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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 / 45
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 / 45
Notes
Notes
Notes
1
Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
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 / 45
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 / 45
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 / 45
Notes
Notes
Notes
2
Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
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 / 45
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 / 45
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 / 45
Notes
Notes
Notes
3
Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
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 / 45
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 / 45
Step 3: One sign chart to rule them all
Remember, f (x) = 2x3
− 3x2
− 12x.
f (x)
monotonicity−1 2
+− −+
f (x)
concavity1/2
−− −− ++ ++
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 / 45
Notes
Notes
Notes
4
Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
Combinations of monotonicity and concavity
III
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 / 45
Step 3: One sign chart to rule them all
Remember, f (x) = 2x3
− 3x2
− 12x.
f (x)
monotonicity−1 2
+− −+
f (x)
concavity1/2
−− −− ++ ++
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 / 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 / 45
Notes
Notes
Notes
5
Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
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 / 45
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 / 45
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 / 45
Notes
Notes
Notes
6
Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
Step 3: Grand Unified Sign Chart
Remember, f (x) = x4
− 4x3
+ 10.
f (x)
monotonicity3
0
0
0− − − +
f (x)
concavity0
0
2
0++ −− ++ ++
f (x)
shape0
10
IP
2
−6
IP
3
−17
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 19 / 45
Step 4: Graph
f (x) = x4
− 4x3
+ 10
x
y
f (x)
shape0
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 / 45
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 / 45
Notes
Notes
Notes
7
Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
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 / 45
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 / 45
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 / 45
Notes
Notes
Notes
8
Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
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 / 45
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 / 45
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 / 45
Notes
Notes
Notes
9
Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
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 / 45
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)
concavity0
−∞−− −− −−−∞ −∞
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 / 45
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 / 45
Notes
Notes
Notes
10
Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
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 / 45
Step 1: Monotonicity
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 45
Step 2: Concavity
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 45
Notes
Notes
Notes
11
Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
Step 3: Synthesis
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 45
Step 4: Graph
x
f (x)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 45
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 / 45
Notes
Notes
Notes
12
Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
Step 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 37 / 45
Step 1: Monotonicity
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 45
Step 2: Concavity
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 45
Notes
Notes
Notes
13
Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
Step 3: Synthesis
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 45
Step 4: Graph
x
y
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 42 / 45
Problem
Graph f (x) = cos x − x
x
y
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 43 / 45
Notes
Notes
Notes
14
Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
Problem
Graph f (x) = x ln x2
x
y
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 44 / 45
Summary
Graphing is a procedure that gets easier with practice.
Remember to follow the checklist.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 45 / 45
Notes
Notes
Notes
15
Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010

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Lesson 21: Curve Sketching (Section 041 handout)

  • 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 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 / 45 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 / 45 Notes Notes Notes 1 Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
  • 2. 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 / 45 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 / 45 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 / 45 Notes Notes Notes 2 Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
  • 3. 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 / 45 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 / 45 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 / 45 Notes Notes Notes 3 Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
  • 4. 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 / 45 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 / 45 Step 3: One sign chart to rule them all Remember, f (x) = 2x3 − 3x2 − 12x. f (x) monotonicity−1 2 +− −+ f (x) concavity1/2 −− −− ++ ++ 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 / 45 Notes Notes Notes 4 Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
  • 5. Combinations of monotonicity and concavity III 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 / 45 Step 3: One sign chart to rule them all Remember, f (x) = 2x3 − 3x2 − 12x. f (x) monotonicity−1 2 +− −+ f (x) concavity1/2 −− −− ++ ++ 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 / 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 / 45 Notes Notes Notes 5 Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
  • 6. 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 / 45 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 / 45 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 / 45 Notes Notes Notes 6 Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
  • 7. Step 3: Grand Unified Sign Chart Remember, f (x) = x4 − 4x3 + 10. f (x) monotonicity3 0 0 0− − − + f (x) concavity0 0 2 0++ −− ++ ++ f (x) shape0 10 IP 2 −6 IP 3 −17 min V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 19 / 45 Step 4: Graph f (x) = x4 − 4x3 + 10 x y f (x) shape0 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 / 45 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 / 45 Notes Notes Notes 7 Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
  • 8. 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 / 45 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 / 45 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 / 45 Notes Notes Notes 8 Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
  • 9. 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 / 45 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 / 45 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 / 45 Notes Notes Notes 9 Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
  • 10. 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 / 45 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) concavity0 −∞−− −− −−−∞ −∞ 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 / 45 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 / 45 Notes Notes Notes 10 Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
  • 11. 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 / 45 Step 1: Monotonicity V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 45 Step 2: Concavity V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 45 Notes Notes Notes 11 Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
  • 12. Step 3: Synthesis V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 45 Step 4: Graph x f (x) V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 45 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 / 45 Notes Notes Notes 12 Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
  • 13. Step 0 V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 37 / 45 Step 1: Monotonicity V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 45 Step 2: Concavity V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 45 Notes Notes Notes 13 Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
  • 14. Step 3: Synthesis V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 41 / 45 Step 4: Graph x y V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 42 / 45 Problem Graph f (x) = cos x − x x y V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 43 / 45 Notes Notes Notes 14 Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010
  • 15. Problem Graph f (x) = x ln x2 x y V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 44 / 45 Summary Graphing is a procedure that gets easier with practice. Remember to follow the checklist. V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 45 / 45 Notes Notes Notes 15 Section 4.4 : Curve SketchingV63.0121.041, Calculus I November 17, 2010