1.
Section 4.2
Derivatives and the Shapes of Curves
V63.0121.006/016, Calculus I
New York University
March 30, 2010
Announcements
Quiz 3 on Friday (Sections 2.6–3.5)
Midterm Exam scores have been updated
If your Midterm Letter Grade on Blackboard diﬀers from your
midterm grade on Albert, trust Blackboard
2.
Announcements
Quiz 3 on Friday (Sections 2.6–3.5)
Midterm Exam scores have been updated
If your Midterm Letter Grade on Blackboard diﬀers from your
midterm grade on Albert, trust Blackboard
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 2 / 28
3.
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 3 / 28
4.
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 4 / 28
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 4 / 28
6.
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 4 / 28
7.
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 5 / 28
8.
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 6 / 28
9.
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 7 / 28
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 .
An increasing function “preserves order.”
Write your own deﬁnition (mutatis mutandis) of decreasing,
nonincreasing, nondecreasing
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 7 / 28
11.
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 8 / 28
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).
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 8 / 28
13.
Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f (x) = 2x − 5.
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 9 / 28
14.
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 9 / 28
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 (−∞, ∞).
Example
Describe the monotonicity of f (x) = arctan(x).
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 9 / 28
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).
Solution
1
Since f (x) = is always positive, f (x) is always increasing.
1 + x2
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 9 / 28
17.
Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f (x) = x 2 − 1.
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 10 / 28
18.
Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f (x) = x 2 − 1.
Solution
f (x) = 2x, which is positive when x > 0 and negative when x is.
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 10 / 28
19.
Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f (x) = x 2 − 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 10 / 28
20.
Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f (x) = x 2 − 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 10 / 28
21.
Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f (x) = x 2 − 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 10 / 28
22.
Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f (x) = x 2 − 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 10 / 28
23.
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f (x) = x 2/3 (x + 2).
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
24.
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f (x) = x 2/3 (x + 2).
Solution
f (x) = 3 x −1/3 (x + 2) + x 2/3 = 3 x −1/3 (5x + 4)
2 1
The critical points are 0 and and −4/5.
− × +
x −1/3
0
− 0 +
5x + 4
−4/5
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
25.
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f (x) = x 2/3 (x + 2).
Solution
f (x) = 3 x −1/3 (x + 2) + x 2/3 = 3 x −1/3 (5x + 4)
2 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
26.
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f (x) = x 2/3 (x + 2).
Solution
f (x) = 3 x −1/3 (x + 2) + x 2/3 = 3 x −1/3 (5x + 4)
2 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
27.
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f (x) = x 2/3 (x + 2).
Solution
f (x) = 3 x −1/3 (x + 2) + x 2/3 = 3 x −1/3 (5x + 4)
2 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
28.
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f (x) = x 2/3 (x + 2).
Solution
f (x) = 3 x −1/3 (x + 2) + x 2/3 = 3 x −1/3 (5x + 4)
2 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
29.
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f (x) = x 2/3 (x + 2).
Solution
f (x) = 3 x −1/3 (x + 2) + x 2/3 = 3 x −1/3 (5x + 4)
2 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
30.
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f (x) = x 2/3 (x + 2).
Solution
f (x) = 3 x −1/3 (x + 2) + x 2/3 = 3 x −1/3 (5x + 4)
2 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
31.
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f (x) = x 2/3 (x + 2).
Solution
f (x) = 3 x −1/3 (x + 2) + x 2/3 = 3 x −1/3 (5x + 4)
2 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
32.
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 12 / 28
33.
Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f (x) = x 2 − 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 13 / 28
34.
Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f (x) = x 2 − 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 13 / 28
35.
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f (x) = x 2/3 (x + 2).
Solution
f (x) = 3 x −1/3 (x + 2) + x 2/3 = 3 x −1/3 (5x + 4)
2 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 14 / 28
36.
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f (x) = x 2/3 (x + 2).
Solution
f (x) = 3 x −1/3 (x + 2) + x 2/3 = 3 x −1/3 (5x + 4)
2 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 14 / 28
37.
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f (x) = x 2/3 (x + 2).
Solution
f (x) = 3 x −1/3 (x + 2) + x 2/3 = 3 x −1/3 (5x + 4)
2 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 14 / 28
38.
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 15 / 28
39.
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 16 / 28
40.
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 17 / 28
41.
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 18 / 28
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 .
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 18 / 28
43.
Finding Intervals of Concavity I
Example
Find the intervals of concavity for the graph of f (x) = x 3 + x 2 .
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 19 / 28
44.
Finding Intervals of Concavity I
Example
Find the intervals of concavity for the graph of f (x) = x 3 + x 2 .
Solution
We have f (x) = 3x 2 + 2x, so f (x) = 6x + 2.
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 19 / 28
45.
Finding Intervals of Concavity I
Example
Find the intervals of concavity for the graph of f (x) = x 3 + x 2 .
Solution
We have f (x) = 3x 2 + 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 19 / 28
46.
Finding Intervals of Concavity I
Example
Find the intervals of concavity for the graph of f (x) = x 3 + x 2 .
Solution
We have f (x) = 3x 2 + 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 19 / 28
47.
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f (x) = x 2/3 (x + 2).
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 20 / 28
48.
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f (x) = x 2/3 (x + 2).
Solution
10 −1/3 4 −4/3 2 −4/3
f (x) = x − x = x (5x − 2)
9 9 9
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 20 / 28
49.
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f (x) = x 2/3 (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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 20 / 28
50.
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f (x) = x 2/3 (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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 20 / 28
51.
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f (x) = x 2/3 (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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 20 / 28
52.
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 21 / 28
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.
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 21 / 28
54.
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. So f is increasing on an interval containing 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 22 / 28
55.
Using the Second Derivative Test I
Example
Find the local extrema of f (x) = x 3 + x 2 .
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 23 / 28
56.
Using the Second Derivative Test I
Example
Find the local extrema of f (x) = x 3 + x 2 .
Solution
f (x) = 3x 2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 23 / 28
57.
Using the Second Derivative Test I
Example
Find the local extrema of f (x) = x 3 + x 2 .
Solution
f (x) = 3x 2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.
Remember f (x) = 6x + 2
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 23 / 28
58.
Using the Second Derivative Test I
Example
Find the local extrema of f (x) = x 3 + x 2 .
Solution
f (x) = 3x 2 + 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 23 / 28
59.
Using the Second Derivative Test I
Example
Find the local extrema of f (x) = x 3 + x 2 .
Solution
f (x) = 3x 2 + 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 23 / 28
60.
Using the Second Derivative Test II
Example
Find the local extrema of f (x) = x 2/3 (x + 2)
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 24 / 28
61.
Using the Second Derivative Test II
Example
Find the local extrema of f (x) = x 2/3 (x + 2)
Solution
1
Remember f (x) = x −1/3 (5x + 4) which is zero when x = −4/5
3
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 24 / 28
62.
Using the Second Derivative Test II
Example
Find the local extrema of f (x) = x 2/3 (x + 2)
Solution
1
Remember f (x) = x −1/3 (5x + 4) which is zero when x = −4/5
3
10
Remember f (x) = x −4/3 (5x − 2), which is negative when
9
x = −4/5
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 24 / 28
63.
Using the Second Derivative Test II
Example
Find the local extrema of f (x) = x 2/3 (x + 2)
Solution
1
Remember f (x) = x −1/3 (5x + 4) which is zero when x = −4/5
3
10
Remember f (x) = x −4/3 (5x − 2), which is negative when
9
x = −4/5
So x = −4/5 is a local maximum.
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 24 / 28
64.
Using the Second Derivative Test II
Example
Find the local extrema of f (x) = x 2/3 (x + 2)
Solution
1
Remember f (x) = x −1/3 (5x + 4) which is zero when x = −4/5
3
10
Remember f (x) = x −4/3 (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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 24 / 28
65.
Using the Second Derivative Test II: Graph
Graph of f (x) = x 2/3 (x + 2):
y
(−4/5, 1.03413)
(2/5, 1.30292)
x
(−2, 0) (0, 0)
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 25 / 28
66.
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 26 / 28
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?
Consider these examples:
f (x) = x 4 g (x) = −x 4 h(x) = x 3
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 26 / 28
68.
When ﬁrst and second derivative are zero
function derivatives graph type
f (x) = 4x 3 , f (0) = 0
f (x) = x 4 min
f (x) = 12x 2 , f (0) = 0
g (x) = −4x 3 , g (0) = 0
g (x) = −x 4 max
g (x) = −12x 2 , g (0) = 0
h (x) = 3x 2 , h (0) = 0
h(x) = x 3 inﬂ.
h (x) = 6x, h (0) = 0
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 27 / 28
69.
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) = x 4 g (x) = −x 4 h(x) = x 3
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 28 / 28
70.
What have we learned today?
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 29 / 28
71.
What have we learned today?
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
Next time: Graphing functions!
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 29 / 28
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