Lesson 20: Derivatives and the Shapes of Curves (handout)
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