- The document is a section from a calculus course at NYU that discusses using derivatives to determine the shapes of curves.
- It covers using the first derivative to determine if a function is increasing or decreasing over an interval using the Increasing/Decreasing Test. It also discusses using the second derivative to determine if a function is concave up or down over an interval using the Second Derivative Test.
- Examples are provided to demonstrate finding intervals of monotonicity for functions and classifying critical points as local maxima, minima or neither using the First Derivative Test.
Report
Share
Report
Share
1 of 71
More Related Content
Lesson 20: Derivatives and the shapes of curves
1. Section 4.2
Derivatives and the Shapes of Curves
V63.0121.002.2010Su, Calculus I
New York University
June 9, 2010
Announcements
No office hours tonight
Quiz 3 on Thursday (3.3, 3.4, 3.5, 3.7)
Assignment 4 due Tuesday
. . . . . .
2. Announcements
No office hours tonight
Quiz 3 on Thursday (3.3,
3.4, 3.5, 3.7)
Assignment 4 due Tuesday
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 2 / 30
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.
Use the second derivative
of a function to determine
the intervals along which
the graph of the function is . . . . . .
concave up or concave 4.2 The Shapes of Curves
V63.0121.002.2010Su, Calculus I (NYU) Section June 9, 2010 3 / 30
4. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 4 / 30
5. 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
.
f(b) − f(a) b
.
= f′ (c).
b−a . .
a
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 5 / 30
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
.
f(b) − f(a) b
.
= f′ (c).
b−a . .
a
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 5 / 30
7. 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
.
f(b) − f(a) b
.
= f′ (c).
b−a . .
a
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 5 / 30
8. 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) = 0.
y−x
So f(y) = f(x). Since this is true for all x and y in (a, b), then f is
constant.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 6 / 30
9. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 7 / 30
10. What does it mean for a function to be increasing?
Definition
A function f is increasing on (a, b) if
f(x) < f(y)
whenever x and y are two points in (a, b) with x < y.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 8 / 30
11. What does it mean for a function to be increasing?
Definition
A function f is increasing on (a, b) if
f(x) < f(y)
whenever x and y are two points in (a, b) with x < y.
An increasing function “preserves order.”
Write your own definition (mutatis mutandis) of decreasing,
nonincreasing, nondecreasing
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 8 / 30
12. 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).
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 9 / 30
13. 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).
Proof.
It works the same as the last theorem. Pick two points x and y in (a, b)
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) > 0.
y−x
So
f(y) − f(x) = f′ (c)(y − x) > 0.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 9 / 30
14. Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f(x) = 2x − 5.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 10 / 30
15. 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 (−∞, ∞).
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 10 / 30
16. 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).
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 10 / 30
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 (−∞, ∞).
Example
Describe the monotonicity of f(x) = arctan(x).
Solution
1
Since f′ (x) = is always positive, f(x) is always increasing.
1 + x2
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 10 / 30
18. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 11 / 30
19. 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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 11 / 30
20. 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
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 11 / 30
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.
We can draw a number line:
−
. 0
.. .
+ .′
f
↘
. 0
. ↗
. f
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 11 / 30
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
. ↗
. f
.
So f is decreasing on (−∞, 0) and increasing on (0, ∞).
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 11 / 30
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
.
So f is decreasing on (−∞, 0) and increasing on (0, ∞).
In fact we can say f is decreasing on (−∞, 0] and increasing on
[0, ∞)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 11 / 30
24. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 12 / 30
25. 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.
−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
5
.x+4
−
. 4/5
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 12 / 30
26. 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.
−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
5
.x+4
−
. 4/5
0
.. ×
.. .′ (x)
f
−
. 4/5 0
. f
.(x)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 12 / 30
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.
−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
5
.x+4
−
. 4/5
..
+ 0
.. ×
.. .′ (x)
f
−
. 4/5 0
. f
.(x)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 12 / 30
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.
−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
5
.x+4
−
. 4/5
..
+ 0 − ×
.. . . . . .′ (x)
f
−
. 4/5 0
. f
.(x)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 12 / 30
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.
−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
5
.x+4
−
. 4/5
..
+ 0 − ×
.. . . . . ..
+ .′ (x)
f
−
. 4/5 0
. f
.(x)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 12 / 30
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.
−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
5
.x+4
−
. 4/5
..
+ 0 − ×
.. . . . . ..
+ .′ (x)
f
↗
. −
. 4/5 0
. f
.(x)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 12 / 30
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.
−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
5
.x+4
−
. 4/5
..
+ 0 − ×
.. . . . . ..
+ .′ (x)
f
↗
. − ↘ .
. 4/5 . 0 f
.(x)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 12 / 30
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.
−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
5
.x+4
−
. 4/5
..
+ 0 − ×
.. . . . . ..
+ .′ (x)
f
↗
. − ↘ .
. 4/5 . 0 ↗
. f
.(x)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 12 / 30
33. 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′ (x) > 0 on (a, c) and f′ (x) < 0 on (c, b), then c is a local
maximum.
If f′ (x) < 0 on (a, c) and f′ (x) > 0 on (c, b), then c is a local
minimum.
If f′ (x) has the same sign on (a, c) and (c, b), then c is not a local
extremum.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 13 / 30
34. 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, ∞)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 14 / 30
35. 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
.
m
. in
So f is decreasing on (−∞, 0) and increasing on (0, ∞).
In fact we can say f is decreasing on (−∞, 0] and increasing on
[0, ∞)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 14 / 30
36. 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.
−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
5
.x+4
−
. 4/5
..
+ 0 − ×
.. . . . . ..
+ .′ (x)
f
↗
. − ↘ .
. 4/5 . 0 ↗
. f
.(x)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 15 / 30
37. 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.
−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
5
.x+4
−
. 4/5
..
+ 0 − ×
.. . . . . ..
+ .′ (x)
f
↗
. − ↘ .
. 4/5 . 0 ↗
. f
.(x)
m
. ax
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 15 / 30
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.
−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
5
.x+4
−
. 4/5
..
+ 0 − ×
.. . . . . ..
+ .′ (x)
f
↗
. − ↘ .
. 4/5 . 0 ↗
. f
.(x)
m
. ax . inm
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 15 / 30
39. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 16 / 30
40. Concavity
Definition
The graph of f is called concave up on an interval I if it lies above all
its tangents on I. The graph of f is called concave down on I if it lies
below all its tangents on I.
. .
concave up concave down
We sometimes say a concave up graph “holds water” and a concave
down graph “spills water”.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 17 / 30
41. 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
i
.nflection point
. .
.
concave
down
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 18 / 30
42. Theorem (Concavity Test)
If f′′ (x) > 0 for all x in an interval I, then the graph of f is concave
upward on I.
If f′′ (x) < 0 for all x in I, then the graph of f is concave downward
on I.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 19 / 30
43. Theorem (Concavity Test)
If f′′ (x) > 0 for all x in an interval I, then the graph of f is concave
upward on I.
If f′′ (x) < 0 for all x in I, then the graph of f is concave downward
on I.
Proof.
Suppose f′′ (x) > 0 on I. 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)
f(x) − f(a)
By MVT, there exists a c between a and x with = f′ (c). So
x−a
f(x) = f(a) + f′ (c)(x − a) ≥ f(a) + f′ (a)(x − a) = L(x) .
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 19 / 30
44. Finding Intervals of Concavity I
Example
Find the intervals of concavity for the graph of f(x) = x3 + x2 .
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 20 / 30
45. 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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 20 / 30
46. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 20 / 30
47. 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 (−∞, −1/3), concave up on (−1/3, ∞),
and has an inflection point at (−1/3, 2/27)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 20 / 30
48. Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 21 / 30
49. 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
f′′ (x) = x − x = x (5x − 2)
9 9 9
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 21 / 30
50. 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
f′′ (x) =x − x = x (5x − 2)
9 9 9
The second derivative f′′ (x) is not defined at 0
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 21 / 30
51. 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
f′′ (x) =x − x = x (5x − 2)
9 9 9
The second derivative f′′ (x) is not defined at 0
Otherwise, x−4/3 is always positive, so the concavity is
determined by the 5x − 2 factor
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 21 / 30
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
f′′ (x) =x − x = x (5x − 2)
9 9 9
The second derivative f′′ (x) is not defined at 0
Otherwise, x−4/3 is always positive, so the concavity is
determined by the 5x − 2 factor
So f is concave down on (−∞, 0], concave down on [0, 2/5),
concave up on (2/5, ∞), and has an inflection point when x = 2/5
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 21 / 30
53. 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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 22 / 30
54. 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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 22 / 30
55. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0. Since f′′ is continuous, f′′ (x) > 0 for
all x sufficiently close to c. Since f′′ = (f′ )′ , we know 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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 23 / 30
56. Using the Second Derivative Test I
Example
Find the local extrema of f(x) = x3 + x2 .
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 24 / 30
57. 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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 24 / 30
58. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 24 / 30
59. 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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 24 / 30
60. 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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 24 / 30
61. Using the Second Derivative Test II
Example
Find the local extrema of f(x) = x2/3 (x + 2)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 25 / 30
62. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 25 / 30
63. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 25 / 30
64. 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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 25 / 30
65. 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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 25 / 30
66. 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)
(
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 26 / 30
67. When the second derivative is zero
At inflection points c, if f′ is differentiable at c, then f′′ (c) = 0
Is it necessarily true, though?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 27 / 30
68. When the second derivative is zero
At inflection points c, if f′ is differentiable at c, then f′′ (c) = 0
Is it necessarily true, though?
Consider these examples:
f(x) = x4 g(x) = −x4 h(x) = x3
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 27 / 30
69. 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 .
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 28 / 30
70. When the second derivative is zero
At inflection points c, if f′ is differentiable at c, then f′′ (c) = 0
Is it necessarily true, though?
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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 29 / 30
71. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 30 / 30