Contributed by:

In this section, we will learn:

How the derivative of a function gives us the direction in which the curve proceeds at each point.

How the derivative of a function gives us the direction in which the curve proceeds at each point.

1.
4

APPLICATIONS OF DIFFERENTIATION

APPLICATIONS OF DIFFERENTIATION

2.
APPLICATIONS OF DIFFERENTIATION

Many applications of calculus depend

on our ability to deduce facts about

a function f from information concerning

its derivatives.

Many applications of calculus depend

on our ability to deduce facts about

a function f from information concerning

its derivatives.

3.
APPLICATIONS OF DIFFERENTIATION

4.3

How Derivatives Affect

the Shape of a Graph

In this section, we will learn:

How the derivative of a function gives us the direction

in which the curve proceeds at each point.

4.3

How Derivatives Affect

the Shape of a Graph

In this section, we will learn:

How the derivative of a function gives us the direction

in which the curve proceeds at each point.

4.
DERIVATIVES AND GRAPH SHAPE

As f’(x) represents the slope of the curve

y = f(x) at the point (x, f(x)), it tells us

the direction in which the curve proceeds

at each point.

Thus, it is reasonable to expect that

information about f’(x) will provide us

with information about f(x).

As f’(x) represents the slope of the curve

y = f(x) at the point (x, f(x)), it tells us

the direction in which the curve proceeds

at each point.

Thus, it is reasonable to expect that

information about f’(x) will provide us

with information about f(x).

5.
WHAT DOES f’ SAY ABOUT f ?

To see how the derivative of f can

tell us where a function is increasing

or decreasing, look at the figure.

Increasing functions

and decreasing

functions were

defined in Section 1.1

To see how the derivative of f can

tell us where a function is increasing

or decreasing, look at the figure.

Increasing functions

and decreasing

functions were

defined in Section 1.1

6.
WHAT DOES f’ SAY ABOUT f ?

Between A and B and between C and D,

the tangent lines have positive slope.

So, f’(x) > 0.

Between A and B and between C and D,

the tangent lines have positive slope.

So, f’(x) > 0.

7.
WHAT DOES f’ SAY ABOUT f ?

Between B and C, the tangent lines

have negative slope.

So, f’(x) < 0.

Between B and C, the tangent lines

have negative slope.

So, f’(x) < 0.

8.
WHAT DOES f’ SAY ABOUT f ?

Thus, it appears that f increases when

f’(x) is positive and decreases when f’(x)

is negative.

To prove that this

is always the case,

we use the Mean

Value Theorem.

Thus, it appears that f increases when

f’(x) is positive and decreases when f’(x)

is negative.

To prove that this

is always the case,

we use the Mean

Value Theorem.

9.
INCREASING/DECREASING TEST (I/D TEST)

a.If f’(x) > 0 on an interval, then f is

increasing on that interval.

b.If f’(x) < 0 on an interval, then f is

decreasing on that interval.

a.If f’(x) > 0 on an interval, then f is

increasing on that interval.

b.If f’(x) < 0 on an interval, then f is

decreasing on that interval.

10.
I/D TEST Proof a

Let x1 and x2 be any two numbers in

the interval with x1 < x2.

According to the definition of an increasing

function, we have to show that f(x1) < f(x2).

Let x1 and x2 be any two numbers in

the interval with x1 < x2.

According to the definition of an increasing

function, we have to show that f(x1) < f(x2).

11.
I/D TEST Proof a (Equation 1)

Since we are given that f’(x) > 0, we know that

f is differentiable on [x1, x2].

So, by the Mean Value Theorem, there is

a number c between x1 and x2 such that:

f(x2) – f(x1) = f’(c)(x2 – x1)

Since we are given that f’(x) > 0, we know that

f is differentiable on [x1, x2].

So, by the Mean Value Theorem, there is

a number c between x1 and x2 such that:

f(x2) – f(x1) = f’(c)(x2 – x1)

12.
I/D TEST Proof a and b

Now, f’(c) > 0 by assumption and x2 – x1 > 0

because x1 < x2 .

Thus, the right side of Equation 1 is positive.

So, f(x2) – f(x1) > 0 or f(x1) < f(x2)

This shows that f is increasing.

Part (b) is proved similarly.

Now, f’(c) > 0 by assumption and x2 – x1 > 0

because x1 < x2 .

Thus, the right side of Equation 1 is positive.

So, f(x2) – f(x1) > 0 or f(x1) < f(x2)

This shows that f is increasing.

Part (b) is proved similarly.

13.
I/D TEST Example 1

Find where the function

f(x) = 3x4 – 4x3 – 12x2 + 5

is increasing and where it is decreasing.

Find where the function

f(x) = 3x4 – 4x3 – 12x2 + 5

is increasing and where it is decreasing.

14.
I/D TEST Example 1

f’(x) = 12x3 - 12x2 - 24x = 12x(x – 2)(x + 1)

To use the ID Test, we have to know where

f’(x) > 0 and where f’(x) < 0.

This depends on the signs of the three factors

of f’(x)—namely, 12x, x – 2, and x + 1.

f’(x) = 12x3 - 12x2 - 24x = 12x(x – 2)(x + 1)

To use the ID Test, we have to know where

f’(x) > 0 and where f’(x) < 0.

This depends on the signs of the three factors

of f’(x)—namely, 12x, x – 2, and x + 1.

15.
I/D TEST Example 1

We divide the real line into intervals

whose endpoints are the critical numbers

-1, 0, and 2 and arrange our work in

a chart.

We divide the real line into intervals

whose endpoints are the critical numbers

-1, 0, and 2 and arrange our work in

a chart.

16.
I/D TEST Example 1

A plus sign indicates the given expression

is positive.

A minus sign indicates it is negative.

The last column gives the conclusion

based on the I/D Test.

A plus sign indicates the given expression

is positive.

A minus sign indicates it is negative.

The last column gives the conclusion

based on the I/D Test.

17.
I/D TEST Example 1

For instance, f’(x) < 0 for 0 < x < 2.

So, f is decreasing on (0, 2).

It would also be true to say that f is decreasing

on the closed interval.

For instance, f’(x) < 0 for 0 < x < 2.

So, f is decreasing on (0, 2).

It would also be true to say that f is decreasing

on the closed interval.

18.
I/D TEST Example 1

The graph of f

the information

in the chart.

The graph of f

the information

in the chart.

19.
WHAT DOES f’ SAY ABOUT f ?

Recall from Section 4.1 that, if f has a local

maximum or minimum at c, then c must be

a critical number of f (by Fermat’s Theorem).

However, not every critical number gives rise

to a maximum or a minimum.

So, we need a test that will tell us whether or not f

has a local maximum or minimum at a critical number.

Recall from Section 4.1 that, if f has a local

maximum or minimum at c, then c must be

a critical number of f (by Fermat’s Theorem).

However, not every critical number gives rise

to a maximum or a minimum.

So, we need a test that will tell us whether or not f

has a local maximum or minimum at a critical number.

20.
WHAT DOES f’ SAY ABOUT f ?

You can see from the figure that f(0) = 5 is

a local maximum value of f because f

increases on (-1, 0) and decreases on (0, 2).

In terms of derivatives,

f’(x) > 0 for -1 < x < 0

and f’(x) < 0 for 0 < x < 2.

You can see from the figure that f(0) = 5 is

a local maximum value of f because f

increases on (-1, 0) and decreases on (0, 2).

In terms of derivatives,

f’(x) > 0 for -1 < x < 0

and f’(x) < 0 for 0 < x < 2.

21.
WHAT DOES f’ SAY ABOUT f ?

In other words, the sign of f’(x)

changes from positive to negative at 0.

This observation is the basis of the following test.

In other words, the sign of f’(x)

changes from positive to negative at 0.

This observation is the basis of the following test.

22.
FIRST DERIVATIVE TEST

Suppose that c is a critical number of

a continuous function f.

a. If f’ changes from

positive to negative

at c, then f has

a local maximum at c.

Suppose that c is a critical number of

a continuous function f.

a. If f’ changes from

positive to negative

at c, then f has

a local maximum at c.

23.
FIRST DERIVATIVE TEST

b. If f’ changes from negative to

positive at c, then f has a local minimum

at c.

b. If f’ changes from negative to

positive at c, then f has a local minimum

at c.

24.
FIRST DERIVATIVE TEST

c. If f’ does not change sign at c—for example,

if f’ is positive on both sides of c or negative

on both sides—then f has no local maximum

or minimum at c.

c. If f’ does not change sign at c—for example,

if f’ is positive on both sides of c or negative

on both sides—then f has no local maximum

or minimum at c.

25.
FIRST DERIVATIVE TEST

The First Derivative Test is a consequence

of the I/D Test.

For instance, in (a), since the sign of f’(x) changes

from positive to negative at c, f is increasing to the left

of c and decreasing to the right of c.

It follows that f has

a local maximum at c.

The First Derivative Test is a consequence

of the I/D Test.

For instance, in (a), since the sign of f’(x) changes

from positive to negative at c, f is increasing to the left

of c and decreasing to the right of c.

It follows that f has

a local maximum at c.

26.
FIRST DERIVATIVE TEST

It is easy to remember the test by

visualizing diagrams.

It is easy to remember the test by

visualizing diagrams.

27.
WHAT DOES f’ SAY ABOUT f ? Example 2

Find the local minimum and

maximum values of the function f

in Example 1.

Find the local minimum and

maximum values of the function f

in Example 1.

28.
WHAT DOES f’ SAY ABOUT f ? Example 2

From the chart in the solution to Example 1,

we see that f’(x) changes from negative to

positive at -1.

So, f(-1) = 0 is a local minimum value by

the First Derivative Test.

From the chart in the solution to Example 1,

we see that f’(x) changes from negative to

positive at -1.

So, f(-1) = 0 is a local minimum value by

the First Derivative Test.

29.
WHAT DOES f’ SAY ABOUT f ? Example 2

Similarly, f’ changes from negative to

positive at 2.

So, f(2) = -27 is also a local minimum value.

Similarly, f’ changes from negative to

positive at 2.

So, f(2) = -27 is also a local minimum value.

30.
WHAT DOES f’ SAY ABOUT f ? Example 2

As previously noted, f(0) = 5 is

a local maximum value because f’(x)

changes from positive to negative at 0.

As previously noted, f(0) = 5 is

a local maximum value because f’(x)

changes from positive to negative at 0.

31.
WHAT DOES f’ SAY ABOUT f ? Example 3

Find the local maximum and minimum

values of the function

g(x) = x + 2 sin x 0 ≤ x ≤ 2π

Find the local maximum and minimum

values of the function

g(x) = x + 2 sin x 0 ≤ x ≤ 2π

32.
WHAT DOES f’ SAY ABOUT f ? Example 3

To find the critical numbers of g,

we differentiate:

g’(x) = 1 + 2 cos x

So, g’(x) = 0 when cos x = - ½.

The solutions of this equation are 2π/3 and 4π/3.

To find the critical numbers of g,

we differentiate:

g’(x) = 1 + 2 cos x

So, g’(x) = 0 when cos x = - ½.

The solutions of this equation are 2π/3 and 4π/3.

33.
WHAT DOES f’ SAY ABOUT f ? Example 3

As g is differentiable everywhere,

the only critical numbers are 2π/3 and 4π/3.

So, we analyze g in the following table.

As g is differentiable everywhere,

the only critical numbers are 2π/3 and 4π/3.

So, we analyze g in the following table.

34.
WHAT DOES f’ SAY ABOUT f ? Example 3

As g’(x) changes from positive to negative

at 2π/3, the First Derivative Test tells us

that there is a local maximum at 2π/3.

The local maximum value is:

2π 2π 2π ⎛ 3 ⎞ 2π

g (2π / 3) = + 2sin = + 2 ⎜⎜ ⎟⎟ = + 3

3 3 3 ⎝ 2 ⎠ 3

≈3.83

As g’(x) changes from positive to negative

at 2π/3, the First Derivative Test tells us

that there is a local maximum at 2π/3.

The local maximum value is:

2π 2π 2π ⎛ 3 ⎞ 2π

g (2π / 3) = + 2sin = + 2 ⎜⎜ ⎟⎟ = + 3

3 3 3 ⎝ 2 ⎠ 3

≈3.83

35.
WHAT DOES f’ SAY ABOUT f ? Example 3

Likewise, g’(x) changes from negative to

positive at 4π/3.

So, a local minimum value is:

4π 4π 4π ⎛ 3 ⎞ 4π

g (4π / 3) = + 2sin = + 2⎜

⎜ − ⎟⎟ = − 3

3 3 3 ⎝ 2 ⎠ 3

≈2.46

Likewise, g’(x) changes from negative to

positive at 4π/3.

So, a local minimum value is:

4π 4π 4π ⎛ 3 ⎞ 4π

g (4π / 3) = + 2sin = + 2⎜

⎜ − ⎟⎟ = − 3

3 3 3 ⎝ 2 ⎠ 3

≈2.46

36.
WHAT DOES f’ SAY ABOUT f ? Example 3

The graph of g supports our

The graph of g supports our

37.
WHAT DOES f’’ SAY ABOUT f ?

The figure shows the graphs of

two increasing functions on (a, b).

The figure shows the graphs of

two increasing functions on (a, b).

38.
WHAT DOES f’’ SAY ABOUT f ?

Both graphs join point A to point B, but

they look different because they bend in

different directions.

How can we distinguish between these two types

of behavior?

Both graphs join point A to point B, but

they look different because they bend in

different directions.

How can we distinguish between these two types

of behavior?

39.
WHAT DOES f’’ SAY ABOUT f ?

Here, tangents to these curves have

been drawn at several points.

Here, tangents to these curves have

been drawn at several points.

40.
CONCAVE UPWARD

In the first figure, the curve lies

above the tangents and f is called

concave upward on (a, b).

In the first figure, the curve lies

above the tangents and f is called

concave upward on (a, b).

41.
CONCAVE DOWNWARD

In the second figure, the curve lies

below the tangents and g is called

concave downward on (a, b).

In the second figure, the curve lies

below the tangents and g is called

concave downward on (a, b).

42.
If the graph of f lies above all of

its tangents on an interval I, it is called

concave upward on I.

If the graph of f lies below all of its tangents

on I, it is called concave downward on I.

its tangents on an interval I, it is called

concave upward on I.

If the graph of f lies below all of its tangents

on I, it is called concave downward on I.

43.
The figure shows the graph of a function that

is concave upward (CU) on the intervals (b, c),

(d, e), and (e, p) and concave downward (CD)

on the intervals (a, b), (c, d), and (p, q).

is concave upward (CU) on the intervals (b, c),

(d, e), and (e, p) and concave downward (CD)

on the intervals (a, b), (c, d), and (p, q).

44.
Let’s see how the second derivative

helps determine the intervals of

helps determine the intervals of

45.
From this figure, you can see that, going from

left to right, the slope of the tangent increases.

This means that the derivative f’ is an increasing

function and therefore its derivative f” is positive.

left to right, the slope of the tangent increases.

This means that the derivative f’ is an increasing

function and therefore its derivative f” is positive.

46.
Likewise, in this figure, the slope of

the tangent decreases from left to right.

So, f’ decreases and therefore f’’ is negative.

This reasoning can be

reversed and suggests

that the following

theorem is true.

the tangent decreases from left to right.

So, f’ decreases and therefore f’’ is negative.

This reasoning can be

reversed and suggests

that the following

theorem is true.

47.
CONCAVITY TEST

a.If f’’(x) > 0 for all x in I, then the graph of f

is concave upward on I.

b.If f’’(x) < 0 for all x in I, then the graph of f

is concave downward on I.

a.If f’’(x) > 0 for all x in I, then the graph of f

is concave upward on I.

b.If f’’(x) < 0 for all x in I, then the graph of f

is concave downward on I.

48.
CONCAVITY Example 4

The figure shows a population graph for

Cyprian honeybees raised in an apiary.

How does the rate of population increase change

over time?

When is this rate highest?

Over what

intervals is P

concave upward or

concave downward?

The figure shows a population graph for

Cyprian honeybees raised in an apiary.

How does the rate of population increase change

over time?

When is this rate highest?

Over what

intervals is P

concave upward or

concave downward?

49.
CONCAVITY Example 4

By looking at the slope of the curve as t

increases, we see that the rate of increase

of the population is initially very small.

Then, it gets larger

until it reaches

a maximum at

about t = 12 weeks,

and decreases as

the population

begins to level off.

By looking at the slope of the curve as t

increases, we see that the rate of increase

of the population is initially very small.

Then, it gets larger

until it reaches

a maximum at

about t = 12 weeks,

and decreases as

the population

begins to level off.

50.
CONCAVITY Example 4

As the population approaches its

maximum value of about 75,000 (called

the carrying capacity), the rate of increase,

P’(t), approaches 0.

The curve

appears to be

concave upward

on (0, 12) and

concave downward

on (12, 18).

As the population approaches its

maximum value of about 75,000 (called

the carrying capacity), the rate of increase,

P’(t), approaches 0.

The curve

appears to be

concave upward

on (0, 12) and

concave downward

on (12, 18).

51.
INFLECTION POINT

In the example, the curve changed from

concave upward to concave downward at

approximately the point (12, 38,000).

This point is called

an inflection point

of the curve.

In the example, the curve changed from

concave upward to concave downward at

approximately the point (12, 38,000).

This point is called

an inflection point

of the curve.

52.
INFLECTION POINT

The significance of this point is that

the rate of population increase has its

maximum value there.

In general, an inflection point is a point where

a curve changes its direction of concavity.

The significance of this point is that

the rate of population increase has its

maximum value there.

In general, an inflection point is a point where

a curve changes its direction of concavity.

53.
INFLECTION POINT—DEFINITION

A point P on a curve y = f(x) is called

an inflection point if f is continuous there

and the curve changes from concave upward

to concave downward or from concave

downward to concave upward at P.

A point P on a curve y = f(x) is called

an inflection point if f is continuous there

and the curve changes from concave upward

to concave downward or from concave

downward to concave upward at P.

54.
INFLECTION POINT

For instance, here, B, C, D, and P are

the points of inflection.

Notice that, if a curve has a tangent at a point of

inflection, then the curve crosses its tangent there.

For instance, here, B, C, D, and P are

the points of inflection.

Notice that, if a curve has a tangent at a point of

inflection, then the curve crosses its tangent there.

55.
INFLECTION POINT

In view of the Concavity Test, there is

a point of inflection at any point where

the second derivative changes sign.

In view of the Concavity Test, there is

a point of inflection at any point where

the second derivative changes sign.

56.
WHAT DOES f’’ SAY ABOUT f ? Example 5

Sketch a possible graph of a function f

that satisfies the following conditions:

(i) f’(x) > 0 on (-∞ , 1), f’(x) < 0 on (1, ∞)

(ii) f’’(x) > 0 on (-∞, -2) and (2, ∞), f’’(x) < 0 on (-2, 2)

(iii) ,lim f ( x) = −2 lim f ( x) = 0

x → −∞ x→ ∞

Sketch a possible graph of a function f

that satisfies the following conditions:

(i) f’(x) > 0 on (-∞ , 1), f’(x) < 0 on (1, ∞)

(ii) f’’(x) > 0 on (-∞, -2) and (2, ∞), f’’(x) < 0 on (-2, 2)

(iii) ,lim f ( x) = −2 lim f ( x) = 0

x → −∞ x→ ∞

57.
WHAT DOES f’’ SAY ABOUT f ? E. g. 5—Condition i

The first condition tells us that f

is increasing on (-∞ , 1) and decreasing

on (1, ∞).

The first condition tells us that f

is increasing on (-∞ , 1) and decreasing

on (1, ∞).

58.
WHAT DOES f’’ SAY ABOUT f ? E. g. 5—Condition ii

The second condition says that f

is concave upward on (-∞, -2) and (2, ∞),

and concave downward on (-2, 2).

The second condition says that f

is concave upward on (-∞, -2) and (2, ∞),

and concave downward on (-2, 2).

59.
WHAT DOES f’’ SAY ABOUT f ? E. g. 5—Condition iii

From the third condition, we know

that the graph of f has two horizontal

y = -2

y=0

From the third condition, we know

that the graph of f has two horizontal

y = -2

y=0

60.
WHAT DOES f’’ SAY ABOUT f ? E. g. 5—Condition iii

We first draw the horizontal asymptote

y = -2 as a dashed line.

We then draw the graph of f approaching this asymptote

at the far left—increasing to its maximum point at x = 1

and decreasing toward the x-axis at the far right.

We first draw the horizontal asymptote

y = -2 as a dashed line.

We then draw the graph of f approaching this asymptote

at the far left—increasing to its maximum point at x = 1

and decreasing toward the x-axis at the far right.

61.
WHAT DOES f’’ SAY ABOUT f ? E. g. 5—Condition iii

We also make sure that the graph has

inflection points when x = -2 and 2.

Notice that we made the curve bend upward for x < -2

and x > 2, and bend downward when x is between -2

and 2.

We also make sure that the graph has

inflection points when x = -2 and 2.

Notice that we made the curve bend upward for x < -2

and x > 2, and bend downward when x is between -2

and 2.

62.
WHAT DOES f’’ SAY ABOUT f ?

Another application of the second

derivative is the following test

for maximum and minimum values.

It is a consequence of the Concavity Test.

Another application of the second

derivative is the following test

for maximum and minimum values.

It is a consequence of the Concavity Test.

63.
SECOND DERIVATIVE TEST

Suppose f’’ is continuous near c.

a.If f’(c) = 0 and f’’(c) > 0, then f has

a local minimum at c.

b.If f’(c) = 0 and f’’(c) < 0, then f has

a local maximum at c.

Suppose f’’ is continuous near c.

a.If f’(c) = 0 and f’’(c) > 0, then f has

a local minimum at c.

b.If f’(c) = 0 and f’’(c) < 0, then f has

a local maximum at c.

64.
SECOND DERIVATIVE TEST

For instance, (a) is true because f’’(x) > 0

near c, and so f is concave upward near c.

This means that

the graph of f lies

above its horizontal

tangent at c, and so

f has a local minimum

at c.

For instance, (a) is true because f’’(x) > 0

near c, and so f is concave upward near c.

This means that

the graph of f lies

above its horizontal

tangent at c, and so

f has a local minimum

at c.

65.
WHAT DOES f’’ SAY ABOUT f ? Example 6

Discuss the curve

y = x4 – 4x3

with respect to concavity, points of inflection,

and local maxima and minima.

Use this information to sketch the curve.

Discuss the curve

y = x4 – 4x3

with respect to concavity, points of inflection,

and local maxima and minima.

Use this information to sketch the curve.

66.
WHAT DOES f’’ SAY ABOUT f ? Example 6

If f(x) = x4 – 4x3, then:

f’(x) = 4x3 – 12x2 = 4x2(x – 3)

f’’(x) = 12x2 – 24x = 12x(x – 2)

If f(x) = x4 – 4x3, then:

f’(x) = 4x3 – 12x2 = 4x2(x – 3)

f’’(x) = 12x2 – 24x = 12x(x – 2)

67.
WHAT DOES f’’ SAY ABOUT f ? Example 6

To find the critical numbers, we set f’(x) = 0

and obtain x = 0 and x = 3.

To use the Second Derivative Test,

we evaluate f’’ at these critical numbers:

f’’(0) = 0 f’’(3) = 36 > 0

To find the critical numbers, we set f’(x) = 0

and obtain x = 0 and x = 3.

To use the Second Derivative Test,

we evaluate f’’ at these critical numbers:

f’’(0) = 0 f’’(3) = 36 > 0

68.
WHAT DOES f’’ SAY ABOUT f ? Example 6

As f’(3) = 0 and f’’(3) > 0, f(3) = -27 is

a local minimum.

As f’’(0) = 0, the Second Derivative Test gives

no information about the critical number 0.

As f’(3) = 0 and f’’(3) > 0, f(3) = -27 is

a local minimum.

As f’’(0) = 0, the Second Derivative Test gives

no information about the critical number 0.

69.
WHAT DOES f’’ SAY ABOUT f ? Example 6

However, since f’(x) < 0 for x < 0 and also

for 0 < x < 3, the First Derivative Test tells

us that f does not have a local maximum or

minimum at 0.

In fact, the expression for f’(x) shows that

f decreases to the left of 3 and increases to

the right of 3.

However, since f’(x) < 0 for x < 0 and also

for 0 < x < 3, the First Derivative Test tells

us that f does not have a local maximum or

minimum at 0.

In fact, the expression for f’(x) shows that

f decreases to the left of 3 and increases to

the right of 3.

70.
WHAT DOES f’’ SAY ABOUT f ? Example 6

As f’’(x) = 0 when x = 0 or 2, we divide

the real line into intervals with those numbers

as endpoints and complete the following chart.

As f’’(x) = 0 when x = 0 or 2, we divide

the real line into intervals with those numbers

as endpoints and complete the following chart.

71.
WHAT DOES f’’ SAY ABOUT f ? Example 6

The point (0, 0) is an inflection point—since

the curve changes from concave upward to

concave downward there.

The point (0, 0) is an inflection point—since

the curve changes from concave upward to

concave downward there.

72.
WHAT DOES f’’ SAY ABOUT f ? Example 6

Also, (2, -16) is an inflection point—since

the curve changes from concave downward

to concave upward there.

Also, (2, -16) is an inflection point—since

the curve changes from concave downward

to concave upward there.

73.
WHAT DOES f’’ SAY ABOUT f ? Example 6

Using the local minimum, the intervals

of concavity, and the inflection points,

we sketch the curve.

Using the local minimum, the intervals

of concavity, and the inflection points,

we sketch the curve.

74.
The Second Derivative Test is

inconclusive when f’’(c) = 0.

In other words, at such a point, there might be

a maximum, a minimum, or neither (as in the example).

inconclusive when f’’(c) = 0.

In other words, at such a point, there might be

a maximum, a minimum, or neither (as in the example).

75.
The test also fails when f’’(c) does not exist.

In such cases, the First Derivative Test

must be used.

In fact, even when both tests apply, the First Derivative

Test is often the easier one to use.

In such cases, the First Derivative Test

must be used.

In fact, even when both tests apply, the First Derivative

Test is often the easier one to use.

76.
WHAT DOES f’’ SAY ABOUT f ? Example 7

Sketch the graph of the function

f(x) = x2/3(6 – x)1/3

You can use the differentiation rules to check that

the first two derivatives are:

4 −x −8

f '( x) = 1/ 3 f ''( x) = 4 / 3

x (6 −x) 2 / 3 x (6 −x)5 / 3

As f’(x) = 0 when x = 4 and f’(x) does not exist when

x = 0 or x = 6, the critical numbers are 0, 4, and 6.

Sketch the graph of the function

f(x) = x2/3(6 – x)1/3

You can use the differentiation rules to check that

the first two derivatives are:

4 −x −8

f '( x) = 1/ 3 f ''( x) = 4 / 3

x (6 −x) 2 / 3 x (6 −x)5 / 3

As f’(x) = 0 when x = 4 and f’(x) does not exist when

x = 0 or x = 6, the critical numbers are 0, 4, and 6.

77.
WHAT DOES f’’ SAY ABOUT f ? Example 7

To find the local extreme values, we

use the First Derivative Test.

As f’ changes from negative to positive at 0,

f(0) = 0 is a local minimum.

To find the local extreme values, we

use the First Derivative Test.

As f’ changes from negative to positive at 0,

f(0) = 0 is a local minimum.

78.
WHAT DOES f’’ SAY ABOUT f ? Example 7

Since f’ changes from positive to negative at 4,

f(4) = 25/3 is a local maximum.

The sign of f’ does not change at 6, so there is

no minimum or maximum there.

Since f’ changes from positive to negative at 4,

f(4) = 25/3 is a local maximum.

The sign of f’ does not change at 6, so there is

no minimum or maximum there.

79.
WHAT DOES f’’ SAY ABOUT f ? Example 7

The Second Derivative Test could be

used at 4, but not at 0 or 6—since f’’ does

not exist at either of these numbers.

The Second Derivative Test could be

used at 4, but not at 0 or 6—since f’’ does

not exist at either of these numbers.

80.
WHAT DOES f’’ SAY ABOUT f ? Example 7

Looking at the expression for f’’(x)

and noting that x4/3 ≥ 0 for all x,

we have:

f’’(x) < 0 for x < 0 and for 0 < x < 6

f’’(x) > 0 for x > 6

Looking at the expression for f’’(x)

and noting that x4/3 ≥ 0 for all x,

we have:

f’’(x) < 0 for x < 0 and for 0 < x < 6

f’’(x) > 0 for x > 6

81.
WHAT DOES f’’ SAY ABOUT f ? Example 7

So, f is concave downward on (-∞, 0) and

(0, 6) and concave upward on (6, ∞), and

the only inflection point is (6, 0).

Note that the curve

has vertical tangents

at (0, 0) and (6, 0)

because |f’(x)| → ∞

as x → 0 and as x → 6.

So, f is concave downward on (-∞, 0) and

(0, 6) and concave upward on (6, ∞), and

the only inflection point is (6, 0).

Note that the curve

has vertical tangents

at (0, 0) and (6, 0)

because |f’(x)| → ∞

as x → 0 and as x → 6.

82.
WHAT DOES f’’ SAY ABOUT f ? Example 8

Use the first and second derivatives

of f(x) = e1/x, together with asymptotes,

to sketch its graph.

Notice that the domain of f is {x | x ≠ 0}.

So, we check for vertical asymptotes by

computing the left and right limits as x → 0.

Use the first and second derivatives

of f(x) = e1/x, together with asymptotes,

to sketch its graph.

Notice that the domain of f is {x | x ≠ 0}.

So, we check for vertical asymptotes by

computing the left and right limits as x → 0.

83.
WHAT DOES f’’ SAY ABOUT f ?

As x → 0+, we know that t = 1/x → ∞.

1/ x t

So, lim+ e = lim e = ∞

x→ 0 t→ ∞

This shows that x = 0 is a vertical asymptote.

As x → 0+, we know that t = 1/x → ∞.

1/ x t

So, lim+ e = lim e = ∞

x→ 0 t→ ∞

This shows that x = 0 is a vertical asymptote.

84.
WHAT DOES f’’ SAY ABOUT f ?

As x → 0-, we know that t = 1/x → -∞.

1/ x t

So, lim− e = lim e = 0

t → −∞

x→ 0

As x → 0-, we know that t = 1/x → -∞.

1/ x t

So, lim− e = lim e = 0

t → −∞

x→ 0

85.
WHAT DOES f’’ SAY ABOUT f ?

As x → ±∞, we have 1/x → 0.

1/ x 0

So, lim e = e =1

x → ±∞

This shows that y = 1 is a horizontal asymptote.

As x → ±∞, we have 1/x → 0.

1/ x 0

So, lim e = e =1

x → ±∞

This shows that y = 1 is a horizontal asymptote.

86.
WHAT DOES f’’ SAY ABOUT f ?

Now, let’s compute the derivative.

1/ x

e

The Chain Rule gives: f '( x ) = − 2

x

Since e1/x > 0 and x2 > 0 for all x ≠ 0, we have

f’(x) < 0 for all x ≠ 0.

Thus, f is decreasing on (-∞, 0) and on (0, ∞) .

Now, let’s compute the derivative.

1/ x

e

The Chain Rule gives: f '( x ) = − 2

x

Since e1/x > 0 and x2 > 0 for all x ≠ 0, we have

f’(x) < 0 for all x ≠ 0.

Thus, f is decreasing on (-∞, 0) and on (0, ∞) .

87.
WHAT DOES f’’ SAY ABOUT f ?

There is no critical number.

So, the function has no maximum

or minimum.

There is no critical number.

So, the function has no maximum

or minimum.

88.
WHAT DOES f’’ SAY ABOUT f ?

The second derivative is:

2 1/ x 2 1/ x

x e (−1/ x ) −e (2 x)

f ''( x) = − 4

x

1/ x

e (2 x + 1)

= 4

x

The second derivative is:

2 1/ x 2 1/ x

x e (−1/ x ) −e (2 x)

f ''( x) = − 4

x

1/ x

e (2 x + 1)

= 4

x

89.
WHAT DOES f’’ SAY ABOUT f ?

As e1/x > 0 and x4 > 0, we have:

f’’(x) > 0 when x > -½ (x ≠ 0)

f’’(x) < 0 when x < -½

So, the curve is concave downward on (-∞, -½)

and concave upward on (-½, 0) and on (0, ∞).

The inflection point is (-½, e-2).

As e1/x > 0 and x4 > 0, we have:

f’’(x) > 0 when x > -½ (x ≠ 0)

f’’(x) < 0 when x < -½

So, the curve is concave downward on (-∞, -½)

and concave upward on (-½, 0) and on (0, ∞).

The inflection point is (-½, e-2).

90.
WHAT DOES f’’ SAY ABOUT f ?

To sketch the graph of f, we first draw

the horizontal asymptote y = 1 (as a dashed

line), together with the parts of the curve

near the asymptotes

in a preliminary sketch.

To sketch the graph of f, we first draw

the horizontal asymptote y = 1 (as a dashed

line), together with the parts of the curve

near the asymptotes

in a preliminary sketch.

91.
WHAT DOES f’’ SAY ABOUT f ?

These parts reflect the information concerning

limits and the fact that f is decreasing on both

(-∞, 0) and (0, ∞).

Notice that we have

indicated that f(x) → 0

as x → 0- even though

f(0) does not exist.

These parts reflect the information concerning

limits and the fact that f is decreasing on both

(-∞, 0) and (0, ∞).

Notice that we have

indicated that f(x) → 0

as x → 0- even though

f(0) does not exist.

92.
WHAT DOES f’’ SAY ABOUT f ?

Here, we finish the sketch by

incorporating the information concerning

concavity and the inflection point.

Here, we finish the sketch by

incorporating the information concerning

concavity and the inflection point.

93.
WHAT DOES f’’ SAY ABOUT f ?

Finally, we check our work with

a graphing device.

Finally, we check our work with

a graphing device.