 .. 0 . 0 . − . + . f′ (x) . f(x) .. ∞ . 0 .. 0 . −2 . − . + . − . ↘ . ↗ . ↘ . min V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 55
  • 208. . . . . . . Step 1: Monotonicity We have f′ (x) = − 1 x2 − 2 x3 = − x + 2 x3 . The critical points are x = −2 and x = 0. We have the following sign chart: .. −(x + 2)..0 . −2 .+ . −. x3 .. 0 . 0 . − . + . f′ (x) . f(x) .. ∞ . 0 .. 0 . −2 . − . + . − . ↘ . ↗ . ↘ . min . VA V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 55
  • 209. . . . . . . Step 2: Concavity We have f′′ (x) = 2 x3 + 6 x4 = 2(x + 3) x4 . The critical points of f′ are −3 and 0. Sign chart: .. (x + 3)..0 . −3 .− . +. x4 .. 0 . 0 . + . + . f′′ (x) . f(x) .. ∞ . 0 .. 0 . −3 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
  • 210. . . . . . . Step 2: Concavity We have f′′ (x) = 2 x3 + 6 x4 = 2(x + 3) x4 . The critical points of f′ are −3 and 0. Sign chart: .. (x + 3)..0 . −3 .− . +. x4 .. 0 . 0 . + . + . f′′ (x) . f(x) .. ∞ . 0 .. 0 . −3 . −− V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
  • 211. . . . . . . Step 2: Concavity We have f′′ (x) = 2 x3 + 6 x4 = 2(x + 3) x4 . The critical points of f′ are −3 and 0. Sign chart: .. (x + 3)..0 . −3 .− . +. x4 .. 0 . 0 . + . + . f′′ (x) . f(x) .. ∞ . 0 .. 0 . −3 . −− . ++ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
  • 212. . . . . . . Step 2: Concavity We have f′′ (x) = 2 x3 + 6 x4 = 2(x + 3) x4 . The critical points of f′ are −3 and 0. Sign chart: .. (x + 3)..0 . −3 .− . +. x4 .. 0 . 0 . + . + . f′′ (x) . f(x) .. ∞ . 0 .. 0 . −3 . −− . ++ . ++ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
  • 213. . . . . . . Step 2: Concavity We have f′′ (x) = 2 x3 + 6 x4 = 2(x + 3) x4 . The critical points of f′ are −3 and 0. Sign chart: .. (x + 3)..0 . −3 .− . +. x4 .. 0 . 0 . + . + . f′′ (x) . f(x) .. ∞ . 0 .. 0 . −3 . −− . ++ . ++ . ⌢ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
  • 214. . . . . . . Step 2: Concavity We have f′′ (x) = 2 x3 + 6 x4 = 2(x + 3) x4 . The critical points of f′ are −3 and 0. Sign chart: .. (x + 3)..0 . −3 .− . +. x4 .. 0 . 0 . + . + . f′′ (x) . f(x) .. ∞ . 0 .. 0 . −3 . −− . ++ . ++ . ⌢ . ⌣ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
  • 215. . . . . . . Step 2: Concavity We have f′′ (x) = 2 x3 + 6 x4 = 2(x + 3) x4 . The critical points of f′ are −3 and 0. Sign chart: .. (x + 3)..0 . −3 .− . +. x4 .. 0 . 0 . + . + . f′′ (x) . f(x) .. ∞ . 0 .. 0 . −3 . −− . ++ . ++ . ⌢ . ⌣ . ⌣ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
  • 216. . . . . . . Step 2: Concavity We have f′′ (x) = 2 x3 + 6 x4 = 2(x + 3) x4 . The critical points of f′ are −3 and 0. Sign chart: .. (x + 3)..0 . −3 .− . +. x4 .. 0 . 0 . + . + . f′′ (x) . f(x) .. ∞ . 0 .. 0 . −3 . −− . ++ . ++ . ⌢ . ⌣ . ⌣ . IP V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
  • 217. . . . . . . Step 2: Concavity We have f′′ (x) = 2 x3 + 6 x4 = 2(x + 3) x4 . The critical points of f′ are −3 and 0. Sign chart: .. (x + 3)..0 . −3 .− . +. x4 .. 0 . 0 . + . + . f′′ (x) . f(x) .. ∞ . 0 .. 0 . −3 . −− . ++ . ++ . ⌢ . ⌣ . ⌣ . IP . VA V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
  • 218. . . . . . . Step 3: Synthesis .. f′ . monotonicity .. ∞ . 0 .. 0 . −2 . − . + . − . ↘ . ↗ . ↘ . f′′ . concavity .. ∞ . 0 .. 0 . −3 . −− . ++ . ++ . ⌢ . ⌣ . ⌣ . f . shape of f .. ∞ . 0 .. 0 . −1 .. −2 . −1/4 .. −3 . −2/9 . −∞ . 0 . ∞ . 0 . − . + . + V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
  • 219. . . . . . . Step 3: Synthesis .. f′ . monotonicity .. ∞ . 0 .. 0 . −2 . − . + . − . ↘ . ↗ . ↘ . f′′ . concavity .. ∞ . 0 .. 0 . −3 . −− . ++ . ++ . ⌢ . ⌣ . ⌣ . f . shape of f .. ∞ . 0 .. 0 . −1 .. −2 . −1/4 .. −3 . −2/9 . −∞ . 0 . ∞ . 0 . − . + . + . HA V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
  • 220. . . . . . . Step 3: Synthesis .. f′ . monotonicity .. ∞ . 0 .. 0 . −2 . − . + . − . ↘ . ↗ . ↘ . f′′ . concavity .. ∞ . 0 .. 0 . −3 . −− . ++ . ++ . ⌢ . ⌣ . ⌣ . f . shape of f .. ∞ . 0 .. 0 . −1 .. −2 . −1/4 .. −3 . −2/9 . −∞ . 0 . ∞ . 0 . − . + . + . HA . V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
  • 221. . . . . . . Step 3: Synthesis .. f′ . monotonicity .. ∞ . 0 .. 0 . −2 . − . + . − . ↘ . ↗ . ↘ . f′′ . concavity .. ∞ . 0 .. 0 . −3 . −− . ++ . ++ . ⌢ . ⌣ . ⌣ . f . shape of f .. ∞ . 0 .. 0 . −1 .. −2 . −1/4 .. −3 . −2/9 . −∞ . 0 . ∞ . 0 . − . + . + . HA .. IP V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
  • 222. . . . . . . Step 3: Synthesis .. f′ . monotonicity .. ∞ . 0 .. 0 . −2 . − . + . − . ↘ . ↗ . ↘ . f′′ . concavity .. ∞ . 0 .. 0 . −3 . −− . ++ . ++ . ⌢ . ⌣ . ⌣ . f . shape of f .. ∞ . 0 .. 0 . −1 .. −2 . −1/4 .. −3 . −2/9 . −∞ . 0 . ∞ . 0 . − . + . + . HA .. IP . V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
  • 223. . . . . . . Step 3: Synthesis .. f′ . monotonicity .. ∞ . 0 .. 0 . −2 . − . + . − . ↘ . ↗ . ↘ . f′′ . concavity .. ∞ . 0 .. 0 . −3 . −− . ++ . ++ . ⌢ . ⌣ . ⌣ . f . shape of f .. ∞ . 0 .. 0 . −1 .. −2 . −1/4 .. −3 . −2/9 . −∞ . 0 . ∞ . 0 . − . + . + . HA .. IP .. min V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
  • 224. . . . . . . Step 3: Synthesis .. f′ . monotonicity .. ∞ . 0 .. 0 . −2 . − . + . − . ↘ . ↗ . ↘ . f′′ . concavity .. ∞ . 0 .. 0 . −3 . −− . ++ . ++ . ⌢ . ⌣ . ⌣ . f . shape of f .. ∞ . 0 .. 0 . −1 .. −2 . −1/4 .. −3 . −2/9 . −∞ . 0 . ∞ . 0 . − . + . + . HA .. IP .. min . V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
  • 225. . . . . . . Step 3: Synthesis .. f′ . monotonicity .. ∞ . 0 .. 0 . −2 . − . + . − . ↘ . ↗ . ↘ . f′′ . concavity .. ∞ . 0 .. 0 . −3 . −− . ++ . ++ . ⌢ . ⌣ . ⌣ . f . shape of f .. ∞ . 0 .. 0 . −1 .. −2 . −1/4 .. −3 . −2/9 . −∞ . 0 . ∞ . 0 . − . + . + . HA .. IP .. min .. 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
  • 226. . . . . . . Step 3: Synthesis .. f′ . monotonicity .. ∞ . 0 .. 0 . −2 . − . + . − . ↘ . ↗ . ↘ . f′′ . concavity .. ∞ . 0 .. 0 . −3 . −− . ++ . ++ . ⌢ . ⌣ . ⌣ . f . shape of f .. ∞ . 0 .. 0 . −1 .. −2 . −1/4 .. −3 . −2/9 . −∞ . 0 . ∞ . 0 . − . + . + . HA .. IP .. min .. 0 . V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
  • 227. . . . . . . Step 3: Synthesis .. f′ . monotonicity .. ∞ . 0 .. 0 . −2 . − . + . − . ↘ . ↗ . ↘ . f′′ . concavity .. ∞ . 0 .. 0 . −3 . −− . ++ . ++ . ⌢ . ⌣ . ⌣ . f . shape of f .. ∞ . 0 .. 0 . −1 .. −2 . −1/4 .. −3 . −2/9 . −∞ . 0 . ∞ . 0 . − . + . + . HA .. IP .. min .. 0 .. VA V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
  • 228. . . . . . . Step 3: Synthesis .. f′ . monotonicity .. ∞ . 0 .. 0 . −2 . − . + . − . ↘ . ↗ . ↘ . f′′ . concavity .. ∞ . 0 .. 0 . −3 . −− . ++ . ++ . ⌢ . ⌣ . ⌣ . f . shape of f .. ∞ . 0 .. 0 . −1 .. −2 . −1/4 .. −3 . −2/9 . −∞ . 0 . ∞ . 0 . − . + . + . HA .. IP .. min .. 0 .. VA . V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
  • 229. . . . . . . Step 3: Synthesis .. f′ . monotonicity .. ∞ . 0 .. 0 . −2 . − . + . − . ↘ . ↗ . ↘ . f′′ . concavity .. ∞ . 0 .. 0 . −3 . −− . ++ . ++ . ⌢ . ⌣ . ⌣ . f . shape of f .. ∞ . 0 .. 0 . −1 .. −2 . −1/4 .. −3 . −2/9 . −∞ . 0 . ∞ . 0 . − . + . + . HA .. IP .. min .. 0 .. VA .. HA V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 55
  • 230. . . . . . . Step 4: Graph .. x. y .. (−3, −2/9) .. (−2, −1/4) . f . shape of f .. ∞ . 0 .. 0 . −1 .. −2 . −1/4 .. −3 . −2/9 . −∞ . 0 . ∞ . 0 . − . + . + . HA .. IP .. min .. 0 .. VA .. HA V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 42 / 55
  • 231. . . . . . . Trigonometric and polynomial together Problem Graph f(x) = cos x − x V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 43 / 55
  • 232. . . . . . . Step 0: intercepts and asymptotes f(0) = 1 and f(−π/2) = −π/2. So by the Intermediate Value Theorem there is a zero in between. We don’t know it’s precise value, though. Since −1 ≤ cos x ≤ 1 for all x, we have −1 − x ≤ cos x − x ≤ 1 − x for all x. This means that lim x→∞ f(x) = −∞ and lim x→−∞ f(x) = ∞. V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 44 / 55
  • 233. . . . . . . Step 1: Monotonicity If f(x) = cos x − x, then f′ (x) = − sin x − 1 = (−1)(sin x + 1). f′ (x) = 0 if x = 3π/2 + 2πk, where k is any integer f′ (x) is periodic with period 2π Since −1 ≤ sin x ≤ 1 for all x, we have 0 ≤ sin x + 1 ≤ 2 =⇒ −2 ≤ (−1)(sin x + 1) ≤ 0 for all x. This means f′ (x) is negative at all other points. .. f′ (x) . f(x) .. −π/2 .0 .. 3π/2 . 0.. 7π/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 45 / 55
  • 234. . . . . . . Step 1: Monotonicity If f(x) = cos x − x, then f′ (x) = − sin x − 1 = (−1)(sin x + 1). f′ (x) = 0 if x = 3π/2 + 2πk, where k is any integer f′ (x) is periodic with period 2π Since −1 ≤ sin x ≤ 1 for all x, we have 0 ≤ sin x + 1 ≤ 2 =⇒ −2 ≤ (−1)(sin x + 1) ≤ 0 for all x. This means f′ (x) is negative at all other points. .. f′ (x) . f(x) .. −π/2 .0 .. 3π/2 . 0.. 7π/2 . 0. − V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 45 / 55
  • 235. . . . . . . Step 1: Monotonicity If f(x) = cos x − x, then f′ (x) = − sin x − 1 = (−1)(sin x + 1). f′ (x) = 0 if x = 3π/2 + 2πk, where k is any integer f′ (x) is periodic with period 2π Since −1 ≤ sin x ≤ 1 for all x, we have 0 ≤ sin x + 1 ≤ 2 =⇒ −2 ≤ (−1)(sin x + 1) ≤ 0 for all x. This means f′ (x) is negative at all other points. .. f′ (x) . f(x) .. −π/2 .0 .. 3π/2 . 0.. 7π/2 . 0. −. − V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 45 / 55
  • 236. . . . . . . Step 1: Monotonicity If f(x) = cos x − x, then f′ (x) = − sin x − 1 = (−1)(sin x + 1). f′ (x) = 0 if x = 3π/2 + 2πk, where k is any integer f′ (x) is periodic with period 2π Since −1 ≤ sin x ≤ 1 for all x, we have 0 ≤ sin x + 1 ≤ 2 =⇒ −2 ≤ (−1)(sin x + 1) ≤ 0 for all x. This means f′ (x) is negative at all other points. .. f′ (x) . f(x) .. −π/2 .0 .. 3π/2 . 0.. 7π/2 . 0. −. ↘ . − V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 45 / 55
  • 237. . . . . . . Step 1: Monotonicity If f(x) = cos x − x, then f′ (x) = − sin x − 1 = (−1)(sin x + 1). f′ (x) = 0 if x = 3π/2 + 2πk, where k is any integer f′ (x) is periodic with period 2π Since −1 ≤ sin x ≤ 1 for all x, we have 0 ≤ sin x + 1 ≤ 2 =⇒ −2 ≤ (−1)(sin x + 1) ≤ 0 for all x. This means f′ (x) is negative at all other points. .. f′ (x) . f(x) .. −π/2 .0 .. 3π/2 . 0.. 7π/2 . 0. −. ↘ . −. ↘ V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 45 / 55
  • 238. . . . . . . Step 2: Concavity If f′ (x) = − sin x − 1, then f′′ (x) = − cos x. This is 0 when x = π/2 + πk, where k is any integer. This is periodic with period 2π .. f′′ (x) . f(x) .. −π/2 .0 .. π/2 . 0.. 3π/2 . 0.. 5π/2 . 0.. 7π/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
  • 239. . . . . . . Step 2: Concavity If f′ (x) = − sin x − 1, then f′′ (x) = − cos x. This is 0 when x = π/2 + πk, where k is any integer. This is periodic with period 2π .. f′′ (x) . f(x) .−−.. −π/2 .0 .. π/2 . 0.. 3π/2 . 0.. 5π/2 . 0.. 7π/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
  • 240. . . . . . . Step 2: Concavity If f′ (x) = − sin x − 1, then f′′ (x) = − cos x. This is 0 when x = π/2 + πk, where k is any integer. This is periodic with period 2π .. f′′ (x) . f(x) .−−. ++.. −π/2 .0 .. π/2 . 0.. 3π/2 . 0.. 5π/2 . 0.. 7π/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
  • 241. . . . . . . Step 2: Concavity If f′ (x) = − sin x − 1, then f′′ (x) = − cos x. This is 0 when x = π/2 + πk, where k is any integer. This is periodic with period 2π .. f′′ (x) . f(x) .−−. ++. −−.. −π/2 .0 .. π/2 . 0.. 3π/2 . 0.. 5π/2 . 0.. 7π/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
  • 242. . . . . . . Step 2: Concavity If f′ (x) = − sin x − 1, then f′′ (x) = − cos x. This is 0 when x = π/2 + πk, where k is any integer. This is periodic with period 2π .. f′′ (x) . f(x) .−−. ++. −−. ++.. −π/2 .0 .. π/2 . 0.. 3π/2 . 0.. 5π/2 . 0.. 7π/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
  • 243. . . . . . . Step 2: Concavity If f′ (x) = − sin x − 1, then f′′ (x) = − cos x. This is 0 when x = π/2 + πk, where k is any integer. This is periodic with period 2π .. f′′ (x) . f(x) .−−. ⌢ . ++. −−. ++.. −π/2 .0 .. π/2 . 0.. 3π/2 . 0.. 5π/2 . 0.. 7π/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
  • 244. . . . . . . Step 2: Concavity If f′ (x) = − sin x − 1, then f′′ (x) = − cos x. This is 0 when x = π/2 + πk, where k is any integer. This is periodic with period 2π .. f′′ (x) . f(x) .−−. ⌢ . ++. ⌣ . −−. ++.. −π/2 .0 .. π/2 . 0.. 3π/2 . 0.. 5π/2 . 0.. 7π/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
  • 245. . . . . . . Step 2: Concavity If f′ (x) = − sin x − 1, then f′′ (x) = − cos x. This is 0 when x = π/2 + πk, where k is any integer. This is periodic with period 2π .. f′′ (x) . f(x) .−−. ⌢ . ++. ⌣ . −−. ⌢ . ++.. −π/2 .0 .. π/2 . 0.. 3π/2 . 0.. 5π/2 . 0.. 7π/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
  • 246. . . . . . . Step 2: Concavity If f′ (x) = − sin x − 1, then f′′ (x) = − cos x. This is 0 when x = π/2 + πk, where k is any integer. This is periodic with period 2π .. f′′ (x) . f(x) .−−. ⌢ . ++. ⌣ . −−. ⌢ . ++. ⌣ .. −π/2 .0 .. π/2 . 0.. 3π/2 . 0.. 5π/2 . 0.. 7π/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
  • 247. . . . . . . Step 2: Concavity If f′ (x) = − sin x − 1, then f′′ (x) = − cos x. This is 0 when x = π/2 + πk, where k is any integer. This is periodic with period 2π .. f′′ (x) . f(x) .−−. ⌢ . ++. ⌣ . −−. ⌢ . ++. ⌣ .. −π/2 .0 . IP .. π/2 . 0.. 3π/2 . 0.. 5π/2 . 0.. 7π/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
  • 248. . . . . . . Step 2: Concavity If f′ (x) = − sin x − 1, then f′′ (x) = − cos x. This is 0 when x = π/2 + πk, where k is any integer. This is periodic with period 2π .. f′′ (x) . f(x) .−−. ⌢ . ++. ⌣ . −−. ⌢ . ++. ⌣ .. −π/2 .0 . IP .. π/2 . 0. IP .. 3π/2 . 0.. 5π/2 . 0.. 7π/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
  • 249. . . . . . . Step 2: Concavity If f′ (x) = − sin x − 1, then f′′ (x) = − cos x. This is 0 when x = π/2 + πk, where k is any integer. This is periodic with period 2π .. f′′ (x) . f(x) .−−. ⌢ . ++. ⌣ . −−. ⌢ . ++. ⌣ .. −π/2 .0 . IP .. π/2 . 0. IP .. 3π/2 . 0. IP .. 5π/2 . 0.. 7π/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
  • 250. . . . . . . Step 2: Concavity If f′ (x) = − sin x − 1, then f′′ (x) = − cos x. This is 0 when x = π/2 + πk, where k is any integer. This is periodic with period 2π .. f′′ (x) . f(x) .−−. ⌢ . ++. ⌣ . −−. ⌢ . ++. ⌣ .. −π/2 .0 . IP .. π/2 . 0. IP .. 3π/2 . 0. IP .. 5π/2 . 0. IP .. 7π/2 . 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
  • 251. . . . . . . Step 2: Concavity If f′ (x) = − sin x − 1, then f′′ (x) = − cos x. This is 0 when x = π/2 + πk, where k is any integer. This is periodic with period 2π .. f′′ (x) . f(x) .−−. ⌢ . ++. ⌣ . −−. ⌢ . ++. ⌣ .. −π/2 .0 . IP .. π/2 . 0. IP .. 3π/2 . 0. IP .. 5π/2 . 0. IP .. 7π/2 . 0. IP V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
  • 252. . . . . . . Step 3: Synthesis .. f′ (x) . mono .. −π/2 .0 .. 3π/2 . 0.. 7π/2 . 0. −. ↘ . −. ↘ . f′′ (x) . conc .. −π/2 . 0 .. π/2 . 0 .. 3π/2 . 0 .. 5π/2 . 0 .. 7π/2 . 0 . −− . ⌢ . ++ . ⌣ . −− . ⌢ . ++ . ⌣ . f(x) . shape .. −π/2 . π/2 . IP .. π/2 . −π/2 . IP .. 3π/2 . −3π/2 . IP .. 5π/2 . −5π/2 . IP .. 7π/2 . −7π/2 . IP V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 47 / 55
  • 253. . . . . . . Step 3: Synthesis .. f′ (x) . mono .. −π/2 .0 .. 3π/2 . 0.. 7π/2 . 0. −. ↘ . −. ↘ . f′′ (x) . conc .. −π/2 . 0 .. π/2 . 0 .. 3π/2 . 0 .. 5π/2 . 0 .. 7π/2 . 0 . −− . ⌢ . ++ . ⌣ . −− . ⌢ . ++ . ⌣ . f(x) . shape .. −π/2 . π/2 . IP .. π/2 . −π/2 . IP .. 3π/2 . −3π/2 . IP .. 5π/2 . −5π/2 . IP .. 7π/2 . −7π/2 . IP . V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 47 / 55
  • 254. . . . . . . Step 3: Synthesis .. f′ (x) . mono .. −π/2 .0 .. 3π/2 . 0.. 7π/2 . 0. −. ↘ . −. ↘ . f′′ (x) . conc .. −π/2 . 0 .. π/2 . 0 .. 3π/2 . 0 .. 5π/2 . 0 .. 7π/2 . 0 . −− . ⌢ . ++ . ⌣ . −− . ⌢ . ++ . ⌣ . f(x) . shape .. −π/2 . π/2 . IP .. π/2 . −π/2 . IP .. 3π/2 . −3π/2 . IP .. 5π/2 . −5π/2 . IP .. 7π/2 . −7π/2 . IP .. V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 47 / 55
  • 255. . . . . . . Step 3: Synthesis .. f′ (x) . mono .. −π/2 .0 .. 3π/2 . 0.. 7π/2 . 0. −. ↘ . −. ↘ . f′′ (x) . conc .. −π/2 . 0 .. π/2 . 0 .. 3π/2 . 0 .. 5π/2 . 0 .. 7π/2 . 0 . −− . ⌢ . ++ . ⌣ . −− . ⌢ . ++ . ⌣ . f(x) . shape .. −π/2 . π/2 . IP .. π/2 . −π/2 . IP .. 3π/2 . −3π/2 . IP .. 5π/2 . −5π/2 . IP .. 7π/2 . −7π/2 . IP ... V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 47 / 55
  • 256. . . . . . . Step 3: Synthesis .. f′ (x) . mono .. −π/2 .0 .. 3π/2 . 0.. 7π/2 . 0. −. ↘ . −. ↘ . f′′ (x) . conc .. −π/2 . 0 .. π/2 . 0 .. 3π/2 . 0 .. 5π/2 . 0 .. 7π/2 . 0 . −− . ⌢ . ++ . ⌣ . −− . ⌢ . ++ . ⌣ . f(x) . shape .. −π/2 . π/2 . IP .. π/2 . −π/2 . IP .. 3π/2 . −3π/2 . IP .. 5π/2 . −5π/2 . IP .. 7π/2 . −7π/2 . IP .... V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 47 / 55