Lesson 20: Derivatives and the Shape of Curves (Section 041 slides)

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

. . . . . .

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

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.

. . . . . .


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)

. . . . . .


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

. . . . . .



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

. . . . . .



c
(a, b) such that
b
f(b) − f(a)
= f′ (c). .
b−a a

. . . . . .



c
(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)

. . . . . .


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.

. . . . . .


Outline


Monotonicity

Concavity
Definitions

. . . . . .


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.

. . . . . .


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

. . . . . .



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.

. . . . . .



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).

. . . . . .


Finding intervals of monotonicity I

Example
Find the intervals of monotonicity of f(x) = 2x − 5.

. . . . . .



Example

Solution
f′ (x) = 2 is always positive, so f is increasing on (−∞, ∞).

. . . . . .



Example

Solution

Example
Describe the monotonicity of f(x) = arctan(x).

. . . . . .



Example

Solution

Example
Describe the monotonicity of f(x) = arctan(x).

Solution
1
Since f′ (x) = is always positive, f(x) is always increasing.
1 + x2

. . . . . .


Finding intervals of monotonicity II

Example
Find the intervals of monotonicity of f(x) = x2 − 1.

. . . . . .



Example

Solution

f′ (x) = 2x, which is positive when x > 0 and negative when x is.

. . . . . .



Example

Solution

We can draw a number line:
− 0. + f′
0

. . . . . .



Example

Solution

− 0. + f′
↘ 0 ↗ f

. . . . . .



Example

Solution

− 0. + f′
↘ 0 ↗ f

So f is decreasing on (−∞, 0) and increasing on (0, ∞).

. . . . . .



Example

Solution

− 0. + f′
↘ 0 ↗ f

In fact we can say f is decreasing on (−∞, 0] and increasing on
[0, ∞)
. . . . . .


Finding intervals of monotonicity III

Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).

. . . . . .



Example

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

. . . . . .



Example

Solution

f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1

− ×. +
x−1/3
0
− 0 +
5x + 4
−4/5
0 × f′ (x)
−4/5 0 f(x)

. . . . . .



Example

Solution

f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1

− ×. +
x−1/3
0
− 0 +
5x + 4
−4/5
+ 0 × f′ (x)
−4/5 0 f(x)

. . . . . .



Example

Solution

f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1

− ×. +
x−1/3
0
− 0 +
5x + 4
−4/5
+ 0 − × f′ (x)
−4/5 0 f(x)

. . . . . .



Example

Solution

f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1

− ×. +
x−1/3
0
− 0 +
5x + 4
−4/5
+ 0 − × + f′ (x)
−4/5 0 f(x)

. . . . . .



Example

Solution

f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1

− ×. +
x−1/3
0
− 0 +
5x + 4
−4/5
+ 0 − × + f′ (x)
↗ −4/5 0 f(x)

. . . . . .



Example

Solution

f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1

− ×. +
x−1/3
0
− 0 +
5x + 4
−4/5
+ 0 − × + f′ (x)
↗ −4/5 ↘ 0 f(x)

. . . . . .



Example

Solution

f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1

− ×. +
x−1/3
0
− 0 +
5x + 4
−4/5
+ 0 − × + f′ (x)
↗ −4/5 ↘ 0 ↗ f(x)

. . . . . .



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.

. . . . . .



Example

Solution

− 0. + f′
↘ 0 ↗ f

[0, ∞)
. . . . . .



Example

Solution

− 0. + f′
↘ 0 ↗ f
min
[0, ∞)
. . . . . .



Example

Solution

f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1

− ×. +
x−1/3
0
− 0 +
5x + 4
−4/5
+ 0 − × + f′ (x)
↗ −4/5 ↘ 0 ↗ f(x)

. . . . . .



Example

Solution

f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1

− ×. +
x−1/3
0
− 0 +
5x + 4
−4/5
+ 0 − × + f′ (x)
↗ −4/5 ↘ 0 ↗ f(x)
max
. . . . . .



Example

Solution

f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1

− ×. +
x−1/3
0
− 0 +
5x + 4
−4/5
+ 0 − × + f′ (x)
↗ −4/5 ↘ 0 ↗ f(x)
max min
. . . . . .


Outline


Monotonicity

Concavity
Definitions

. . . . . .


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”.
. . . . . .


Synonyms for concavity

Remark

“concave up” = “concave upwards” = “convex”
“concave down” = “concave downwards” = “concave”

. . . . . .


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

. . . . . .


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.

. . . . . .


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). . . . . . .


Finding Intervals of Concavity I

Example
Find the intervals of concavity for the graph of f(x) = x3 + x2 .

. . . . . .



Example

Solution

We have f′ (x) = 3x2 + 2x, so f′′ (x) = 6x + 2.

. . . . . .



Example

Solution

This is negative when x < −1/3, positive when x > −1/3, and 0
when x = −1/3

. . . . . .



Example

Solution

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)

. . . . . .


Finding Intervals of Concavity II

Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).

. . . . . .



Example

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

. . . . . .



Example

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)

. . . . . .



Example

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)

. . . . . .



Example

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)

. . . . . .



Example

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)

. . . . . .



Example

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)

. . . . . .



Example

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)

. . . . . .



Example

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)

. . . . . .



Example

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

. . . . . .



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.

. . . . . .



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.

. . . . . .


Proof of the Second Derivative Test

Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.

+ f′′ = (f′ )′
.
c f′
0 f′
c f

. . . . . .



Proof.

Since f′′ is continuous, f′′ = (f′ )′
+
f′′ (x) > 0 for all x .
sufficiently close to c.
c f′
0 f′
c f

. . . . . .



Proof.

+ +
f′′ (x) > 0 for all x .
c f′
0 f′
c f

. . . . . .



Proof.

+ + +
f′′ (x) > 0 for all x .
c f′
0 f′
c f

. . . . . .



Proof.

+ + +
f′′ (x) > 0 for all x .
c f′
0 f′
Since f′′ = (f′ )′ , we know f′ c f
is increasing near c.

. . . . . .



Proof.

+ + +
f′′ (x) > 0 for all x .
sufficiently close to c. ↗ c f′
0 f′

. . . . . .



Proof.

+ + +
f′′ (x) > 0 for all x .
sufficiently close to c. ↗ c ↗ f′
0 f′

. . . . . .



Proof.

+ + +
f′′ (x) > 0 for all x .
0 f′

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.

. . . . . .



Proof.

+ + +
f′′ (x) > 0 for all x .
− 0 f′


. . . . . .



Proof.

+ + +
f′′ (x) > 0 for all x .
− 0 + f′


. . . . . .



Proof.

+ + +
f′′ (x) > 0 for all x .
− 0 + f′

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.

. . . . . .



Proof.

+ + +
f′′ (x) > 0 for all x .
− 0 + f′
Since f′′ = (f′ )′ , we know f′
↘ c f

at c.

. . . . . .



Proof.

+ + +
f′′ (x) > 0 for all x .
− 0 + f′
↘ c ↗ f

at c.

. . . . . .



Proof.

+ + +
f′′ (x) > 0 for all x .
− 0 + f′
↘ c ↗ f
is increasing near c. min

at c.

. . . . . .


Using the Second Derivative Test I

Example
Find the local extrema of f(x) = x3 + x2 .

. . . . . .



Example

Solution

f′ (x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.

. . . . . .



Example

Solution

Remember f′′ (x) = 6x + 2

. . . . . .



Example

Solution

Since f′′ (−2/3) = −2 < 0, −2/3 is a local maximum.

. . . . . .



Example

Solution

Since f′′ (−2/3) = −2 < 0, −2/3 is a local maximum.
Since f′′ (0) = 2 > 0, 0 is a local minimum.

. . . . . .


Using the Second Derivative Test II

Example
Find the local extrema of f(x) = x2/3 (x + 2)

. . . . . .



Example

Solution
1 −1/3
Remember f′ (x) = x (5x + 4) which is zero when x = −4/5
3

. . . . . .



Example

Solution
1 −1/3
3
10 −4/3
Remember f′′ (x) = x (5x − 2), which is negative when
9
x = −4/5

. . . . . .



Example

Solution
1 −1/3
3
10 −4/3
9
x = −4/5
So x = −4/5 is a local maximum.

. . . . . .



Example

Solution
1 −1/3
3
10 −4/3
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.

. . . . . .


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)

. . . . . .


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?

. . . . . .



Remark

Consider these examples:

f(x) = x4 g(x) = −x4 h(x) = x3

. . . . . .


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 .

. . . . . .



Remark

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.

. . . . . .


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

. . . . . .


Lesson 20: Derivatives and the Shape of Curves (Section 041 slides)

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