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Lesson 20: Derivatives and the Shape of Curves (Section 041 slides)
1. Section 4.2
Derivatives and the Shapes of Curves
V63.0121.041, Calculus I
New York .
University
November 15, 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 4.2 The Shapes of Curves November 15, 2010 2 / 32
3. Objectives
Use the derivative of a
function to determine the
intervals along which the
function is increasing or
decreasing (The
Increasing/Decreasing
Test)
Use the First Derivative
Test to classify critical
points of a function as local
maxima, local minima, or
neither.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 3 / 32
4. Objectives
Use the second derivative
of a function to determine
the intervals along which
the graph of the function is
concave up or concave
down (The Concavity Test)
Use the first and second
derivative of a function to
classify critical points as
local maxima or local
minima, when applicable
(The Second Derivative
Test)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 4 / 32
5. Outline
Recall: The Mean Value Theorem
Monotonicity
The Increasing/Decreasing Test
Finding intervals of monotonicity
The First Derivative Test
Concavity
Definitions
Testing for Concavity
The Second Derivative Test
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 5 / 32
6. Recall: The Mean Value Theorem
Theorem (The Mean Value Theorem)
Let f be continuous on [a, b]
and differentiable on (a, b).
Then there exists a point c in
(a, b) such that
b
f(b) − f(a)
= f′ (c). .
b−a a
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 6 / 32
7. Recall: The Mean Value Theorem
Theorem (The Mean Value Theorem)
Let f be continuous on [a, b]
and differentiable on (a, b).
Then there exists a point c in
(a, b) such that
b
f(b) − f(a)
= f′ (c). .
b−a a
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 6 / 32
8. Recall: The Mean Value Theorem
Theorem (The Mean Value Theorem)
c
Let f be continuous on [a, b]
and differentiable on (a, b).
Then there exists a point c in
(a, b) such that
b
f(b) − f(a)
= f′ (c). .
b−a a
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 6 / 32
9. Recall: The Mean Value Theorem
Theorem (The Mean Value Theorem)
c
Let f be continuous on [a, b]
and differentiable on (a, b).
Then there exists a point c in
(a, b) such that
b
f(b) − f(a)
= f′ (c). .
b−a a
Another way to put this is that there exists a point c such that
f(b) = f(a) + f′ (c)(b − a)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 6 / 32
10. Why the MVT is the MITC
Most Important Theorem In Calculus!
Theorem
Let f′ = 0 on an interval (a, b). Then f is constant on (a, b).
Proof.
Pick any points x and y in (a, b) with x < y. Then f is continuous on
[x, y] and differentiable on (x, y). By MVT there exists a point z in (x, y)
such that
f(y) = f(x) + f′ (z)(y − x)
So f(y) = f(x). Since this is true for all x and y in (a, b), then f is
constant.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 7 / 32
11. Outline
Recall: The Mean Value Theorem
Monotonicity
The Increasing/Decreasing Test
Finding intervals of monotonicity
The First Derivative Test
Concavity
Definitions
Testing for Concavity
The Second Derivative Test
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 8 / 32
12. What does it mean for a function to be increasing?
Definition
A function f is increasing on the interval I if
f(x) < f(y)
whenever x and y are two points in I with x < y.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 9 / 32
13. What does it mean for a function to be increasing?
Definition
A function f is increasing on the interval I if
f(x) < f(y)
whenever x and y are two points in I with x < y.
An increasing function “preserves order.”
I could be bounded or infinite, open, closed, or
half-open/half-closed.
Write your own definition (mutatis mutandis) of decreasing,
nonincreasing, nondecreasing
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 9 / 32
14. The Increasing/Decreasing Test
Theorem (The Increasing/Decreasing Test)
If f′ > 0 on an interval, then f is increasing on that interval. If f′ < 0 on
an interval, then f is decreasing on that interval.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 10 / 32
15. The Increasing/Decreasing Test
Theorem (The Increasing/Decreasing Test)
If f′ > 0 on an interval, then f is increasing on that interval. If f′ < 0 on
an interval, then f is decreasing on that interval.
Proof.
It works the same as the last theorem. Assume f′ (x) > 0 on an interval
I. Pick two points x and y in I with x < y. We must show f(x) < f(y). By
MVT there exists a point c in (x, y) such that
f(y) − f(x) = f′ (c)(y − x) > 0.
So f(y) > f(x).
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 10 / 32
16. Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f(x) = 2x − 5.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 11 / 32
17. Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f(x) = 2x − 5.
Solution
f′ (x) = 2 is always positive, so f is increasing on (−∞, ∞).
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 11 / 32
18. Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f(x) = 2x − 5.
Solution
f′ (x) = 2 is always positive, so f is increasing on (−∞, ∞).
Example
Describe the monotonicity of f(x) = arctan(x).
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 11 / 32
19. Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f(x) = 2x − 5.
Solution
f′ (x) = 2 is always positive, so f is increasing on (−∞, ∞).
Example
Describe the monotonicity of f(x) = arctan(x).
Solution
1
Since f′ (x) = is always positive, f(x) is always increasing.
1 + x2
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 11 / 32
20. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32
21. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
f′ (x) = 2x, which is positive when x > 0 and negative when x is.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32
22. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
f′ (x) = 2x, which is positive when x > 0 and negative when x is.
We can draw a number line:
− 0. + f′
0
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32
23. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
f′ (x) = 2x, which is positive when x > 0 and negative when x is.
We can draw a number line:
− 0. + f′
↘ 0 ↗ f
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32
24. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
f′ (x) = 2x, which is positive when x > 0 and negative when x is.
We can draw a number line:
− 0. + f′
↘ 0 ↗ f
So f is decreasing on (−∞, 0) and increasing on (0, ∞).
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32
25. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
f′ (x) = 2x, which is positive when x > 0 and negative when x is.
We can draw a number line:
− 0. + f′
↘ 0 ↗ f
So f is decreasing on (−∞, 0) and increasing on (0, ∞).
In fact we can say f is decreasing on (−∞, 0] and increasing on
[0, ∞)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32
26. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
27. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solution
f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1
The critical points are 0 and and −4/5.
− ×. +
x−1/3
0
− 0 +
5x + 4
−4/5
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
28. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solution
f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1
The critical points are 0 and and −4/5.
− ×. +
x−1/3
0
− 0 +
5x + 4
−4/5
0 × f′ (x)
−4/5 0 f(x)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
29. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solution
f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1
The critical points are 0 and and −4/5.
− ×. +
x−1/3
0
− 0 +
5x + 4
−4/5
+ 0 × f′ (x)
−4/5 0 f(x)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
30. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solution
f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1
The critical points are 0 and and −4/5.
− ×. +
x−1/3
0
− 0 +
5x + 4
−4/5
+ 0 − × f′ (x)
−4/5 0 f(x)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
31. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solution
f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1
The critical points are 0 and and −4/5.
− ×. +
x−1/3
0
− 0 +
5x + 4
−4/5
+ 0 − × + f′ (x)
−4/5 0 f(x)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
32. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solution
f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1
The critical points are 0 and and −4/5.
− ×. +
x−1/3
0
− 0 +
5x + 4
−4/5
+ 0 − × + f′ (x)
↗ −4/5 0 f(x)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
33. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solution
f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1
The critical points are 0 and and −4/5.
− ×. +
x−1/3
0
− 0 +
5x + 4
−4/5
+ 0 − × + f′ (x)
↗ −4/5 ↘ 0 f(x)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
34. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solution
f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1
The critical points are 0 and and −4/5.
− ×. +
x−1/3
0
− 0 +
5x + 4
−4/5
+ 0 − × + f′ (x)
↗ −4/5 ↘ 0 ↗ f(x)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
35. The First Derivative Test
Theorem (The First Derivative Test)
Let f be continuous on [a, b] and c a critical point of f in (a, b).
If f′ changes from positive to negative at c, then c is a local
maximum.
If f′ changes from negative to positive at c, then c is a local
minimum.
If f′ (x) has the same sign on either side of c, then c is not a local
extremum.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 14 / 32
36. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
f′ (x) = 2x, which is positive when x > 0 and negative when x is.
We can draw a number line:
− 0. + f′
↘ 0 ↗ f
So f is decreasing on (−∞, 0) and increasing on (0, ∞).
In fact we can say f is decreasing on (−∞, 0] and increasing on
[0, ∞)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 15 / 32
37. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
f′ (x) = 2x, which is positive when x > 0 and negative when x is.
We can draw a number line:
− 0. + f′
↘ 0 ↗ f
min
So f is decreasing on (−∞, 0) and increasing on (0, ∞).
In fact we can say f is decreasing on (−∞, 0] and increasing on
[0, ∞)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 15 / 32
38. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solution
f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1
The critical points are 0 and and −4/5.
− ×. +
x−1/3
0
− 0 +
5x + 4
−4/5
+ 0 − × + f′ (x)
↗ −4/5 ↘ 0 ↗ f(x)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 16 / 32
39. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solution
f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1
The critical points are 0 and and −4/5.
− ×. +
x−1/3
0
− 0 +
5x + 4
−4/5
+ 0 − × + f′ (x)
↗ −4/5 ↘ 0 ↗ f(x)
max
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 16 / 32
40. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solution
f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1
The critical points are 0 and and −4/5.
− ×. +
x−1/3
0
− 0 +
5x + 4
−4/5
+ 0 − × + f′ (x)
↗ −4/5 ↘ 0 ↗ f(x)
max min
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 16 / 32
41. Outline
Recall: The Mean Value Theorem
Monotonicity
The Increasing/Decreasing Test
Finding intervals of monotonicity
The First Derivative Test
Concavity
Definitions
Testing for Concavity
The Second Derivative Test
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 17 / 32
42. Concavity
Definition
The graph of f is called concave upwards on an interval if it lies above
all its tangents on that interval. The graph of f is called concave
downwards on an interval if it lies below all its tangents on that
interval.
. .
concave up concave down
We sometimes say a concave up graph “holds water” and a concave
down graph “spills water”.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 18 / 32
43. Synonyms for concavity
Remark
“concave up” = “concave upwards” = “convex”
“concave down” = “concave downwards” = “concave”
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 19 / 32
44. Inflection points indicate a change in concavity
Definition
A point P on a curve y = f(x) is called an inflection point if f is
continuous at P and the curve changes from concave upward to
concave downward at P (or vice versa).
concave up
inflection point
.
concave
down
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 20 / 32
45. Theorem (Concavity Test)
If f′′ (x) > 0 for all x in an interval, then the graph of f is concave
upward on that interval.
If f′′ (x) < 0 for all x in an interval, then the graph of f is concave
downward on that interval.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 21 / 32
46. Theorem (Concavity Test)
If f′′ (x) > 0 for all x in an interval, then the graph of f is concave
upward on that interval.
If f′′ (x) < 0 for all x in an interval, then the graph of f is concave
downward on that interval.
Proof.
Suppose f′′ (x) > 0 on the interval I (which could be infinite). This
means f′ is increasing on I. Let a and x be in I. The tangent line
through (a, f(a)) is the graph of
L(x) = f(a) + f′ (a)(x − a)
By MVT, there exists a c between a and x with
f(x) = f(a) + f′ (c)(x − a)
Since f′ is increasing, f(x) > L(x). . . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 21 / 32
47. Finding Intervals of Concavity I
Example
Find the intervals of concavity for the graph of f(x) = x3 + x2 .
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 22 / 32
48. Finding Intervals of Concavity I
Example
Find the intervals of concavity for the graph of f(x) = x3 + x2 .
Solution
We have f′ (x) = 3x2 + 2x, so f′′ (x) = 6x + 2.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 22 / 32
49. Finding Intervals of Concavity I
Example
Find the intervals of concavity for the graph of f(x) = x3 + x2 .
Solution
We have f′ (x) = 3x2 + 2x, so f′′ (x) = 6x + 2.
This is negative when x < −1/3, positive when x > −1/3, and 0
when x = −1/3
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 22 / 32
50. Finding Intervals of Concavity I
Example
Find the intervals of concavity for the graph of f(x) = x3 + x2 .
Solution
We have f′ (x) = 3x2 + 2x, so f′′ (x) = 6x + 2.
This is negative when x < −1/3, positive when x > −1/3, and 0
when x = −1/3
So f is concave down on the open interval (−∞, −1/3), concave up
on the open interval (−1/3, ∞), and has an inflection point at the
point (−1/3, 2/27)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 22 / 32
51. Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
52. Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solution
10 −1/3 4 −4/3 2 −4/3
We have f′′ (x) = x − x = x (5x − 2).
9 9 9
+ ×
. +
x−4/3
0
− 0 +
5x − 2
2/5
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
53. Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solution
10 −1/3 4 −4/3 2 −4/3
We have f′′ (x) = x − x = x (5x − 2).
9 9 9
+ ×
. +
x−4/3
0
− 0 +
5x − 2
2/5
× 0 f′′ (x)
0 2/5 f(x)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
54. Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solution
10 −1/3 4 −4/3 2 −4/3
We have f′′ (x) = x − x = x (5x − 2).
9 9 9
+ ×
. +
x−4/3
0
− 0 +
5x − 2
2/5
−− × 0 f′′ (x)
0 2/5 f(x)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
55. Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solution
10 −1/3 4 −4/3 2 −4/3
We have f′′ (x) = x − x = x (5x − 2).
9 9 9
+ ×. +
x−4/3
0
− 0 +
5x − 2
2/5
−− ×−− 0 f′′ (x)
0 2/5 f(x)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
56. Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solution
10 −1/3 4 −4/3 2 −4/3
We have f′′ (x) = x − x = x (5x − 2).
9 9 9
+ ×. +
x−4/3
0
− 0 +
5x − 2
2/5
−− ×−− 0 ++ f′′ (x)
0 2/5 f(x)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
57. Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solution
10 −1/3 4 −4/3 2 −4/3
We have f′′ (x) = x − x = x (5x − 2).
9 9 9
+ ×. +
x−4/3
0
− 0 +
5x − 2
2/5
−− ×−− 0 ++ f′′ (x)
⌢ 0 2/5 f(x)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
58. Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solution
10 −1/3 4 −4/3 2 −4/3
We have f′′ (x) = x − x = x (5x − 2).
9 9 9
+ ×. +
x−4/3
0
− 0 +
5x − 2
2/5
−− ×−− 0 ++ f′′ (x)
⌢ 0 ⌢ 2/5 f(x)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
59. Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solution
10 −1/3 4 −4/3 2 −4/3
We have f′′ (x) = x − x = x (5x − 2).
9 9 9
+ ×. +
x−4/3
0
− 0 +
5x − 2
2/5
−− ×−− 0 ++ f′′ (x)
⌢ 0 ⌢ 2/5 ⌣ f(x)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
60. Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solution
10 −1/3 4 −4/3 2 −4/3
We have f′′ (x) = x − x = x (5x − 2).
9 9 9
+ ×. +
x−4/3
0
− 0 +
5x − 2
2/5
−− ×−− 0 ++ f′′ (x)
⌢ 0 ⌢ 2/5 ⌣ f(x)
IP
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
61. The Second Derivative Test
Theorem (The Second Derivative Test)
Let f, f′ , and f′′ be continuous on [a, b]. Let c be be a point in (a, b) with
f′ (c) = 0.
If f′′ (c) < 0, then c is a local maximum.
If f′′ (c) > 0, then c is a local minimum.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 24 / 32
62. The Second Derivative Test
Theorem (The Second Derivative Test)
Let f, f′ , and f′′ be continuous on [a, b]. Let c be be a point in (a, b) with
f′ (c) = 0.
If f′′ (c) < 0, then c is a local maximum.
If f′′ (c) > 0, then c is a local minimum.
Remarks
If f′′ (c) = 0, the second derivative test is inconclusive (this does
not mean c is neither; we just don’t know yet).
We look for zeroes of f′ and plug them into f′′ to determine if their f
values are local extreme values.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 24 / 32
63. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
+ f′′ = (f′ )′
.
c f′
0 f′
c f
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
64. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′′ is continuous, f′′ = (f′ )′
+
f′′ (x) > 0 for all x .
sufficiently close to c.
c f′
0 f′
c f
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
65. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′′ is continuous, f′′ = (f′ )′
+ +
f′′ (x) > 0 for all x .
sufficiently close to c.
c f′
0 f′
c f
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
66. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′′ is continuous, f′′ = (f′ )′
+ + +
f′′ (x) > 0 for all x .
sufficiently close to c.
c f′
0 f′
c f
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
67. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′′ is continuous, f′′ = (f′ )′
+ + +
f′′ (x) > 0 for all x .
sufficiently close to c.
c f′
0 f′
Since f′′ = (f′ )′ , we know f′ c f
is increasing near c.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
68. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′′ is continuous, f′′ = (f′ )′
+ + +
f′′ (x) > 0 for all x .
sufficiently close to c. ↗ c f′
0 f′
Since f′′ = (f′ )′ , we know f′ c f
is increasing near c.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
69. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′′ is continuous, f′′ = (f′ )′
+ + +
f′′ (x) > 0 for all x .
sufficiently close to c. ↗ c ↗ f′
0 f′
Since f′′ = (f′ )′ , we know f′ c f
is increasing near c.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
70. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′′ is continuous, f′′ = (f′ )′
+ + +
f′′ (x) > 0 for all x .
sufficiently close to c. ↗ c ↗ f′
0 f′
Since f′′ = (f′ )′ , we know f′ c f
is increasing near c.
Since f′ (c) = 0 and f′ is increasing, f′ (x) < 0 for x close to c and
less than c, and f′ (x) > 0 for x close to c and more than c.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
71. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′′ is continuous, f′′ = (f′ )′
+ + +
f′′ (x) > 0 for all x .
sufficiently close to c. ↗ c ↗ f′
− 0 f′
Since f′′ = (f′ )′ , we know f′ c f
is increasing near c.
Since f′ (c) = 0 and f′ is increasing, f′ (x) < 0 for x close to c and
less than c, and f′ (x) > 0 for x close to c and more than c.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
72. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′′ is continuous, f′′ = (f′ )′
+ + +
f′′ (x) > 0 for all x .
sufficiently close to c. ↗ c ↗ f′
− 0 + f′
Since f′′ = (f′ )′ , we know f′ c f
is increasing near c.
Since f′ (c) = 0 and f′ is increasing, f′ (x) < 0 for x close to c and
less than c, and f′ (x) > 0 for x close to c and more than c.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
73. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′′ is continuous, f′′ = (f′ )′
+ + +
f′′ (x) > 0 for all x .
sufficiently close to c. ↗ c ↗ f′
− 0 + f′
Since f′′ = (f′ )′ , we know f′ c f
is increasing near c.
Since f′ (c) = 0 and f′ is increasing, f′ (x) < 0 for x close to c and
less than c, and f′ (x) > 0 for x close to c and more than c.
This means f′ changes sign from negative to positive at c, which
means (by the First Derivative Test) that f has a local minimum
at c.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
74. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′′ is continuous, f′′ = (f′ )′
+ + +
f′′ (x) > 0 for all x .
sufficiently close to c. ↗ c ↗ f′
− 0 + f′
Since f′′ = (f′ )′ , we know f′
↘ c f
is increasing near c.
Since f′ (c) = 0 and f′ is increasing, f′ (x) < 0 for x close to c and
less than c, and f′ (x) > 0 for x close to c and more than c.
This means f′ changes sign from negative to positive at c, which
means (by the First Derivative Test) that f has a local minimum
at c.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
75. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′′ is continuous, f′′ = (f′ )′
+ + +
f′′ (x) > 0 for all x .
sufficiently close to c. ↗ c ↗ f′
− 0 + f′
Since f′′ = (f′ )′ , we know f′
↘ c ↗ f
is increasing near c.
Since f′ (c) = 0 and f′ is increasing, f′ (x) < 0 for x close to c and
less than c, and f′ (x) > 0 for x close to c and more than c.
This means f′ changes sign from negative to positive at c, which
means (by the First Derivative Test) that f has a local minimum
at c.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
76. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′′ is continuous, f′′ = (f′ )′
+ + +
f′′ (x) > 0 for all x .
sufficiently close to c. ↗ c ↗ f′
− 0 + f′
Since f′′ = (f′ )′ , we know f′
↘ c ↗ f
is increasing near c. min
Since f′ (c) = 0 and f′ is increasing, f′ (x) < 0 for x close to c and
less than c, and f′ (x) > 0 for x close to c and more than c.
This means f′ changes sign from negative to positive at c, which
means (by the First Derivative Test) that f has a local minimum
at c.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
77. Using the Second Derivative Test I
Example
Find the local extrema of f(x) = x3 + x2 .
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 26 / 32
78. Using the Second Derivative Test I
Example
Find the local extrema of f(x) = x3 + x2 .
Solution
f′ (x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 26 / 32
79. Using the Second Derivative Test I
Example
Find the local extrema of f(x) = x3 + x2 .
Solution
f′ (x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.
Remember f′′ (x) = 6x + 2
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 26 / 32
80. Using the Second Derivative Test I
Example
Find the local extrema of f(x) = x3 + x2 .
Solution
f′ (x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.
Remember f′′ (x) = 6x + 2
Since f′′ (−2/3) = −2 < 0, −2/3 is a local maximum.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 26 / 32
81. Using the Second Derivative Test I
Example
Find the local extrema of f(x) = x3 + x2 .
Solution
f′ (x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.
Remember f′′ (x) = 6x + 2
Since f′′ (−2/3) = −2 < 0, −2/3 is a local maximum.
Since f′′ (0) = 2 > 0, 0 is a local minimum.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 26 / 32
82. Using the Second Derivative Test II
Example
Find the local extrema of f(x) = x2/3 (x + 2)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 27 / 32
83. Using the Second Derivative Test II
Example
Find the local extrema of f(x) = x2/3 (x + 2)
Solution
1 −1/3
Remember f′ (x) = x (5x + 4) which is zero when x = −4/5
3
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 27 / 32
84. Using the Second Derivative Test II
Example
Find the local extrema of f(x) = x2/3 (x + 2)
Solution
1 −1/3
Remember f′ (x) = x (5x + 4) which is zero when x = −4/5
3
10 −4/3
Remember f′′ (x) = x (5x − 2), which is negative when
9
x = −4/5
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 27 / 32
85. Using the Second Derivative Test II
Example
Find the local extrema of f(x) = x2/3 (x + 2)
Solution
1 −1/3
Remember f′ (x) = x (5x + 4) which is zero when x = −4/5
3
10 −4/3
Remember f′′ (x) = x (5x − 2), which is negative when
9
x = −4/5
So x = −4/5 is a local maximum.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 27 / 32
86. Using the Second Derivative Test II
Example
Find the local extrema of f(x) = x2/3 (x + 2)
Solution
1 −1/3
Remember f′ (x) = x (5x + 4) which is zero when x = −4/5
3
10 −4/3
Remember f′′ (x) = x (5x − 2), which is negative when
9
x = −4/5
So x = −4/5 is a local maximum.
Notice the Second Derivative Test doesn’t catch the local
minimum x = 0 since f is not differentiable there.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 27 / 32
87. Using the Second Derivative Test II: Graph
Graph of f(x) = x2/3 (x + 2):
y
(−4/5, 1.03413)
(2/5, 1.30292)
. x
(−2, 0) (0, 0)
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 28 / 32
88. When the second derivative is zero
Remark
At inflection points c, if f′ is differentiable at c, then f′′ (c) = 0
If f′′ (c) = 0, must f have an inflection point at c?
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 29 / 32
89. When the second derivative is zero
Remark
At inflection points c, if f′ is differentiable at c, then f′′ (c) = 0
If f′′ (c) = 0, must f have an inflection point at c?
Consider these examples:
f(x) = x4 g(x) = −x4 h(x) = x3
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 29 / 32
90. When first and second derivative are zero
function derivatives graph type
f′ (x) = 4x3 , f′ (0) = 0
f(x) = x4 min
f′′ (x) = 12x2 , f′′ (0) = 0 .
.
g′ (x) = −4x3 , g′ (0) = 0
g(x) = −x4 max
g′′ (x) = −12x2 , g′′ (0) = 0
h′ (x) = 3x2 , h′ (0) = 0
h(x) = x3 infl.
h′′ (x) = 6x, h′′ (0) = 0 .
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 30 / 32
91. When the second derivative is zero
Remark
At inflection points c, if f′ is differentiable at c, then f′′ (c) = 0
If f′′ (c) = 0, must f have an inflection point at c?
Consider these examples:
f(x) = x4 g(x) = −x4 h(x) = x3
All of them have critical points at zero with a second derivative of zero.
But the first has a local min at 0, the second has a local max at 0, and
the third has an inflection point at 0. This is why we say 2DT has
nothing to say when f′′ (c) = 0.
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 31 / 32
92. Summary
Concepts: Mean Value Theorem, monotonicity, concavity
Facts: derivatives can detect monotonicity and concavity
Techniques for drawing curves: the Increasing/Decreasing Test
and the Concavity Test
Techniques for finding extrema: the First Derivative Test and the
Second Derivative Test
. . . . . .
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 32 / 32