1. . V63.0121.001: Calculus I
. Sec on 4.4: Curve Sketching
. April 13, 2011
Notes
Sec on 4.4
Curve Sketching
V63.0121.001: Calculus I
Professor Ma hew Leingang
New York University
April 13, 2011
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Notes
Announcements
Quiz 4 on Sec ons 3.3,
3.4, 3.5, and 3.7 this
week (April 14/15)
Quiz 5 on Sec ons
4.1–4.4 April 28/29
Final Exam Thursday May
12, 2:00–3:50pm
I am teaching Calc II MW
2:00pm and Calc III TR
2:00pm both Fall ’11 and
Spring ’12
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Notes
Objectives
given a func on, graph it
completely, indica ng
zeroes (if easy)
asymptotes if applicable
cri cal points
local/global max/min
inflec on points
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2. . V63.0121.001: Calculus I
. Sec on 4.4: Curve Sketching
. April 13, 2011
Notes
Why?
Graphing func ons is like
dissec on … or diagramming
sentences
You can really know a lot
about a func on when you
know all of its anatomy.
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Notes
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
f(x)
f′ (x)
f(x) = x3 + x2
f′ (x) = 3x2 + 2x .
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Notes
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
f′′ (x) f′ (x) f(x)
f(x) = x3 + x2
f′ (x) = 3x2 + 2x .
f′′ (x) = 6x + 2
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3. . V63.0121.001: Calculus I
. Sec on 4.4: Curve Sketching
. April 13, 2011
Notes
Graphing Checklist
To graph a func on f, follow this plan:
0. Find when f is posi ve, nega ve, zero,
not defined.
1. Find f′ and form its sign chart.
Conclude informa on about
increasing/decreasing and local
max/min.
2. Find f′′ and form its sign chart.
Conclude concave up/concave down
and inflec on.
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Notes
Graphing Checklist
To graph a func on f, follow this plan:
3. Put together a big chart to assemble
monotonicity and concavity data
4. Graph!
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Notes
Outline
Simple examples
A cubic func on
A quar c func on
More Examples
Points of nondifferen ability
Horizontal asymptotes
Ver cal asymptotes
Trigonometric and polynomial together
Logarithmic
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4. . V63.0121.001: Calculus I
. Sec on 4.4: Curve Sketching
. April 13, 2011
Notes
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 quadra c, so we the other two
roots are
√ √
3 ± 32 − 4(2)(−12) 3 ± 105
x= =
4 4
. It’s OK to skip this step for now since the roots are so complicated.
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Notes
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)
+ − +
↗−1 ↘ 2 ↗ f(x)
max min
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Notes
Step 2: Concavity
f′ (x) = 6x2 − 6x − 12
=⇒ f′′ (x) = 12x − 6 = 6(2x − 1)
Another sign chart:
.
−− ++ f′′ (x)
⌢ 1/2 ⌣ f(x)
IP
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5. . V63.0121.001: Calculus I
. Sec on 4.4: Curve Sketching
. April 13, 2011
Notes
Step 3: One sign chart to rule them all
Remember, f(x) = 2x3 − 3x2 − 12x.
+ −. − + f′ (x)
↗−1 ↘ ↘ 2 ↗ monotonicity
−− −− ++ ++ f′′ (x)
⌢ ⌢ 1/2 ⌣ ⌣ concavity
7 −61/2 −20 f(x)
−1 1/2 2 shape of f
max IP min
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Notes
monotonicity and concavity
increasing, decreasing,
concave concave
down down
II I
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III IV
decreasing, increasing,
concave concave
up up
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Notes
Step 3: One sign chart to rule them all
Remember, f(x) = 2x3 − 3x2 − 12x.
+ −. − + f′ (x)
↗−1 ↘ ↘ 2 ↗ monotonicity
−− −− ++ ++ f′′ (x)
⌢ ⌢ 1/2 ⌣ ⌣ concavity
7 −61/2 −20 f(x)
−1 1/2 2 shape of f
max IP min
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6. . V63.0121.001: Calculus I
. Sec on 4.4: Curve Sketching
. April 13, 2011
f(x) Notes
Step 4: Graph
f(x) = 2x3 − 3x2 − 12x
( √ ) (−1, 7)
3− 105
4 ,0 (0, 0)
. ( x√ )
(1/2, −61/2)
3+ 105
4 ,0
(2, −20)
7 −61/2 −20 f(x)
−1 1/2 2 shape of f
max IP min
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Notes
Graphing a quartic
Example
Graph f(x) = x4 − 4x3 + 10
(Step 0) We know f(0) = 10 and lim f(x) = +∞. Not too many
x→±∞
other points on the graph are evident.
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Notes
Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
We make its sign chart.
+0 . + +
4x2
0
− − 0+
(x − 3)
3 ′
−0 − 0 + f (x)
↘0 ↘ 3 ↗ f(x)
min
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7. . V63.0121.001: Calculus I
. Sec on 4.4: Curve Sketching
. April 13, 2011
Notes
Step 2: Concavity
f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
Here is its sign chart:
−0. + +
12x
0
− 0 +−
x−2
2 f′′ (x)
++0 −− 0 ++
⌣0 ⌢ 2 ⌣ f(x)
IP IP
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Notes
Step 3: Grand Unified Sign Chart
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Remember, f(x) = x4 − 4x3 + 10.
−0 − −0+ f′ (x)
↘0 ↘ ↘3↗ monotonicity
f′′ (x)
++0 −− 0++ ++
⌣0 ⌢ 2⌣ ⌣ concavity
10 −6 −17 f(x)
0 2 3 shape
IP IP min
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Notes
Step 4: Graph
f(x) = x4 − 4x3 + 10
(0, 10)
. x
(2, −6)
(3, −17)
10 −6 −17 f(x)
0 2 3 shape
IP IP min
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8. . V63.0121.001: Calculus I
. Sec on 4.4: Curve Sketching
. April 13, 2011
Notes
Outline
Simple examples
A cubic func on
A quar c func on
More Examples
Points of nondifferen ability
Horizontal asymptotes
Ver cal asymptotes
Trigonometric and polynomial together
Logarithmic
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Notes
Graphing a function with a cusp
Example
√
Graph f(x) = x + |x|
This func on looks strange because of the absolute value. But
whenever we become nervous, we can just take cases.
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Notes
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 posi ve.
Are there nega ve numbers which are zeroes for f?
√ √
x + −x = 0 =⇒ −x = −x
−x = x2 =⇒ x2 + x = 0
The only solu ons are x = 0 and x = −1.
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9. . V63.0121.001: Calculus I
. Sec on 4.4: Curve Sketching
. April 13, 2011
Notes
Step 0: Asymptotic behavior
√
f(x) = x + |x|
lim f(x) = ∞, because both terms tend to ∞.
x→∞
lim f(x) is indeterminate of the form −∞ + ∞. It’s the same
x→−∞ √
as lim (−y + y)
y→+∞
√
√ √ y+y
lim (−y + y) = lim ( y − y) · √
y→+∞ y→∞ y+y
y − y2
= lim √ = −∞
y→∞ y+y
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Notes
Step 1: The derivative
√
Remember, f(x) = x + |x|.
To find f′ , first assume x > 0. Then
d ( √ ) 1
f′ (x) = x+ x =1+ √
dx 2 x
No ce
f′ (x) > 0 when x > 0 (so no cri cal points here)
lim+ f′ (x) = ∞ (so 0 is a cri cal point)
x→0
lim f′ (x) = 1 (so the graph is asympto c to a line of slope 1)
x→∞
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Notes
Step 1: The derivative
√
Remember, f(x) = x + |x|.
If x is nega ve, we have
d ( √ ) 1
f′ (x) = x + −x = 1 − √
dx 2 −x
No ce
lim− f′ (x) = −∞ (other side of the cri cal point)
x→0
lim f′ (x) = 1 (asympto c to a line of slope 1)
x→−∞
′
f (x) = 0 when
1 √ 1 1 1
1− √ = 0 =⇒ −x = =⇒ −x = =⇒ x = −
2 −x 2 4 4
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10. . V63.0121.001: Calculus I
. Sec on 4.4: Curve Sketching
. April 13, 2011
Notes
Step 1: Monotonicity
1
1 + √
if x > 0
′
f (x) = 2 x
1 − √ 1
if x < 0
2 −x
We can’t make a mul -factor sign chart because of the absolute
value, but we can test points in between cri cal points.
+ 0− ∞ + f′ (x)
.
↗ − 1↘ 0
4
↗ f(x)
max min
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Notes
Step 2: Concavity
( )
d 1 1
If x > 0, then f′′ (x) = 1 + x−1/2 = − x−3/2 This is
dx 2 4
nega ve whenever x > 0. ( )
d 1 1
If x < 0, then f′′ (x) = 1 − (−x)−1/2 = − (−x)−3/2
dx 2 4
which is also always nega ve for nega ve x.
1
In other words, f′′ (x) = − |x|−3/2 .
4
Here is the sign chart:
−− −∞ −− f′′ (x)
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⌢ 0 ⌢ f(x)
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Notes
Step 3: Synthesis
Now we can put these things together.
√
f(x) = x + |x|
+1 + 0− ∞ + f′
+1 (x)
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↗ ↗ − 1↘ 0 ↗ ↗monotonicity
f′′
−∞ −− −− 4 −∞ −− −∞ (x)
⌢ ⌢ 1 ⌢0 ⌢ ⌢concavity
−∞ 0 4 0 +∞f(x)
−1 −4 0
1 shape
zero max min
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11. . V63.0121.001: Calculus I
. Sec on 4.4: Curve Sketching
. April 13, 2011
Notes
Graph
√
f(x) = x + |x|
(− 1 , 1 )
4 4
(−1, 0)
. x
(0, 0)
1
−∞ 0 4 0 +∞ x
−1 −1 0 shape
4
zero max min
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Example with Horizontal Notes
Asymptotes
Example
Graph f(x) = xe−x
2
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Notes
Step 1: Monotonicity
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12. . V63.0121.001: Calculus I
. Sec on 4.4: Curve Sketching
. April 13, 2011
Notes
Step 2: Concavity
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Notes
Step 3: Synthesis
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Notes
Step 4: Graph
f(x)
x
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13. . V63.0121.001: Calculus I
. Sec on 4.4: Curve Sketching
. April 13, 2011
Notes
Example with Vertical Asymptotes
Example
1 1
Graph f(x) = +
x x2
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Notes
Step 0
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Notes
Step 1: Monotonicity
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14. . V63.0121.001: Calculus I
. Sec on 4.4: Curve Sketching
. April 13, 2011
Notes
Step 2: Concavity
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Notes
Step 3: Synthesis
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Notes
Step 4: Graph
y
. x
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15. . V63.0121.001: Calculus I
. Sec on 4.4: Curve Sketching
. April 13, 2011
Trigonometric and polynomial Notes
together
Problem
Graph f(x) = cos x − x
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Notes
Step 0: intercepts and asymptotes
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Notes
Step 1: Monotonicity
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16. . V63.0121.001: Calculus I
. Sec on 4.4: Curve Sketching
. April 13, 2011
Notes
Step 2: Concavity
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Notes
Step 3: Synthesis
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Notes
Step 4: Graph
f(x) = cos x − x
y
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x
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17. . V63.0121.001: Calculus I
. Sec on 4.4: Curve Sketching
. April 13, 2011
Notes
Logarithmic
Problem
Graph f(x) = x ln x2
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Step 0: Intercepts and Notes
Asymptotes
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Notes
Step 1: Monotonicity
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18. . V63.0121.001: Calculus I
. Sec on 4.4: Curve Sketching
. April 13, 2011
Notes
Step 2: Concavity
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Notes
Step 3: Synthesis
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Notes
Step 4: Graph
y
. x
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19. . V63.0121.001: Calculus I
. Sec on 4.4: Curve Sketching
. April 13, 2011
Notes
Summary
Graphing is a procedure that gets easier with prac ce.
Remember to follow the checklist.
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Notes
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Notes
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