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In this section, we will learn: How to find the maximum and minimum values of a function using derivatives.

1.
4

APPLICATIONS OF DIFFERENTIATION

APPLICATIONS OF DIFFERENTIATION

2.
APPLICATIONS OF DIFFERENTIATION

We have already investigated some

applications of derivatives.

However, now that we know the differentiation

rules, we are in a better position to pursue

the applications of differentiation in greater

We have already investigated some

applications of derivatives.

However, now that we know the differentiation

rules, we are in a better position to pursue

the applications of differentiation in greater

3.
APPLICATIONS OF DIFFERENTIATION

Here, we learn how derivatives affect

the shape of a graph of a function and,

in particular, how they help us locate

maximum and minimum values of functions.

Here, we learn how derivatives affect

the shape of a graph of a function and,

in particular, how they help us locate

maximum and minimum values of functions.

4.
APPLICATIONS OF DIFFERENTIATION

Many practical problems require us to

minimize a cost or maximize an area or

somehow find the best possible outcome

of a situation.

In particular, we will be able to investigate

the optimal shape of a can and to explain

the location of rainbows in the sky.

Many practical problems require us to

minimize a cost or maximize an area or

somehow find the best possible outcome

of a situation.

In particular, we will be able to investigate

the optimal shape of a can and to explain

the location of rainbows in the sky.

5.
APPLICATIONS OF DIFFERENTIATION

4.1

Maximum and

Minimum Values

In this section, we will learn:

How to find the maximum

and minimum values of a function.

4.1

Maximum and

Minimum Values

In this section, we will learn:

How to find the maximum

and minimum values of a function.

6.
OPTIMIZATION PROBLEMS

Some of the most important

applications of differential calculus

are optimization problems.

In these, we are required to find the optimal (best)

way of doing something.

Some of the most important

applications of differential calculus

are optimization problems.

In these, we are required to find the optimal (best)

way of doing something.

7.
Here are some examples of such problems

that we will solve in this chapter.

What is the shape of a can that minimizes

manufacturing costs?

What is the maximum acceleration of a space shuttle?

(This is an important question to the astronauts

who have to withstand the effects of acceleration.)

that we will solve in this chapter.

What is the shape of a can that minimizes

manufacturing costs?

What is the maximum acceleration of a space shuttle?

(This is an important question to the astronauts

who have to withstand the effects of acceleration.)

8.
Here are some more examples.

What is the radius of a contracted windpipe

that expels air most rapidly during a cough?

At what angle should blood vessels branch

so as to minimize the energy expended by

the heart in pumping blood?

What is the radius of a contracted windpipe

that expels air most rapidly during a cough?

At what angle should blood vessels branch

so as to minimize the energy expended by

the heart in pumping blood?

9.
OPTIMIZATION PROBLEMS

These problems can be reduced to

finding the maximum or minimum values

of a function.

Let’s first explain exactly what we mean

by maximum and minimum values.

These problems can be reduced to

finding the maximum or minimum values

of a function.

Let’s first explain exactly what we mean

by maximum and minimum values.

10.
MAXIMUM & MINIMUM VALUES Definition 1

A function f has an absolute maximum

(or global maximum) at c if f(c) ≥ f(x) for

all x in D, where D is the domain of f.

The number f(c) is called the maximum value

of f on D.

A function f has an absolute maximum

(or global maximum) at c if f(c) ≥ f(x) for

all x in D, where D is the domain of f.

The number f(c) is called the maximum value

of f on D.

11.
MAXIMUM & MINIMUM VALUES Definition 1

Similarly, f has an absolute minimum at c

if f(c) ≤ f(x) for all x in D and the number f(c)

is called the minimum value of f on D.

The maximum and minimum values of f

are called the extreme values of f.

Similarly, f has an absolute minimum at c

if f(c) ≤ f(x) for all x in D and the number f(c)

is called the minimum value of f on D.

The maximum and minimum values of f

are called the extreme values of f.

12.
MAXIMUM & MINIMUM VALUES

The figure shows the graph of a function f

with absolute maximum at d and absolute

minimum at a.

Note that (d, f(d)) is

the highest point on

the graph and (a, f(a))

is the lowest point.

The figure shows the graph of a function f

with absolute maximum at d and absolute

minimum at a.

Note that (d, f(d)) is

the highest point on

the graph and (a, f(a))

is the lowest point.

13.
LOCAL MAXIMUM VALUE

If we consider only values of x near b—for

instance, if we restrict our attention to the

interval (a, c)—then f(b) is the largest of those

values of f(x).

It is called a local

maximum value of f.

If we consider only values of x near b—for

instance, if we restrict our attention to the

interval (a, c)—then f(b) is the largest of those

values of f(x).

It is called a local

maximum value of f.

14.
LOCAL MINIMUM VALUE

Likewise, f(c) is called a local minimum value

of f because f(c) ≤ f(x) for x near c—for

instance, in the interval (b, d).

The function f also has

a local minimum at e.

Likewise, f(c) is called a local minimum value

of f because f(c) ≤ f(x) for x near c—for

instance, in the interval (b, d).

The function f also has

a local minimum at e.

15.
MAXIMUM & MINIMUM VALUES Definition 2

In general, we have the following definition.

A function f has a local maximum (or relative

maximum) at c if f(c) ≥ f(x) when x is near c.

This means that f(c) ≥ f(x) for all x in some

open interval containing c.

Similarly, f has a local minimum at c if f(c) ≤ f(x)

when x is near c.

In general, we have the following definition.

A function f has a local maximum (or relative

maximum) at c if f(c) ≥ f(x) when x is near c.

This means that f(c) ≥ f(x) for all x in some

open interval containing c.

Similarly, f has a local minimum at c if f(c) ≤ f(x)

when x is near c.

16.
MAXIMUM & MINIMUM VALUES Example 1

The function f(x) = cos x takes on its (local

and absolute) maximum value of 1 infinitely

many times—since cos 2nπ = 1 for any

integer n and -1 ≤ cos x ≤ 1 for all x.

Likewise, cos (2n + 1)π = -1 is its minimum

value—where n is any integer.

The function f(x) = cos x takes on its (local

and absolute) maximum value of 1 infinitely

many times—since cos 2nπ = 1 for any

integer n and -1 ≤ cos x ≤ 1 for all x.

Likewise, cos (2n + 1)π = -1 is its minimum

value—where n is any integer.

17.
MAXIMUM & MINIMUM VALUES Example 2

If f(x) = x2, then f(x) ≥ f(0) because

x2 ≥ 0 for all x.

Therefore, f(0) = 0 is the absolute (and local)

minimum value of f.

If f(x) = x2, then f(x) ≥ f(0) because

x2 ≥ 0 for all x.

Therefore, f(0) = 0 is the absolute (and local)

minimum value of f.

18.
MAXIMUM & MINIMUM VALUES Example 2

This corresponds to the fact that the origin

is the lowest point on the parabola y = x2.

However, there is no highest point on the parabola.

So, this function has no maximum value.

This corresponds to the fact that the origin

is the lowest point on the parabola y = x2.

However, there is no highest point on the parabola.

So, this function has no maximum value.

19.
MAXIMUM & MINIMUM VALUES Example 3

From the graph of the function f(x) = x3,

we see that this function has neither

an absolute maximum value nor an absolute

minimum value.

In fact, it has no local

extreme values either.

From the graph of the function f(x) = x3,

we see that this function has neither

an absolute maximum value nor an absolute

minimum value.

In fact, it has no local

extreme values either.

20.
MAXIMUM & MINIMUM VALUES Example 4

The graph of the function

f(x) = 3x4 – 16x3 + 18x2 -1 ≤ x ≤ 4

is shown here.

The graph of the function

f(x) = 3x4 – 16x3 + 18x2 -1 ≤ x ≤ 4

is shown here.

21.
MAXIMUM & MINIMUM VALUES Example 4

You can see that f(1) = 5 is a local

maximum, whereas the absolute maximum

is f(-1) = 37.

This absolute maximum

is not a local maximum

because it occurs at

an endpoint.

You can see that f(1) = 5 is a local

maximum, whereas the absolute maximum

is f(-1) = 37.

This absolute maximum

is not a local maximum

because it occurs at

an endpoint.

22.
MAXIMUM & MINIMUM VALUES Example 4

Also, f(0) = 0 is a local minimum and

f(3) = -27 is both a local and an absolute

Note that f has

neither a local nor

an absolute maximum

at x = 4.

Also, f(0) = 0 is a local minimum and

f(3) = -27 is both a local and an absolute

Note that f has

neither a local nor

an absolute maximum

at x = 4.

23.
MAXIMUM & MINIMUM VALUES

We have seen that some functions have

extreme values, whereas others do not.

The following theorem gives conditions

under which a function is guaranteed to

possess extreme values.

We have seen that some functions have

extreme values, whereas others do not.

The following theorem gives conditions

under which a function is guaranteed to

possess extreme values.

24.
EXTREME VALUE THEOREM Theorem 3

If f is continuous on a closed interval [a, b],

then f attains an absolute maximum value f(c)

and an absolute minimum value f(d) at some

numbers c and d in [a, b].

If f is continuous on a closed interval [a, b],

then f attains an absolute maximum value f(c)

and an absolute minimum value f(d) at some

numbers c and d in [a, b].

25.
EXTREME VALUE THEOREM

The theorem is illustrated

in the figures.

Note that an extreme value

can be taken on more than once.

The theorem is illustrated

in the figures.

Note that an extreme value

can be taken on more than once.

26.
EXTREME VALUE THEOREM

Although the theorem is intuitively

very plausible, it is difficult to prove

and so we omit the proof.

Although the theorem is intuitively

very plausible, it is difficult to prove

and so we omit the proof.

27.
EXTREME VALUE THEOREM

The figures show that a function need not

possess extreme values if either hypothesis

(continuity or closed interval) is omitted from

the theorem.

The figures show that a function need not

possess extreme values if either hypothesis

(continuity or closed interval) is omitted from

the theorem.

28.
EXTREME VALUE THEOREM

The function f whose graph is shown is

defined on the closed interval [0, 2] but has

no maximum value.

Notice that the range of f

is [0, 3).

The function takes on values

arbitrarily close to 3, but never

actually attains the value 3.

The function f whose graph is shown is

defined on the closed interval [0, 2] but has

no maximum value.

Notice that the range of f

is [0, 3).

The function takes on values

arbitrarily close to 3, but never

actually attains the value 3.

29.
EXTREME VALUE THEOREM

This does not contradict the theorem

because f is not continuous.

Nonetheless, a discontinuous

function could have maximum

and minimum values.

This does not contradict the theorem

because f is not continuous.

Nonetheless, a discontinuous

function could have maximum

and minimum values.

30.
EXTREME VALUE THEOREM

The function g shown here is continuous

on the open interval (0, 2) but has neither

a maximum nor a minimum value.

The range of g is (1, ∞).

The function takes on

arbitrarily large values.

This does not contradict

the theorem because

the interval (0, 2) is not closed.

The function g shown here is continuous

on the open interval (0, 2) but has neither

a maximum nor a minimum value.

The range of g is (1, ∞).

The function takes on

arbitrarily large values.

This does not contradict

the theorem because

the interval (0, 2) is not closed.

31.
EXTREME VALUE THEOREM

The theorem says that a continuous function

on a closed interval has a maximum value

and a minimum value.

However, it does not tell us how to find these

extreme values.

We start by looking for local extreme values.

The theorem says that a continuous function

on a closed interval has a maximum value

and a minimum value.

However, it does not tell us how to find these

extreme values.

We start by looking for local extreme values.

32.
LOCAL EXTREME VALUES

The figure shows the graph of a function f

with a local maximum at c and a local

minimum at d.

The figure shows the graph of a function f

with a local maximum at c and a local

minimum at d.

33.
LOCAL EXTREME VALUES

It appears that, at the maximum and

minimum points, the tangent lines are

horizontal and therefore each has slope 0.

It appears that, at the maximum and

minimum points, the tangent lines are

horizontal and therefore each has slope 0.

34.
LOCAL EXTREME VALUES

We know that the derivative is the slope

of the tangent line.

So, it appears that f ’(c) = 0 and f ’(d) = 0.

We know that the derivative is the slope

of the tangent line.

So, it appears that f ’(c) = 0 and f ’(d) = 0.

35.
LOCAL EXTREME VALUES

The following theorem says that

this is always true for differentiable

The following theorem says that

this is always true for differentiable

36.
FERMAT’S THEOREM Theorem 4

If f has a local maximum or

minimum at c, and if f ’(c) exists, then

f ’(c) = 0.

If f has a local maximum or

minimum at c, and if f ’(c) exists, then

f ’(c) = 0.

37.
FERMAT’S THEOREM Proof

Suppose, for the sake of definiteness, that f

has a local maximum at c.

Then, according to Definition 2, f(c) ≥ f(x)

if x is sufficiently close to c.

Suppose, for the sake of definiteness, that f

has a local maximum at c.

Then, according to Definition 2, f(c) ≥ f(x)

if x is sufficiently close to c.

38.
FERMAT’S THEOREM Proof (Equation 5)

This implies that, if h is sufficiently close to 0,

with h being positive or negative, then

f(c) ≥ f(c + h)

and therefore

f(c + h) – f(c) ≤ 0

This implies that, if h is sufficiently close to 0,

with h being positive or negative, then

f(c) ≥ f(c + h)

and therefore

f(c + h) – f(c) ≤ 0

39.
FERMAT’S THEOREM Proof

We can divide both sides of an inequality

by a positive number.

Thus, if h > 0 and h is sufficiently small,

we have:

f (c h ) f (c )

0

h

We can divide both sides of an inequality

by a positive number.

Thus, if h > 0 and h is sufficiently small,

we have:

f (c h ) f (c )

0

h

40.
FERMAT’S THEOREM Proof

Taking the right-hand limit of both sides

of this inequality (using Theorem 2 in

Section 2.3), we get:

f (c h ) f ( c )

lim lim 0 0

h 0 h h 0

Taking the right-hand limit of both sides

of this inequality (using Theorem 2 in

Section 2.3), we get:

f (c h ) f ( c )

lim lim 0 0

h 0 h h 0

41.
FERMAT’S THEOREM Proof

However, since f ’(c) exists,

we have: f ( c h ) f (c )

f '(c) lim

h 0 h

f (c h ) f (c )

lim

h 0 h

So, we have shown that f ’(c) ≤ 0.

However, since f ’(c) exists,

we have: f ( c h ) f (c )

f '(c) lim

h 0 h

f (c h ) f (c )

lim

h 0 h

So, we have shown that f ’(c) ≤ 0.

42.
FERMAT’S THEOREM Proof

If h < 0, then the direction of the inequality

in Equation 5 is reversed when we divide by

f (c h ) f (c )

0 h0

h

If h < 0, then the direction of the inequality

in Equation 5 is reversed when we divide by

f (c h ) f (c )

0 h0

h

43.
FERMAT’S THEOREM Proof

So, taking the left-hand limit,

we have: f ( c h ) f (c )

f '(c) lim

h 0 h

f (c h ) f (c )

lim 0

h 0 h

So, taking the left-hand limit,

we have: f ( c h ) f (c )

f '(c) lim

h 0 h

f (c h ) f (c )

lim 0

h 0 h

44.
FERMAT’S THEOREM Proof

We have shown that f ’(c) ≥ 0 and also

that f ’(c) ≤ 0.

Since both these inequalities must be true,

the only possibility is that f ’(c) = 0.

We have shown that f ’(c) ≥ 0 and also

that f ’(c) ≤ 0.

Since both these inequalities must be true,

the only possibility is that f ’(c) = 0.

45.
FERMAT’S THEOREM Proof

We have proved the theorem for

the case of a local maximum.

The case of a local minimum can be proved

in a similar manner.

Alternatively, we could use Exercise 76

to deduce it from the case we have just proved.

We have proved the theorem for

the case of a local maximum.

The case of a local minimum can be proved

in a similar manner.

Alternatively, we could use Exercise 76

to deduce it from the case we have just proved.

46.
FERMAT’S THEOREM

The following examples caution us

against reading too much into the theorem.

We can’t expect to locate extreme values

simply by setting f ’(x) = 0 and solving for x.

The following examples caution us

against reading too much into the theorem.

We can’t expect to locate extreme values

simply by setting f ’(x) = 0 and solving for x.

47.
FERMAT’S THEOREM Example 5

If f(x) = x3, then f ’(x) = 3x2, so f ’(0) = 0.

However, f has no maximum or minimum at 0—as you

can see from the graph.

Alternatively, observe

that x3 > 0 for x > 0

but x3 < 0 for x < 0.

If f(x) = x3, then f ’(x) = 3x2, so f ’(0) = 0.

However, f has no maximum or minimum at 0—as you

can see from the graph.

Alternatively, observe

that x3 > 0 for x > 0

but x3 < 0 for x < 0.

48.
FERMAT’S THEOREM Example 5

The fact that f ’(0) = 0 simply means that

the curve y = x3 has a horizontal tangent

at (0, 0).

Instead of having

a maximum or minimum

at (0, 0), the curve crosses

its horizontal tangent there.

The fact that f ’(0) = 0 simply means that

the curve y = x3 has a horizontal tangent

at (0, 0).

Instead of having

a maximum or minimum

at (0, 0), the curve crosses

its horizontal tangent there.

49.
FERMAT’S THEOREM Example 6

The function f(x) = |x| has its (local and

absolute) minimum value at 0.

However, that value can’t be found by setting f ’(x) = 0.

This is because—as

shown in Example 5

in Section 2.8—f ’(0)

does not exist.

The function f(x) = |x| has its (local and

absolute) minimum value at 0.

However, that value can’t be found by setting f ’(x) = 0.

This is because—as

shown in Example 5

in Section 2.8—f ’(0)

does not exist.

50.
Examples 5 and 6 show that we must

be careful when using the theorem.

Example 5 demonstrates that, even when f ’(c) = 0,

there need not be a maximum or minimum at c.

In other words, the converse of the theorem

is false in general.

Furthermore, there may be an extreme value

even when f ’(c) does not exist (as in Example 6).

be careful when using the theorem.

Example 5 demonstrates that, even when f ’(c) = 0,

there need not be a maximum or minimum at c.

In other words, the converse of the theorem

is false in general.

Furthermore, there may be an extreme value

even when f ’(c) does not exist (as in Example 6).

51.
FERMAT’S THEOREM

The theorem does suggest that we should

at least start looking for extreme values of f

at the numbers c where either:

f ’(c) = 0

f ’(c) does not exist

The theorem does suggest that we should

at least start looking for extreme values of f

at the numbers c where either:

f ’(c) = 0

f ’(c) does not exist

52.
FERMAT’S THEOREM

Such numbers are given a

special name—critical numbers.

Such numbers are given a

special name—critical numbers.

53.
CRITICAL NUMBERS Definition 6

A critical number of a function f is

a number c in the domain of f such that

either f ’(c) = 0 or f ’(c) does not exist.

A critical number of a function f is

a number c in the domain of f such that

either f ’(c) = 0 or f ’(c) does not exist.

54.
CRITICAL NUMBERS Example 7

Find the critical numbers of

f(x) = x3/5(4 - x).

The Product Rule gives:

3/ 5 2/5

f '( x) x ( 1) (4 x)( x

3

5 )

3/ 5 3(4 x)

x 2/5

5x

5 x 3(4 x) 12 8 x

2/5

2/5

5x 5x

Find the critical numbers of

f(x) = x3/5(4 - x).

The Product Rule gives:

3/ 5 2/5

f '( x) x ( 1) (4 x)( x

3

5 )

3/ 5 3(4 x)

x 2/5

5x

5 x 3(4 x) 12 8 x

2/5

2/5

5x 5x

55.
CRITICAL NUMBERS Example 7

The same result could be obtained by

first writing f(x) = 4x3/5 – x8/5.

Therefore, f ’(x) = 0 if 12 – 8x = 0.

That is, x = 32 , and f ’(x) does not exist

when x = 0.

Thus, the critical numbers are 3

2

and 0.

The same result could be obtained by

first writing f(x) = 4x3/5 – x8/5.

Therefore, f ’(x) = 0 if 12 – 8x = 0.

That is, x = 32 , and f ’(x) does not exist

when x = 0.

Thus, the critical numbers are 3

2

and 0.

56.
CRITICAL NUMBERS

In terms of critical numbers, Fermat’s

Theorem can be rephrased as follows

(compare Definition 6 with Theorem 4).

In terms of critical numbers, Fermat’s

Theorem can be rephrased as follows

(compare Definition 6 with Theorem 4).

57.
CRITICAL NUMBERS Theorem 7

If f has a local maximum or

minimum at c, then c is a critical

number of f.

If f has a local maximum or

minimum at c, then c is a critical

number of f.

58.
CLOSED INTERVALS

To find an absolute maximum or minimum

of a continuous function on a closed interval,

we note that either:

It is local (in which case, it occurs at

a critical number by Theorem 7).

It occurs at an endpoint of the interval.

To find an absolute maximum or minimum

of a continuous function on a closed interval,

we note that either:

It is local (in which case, it occurs at

a critical number by Theorem 7).

It occurs at an endpoint of the interval.

59.
CLOSED INTERVALS

Therefore, the following

three-step procedure always

Therefore, the following

three-step procedure always

60.
CLOSED INTERVAL METHOD

To find the absolute maximum and minimum

values of a continuous function f on a closed

interval [a, b]:

1. Find the values of f at the critical numbers of f

in (a, b).

2. Find the values of f at the endpoints of the interval.

3. The largest value from 1 and 2 is the absolute

maximum value. The smallest is the absolute minimum

value.

To find the absolute maximum and minimum

values of a continuous function f on a closed

interval [a, b]:

1. Find the values of f at the critical numbers of f

in (a, b).

2. Find the values of f at the endpoints of the interval.

3. The largest value from 1 and 2 is the absolute

maximum value. The smallest is the absolute minimum

value.

61.
CLOSED INTERVAL METHOD Example 8

Find the absolute maximum

and minimum values of the function

f(x) = x3 – 3x2 + 1 -½ ≤ x ≤ 4

Find the absolute maximum

and minimum values of the function

f(x) = x3 – 3x2 + 1 -½ ≤ x ≤ 4

62.
CLOSED INTERVAL METHOD Example 8

As f is continuous on [-½, 4], we

can use the Closed Interval Method:

f(x) = x3 – 3x2 + 1

f ’(x) = 3x2 – 6x = 3x(x – 2)

As f is continuous on [-½, 4], we

can use the Closed Interval Method:

f(x) = x3 – 3x2 + 1

f ’(x) = 3x2 – 6x = 3x(x – 2)

63.
CLOSED INTERVAL METHOD Example 8

As f ’(x) exists for all x, the only critical

numbers of f occur when f ’(x) = 0, that is,

x = 0 or x = 2.

Notice that each of these numbers lies in

the interval (-½, 4).

As f ’(x) exists for all x, the only critical

numbers of f occur when f ’(x) = 0, that is,

x = 0 or x = 2.

Notice that each of these numbers lies in

the interval (-½, 4).

64.
CLOSED INTERVAL METHOD Example 8

The values of f at these critical numbers

are: f(0) = 1 f(2) = -3

The values of f at the endpoints of the interval

are: f(-½) = 1/8 f(4) = 17

Comparing these four numbers, we see that

the absolute maximum value is f(4) = 17 and

the absolute minimum value is f(2) = -3.

The values of f at these critical numbers

are: f(0) = 1 f(2) = -3

The values of f at the endpoints of the interval

are: f(-½) = 1/8 f(4) = 17

Comparing these four numbers, we see that

the absolute maximum value is f(4) = 17 and

the absolute minimum value is f(2) = -3.

65.
CLOSED INTERVAL METHOD Example 8

Note that the absolute maximum occurs

at an endpoint, whereas the absolute

minimum occurs at a critical number.

Note that the absolute maximum occurs

at an endpoint, whereas the absolute

minimum occurs at a critical number.

66.
CLOSED INTERVAL METHOD Example 8

The graph of f is sketched here.

The graph of f is sketched here.

67.
EXACT VALUES

If you have a graphing calculator or

a computer with graphing software, it is

possible to estimate maximum and minimum

values very easily.

However, as the next example shows,

calculus is needed to find the exact values.

If you have a graphing calculator or

a computer with graphing software, it is

possible to estimate maximum and minimum

values very easily.

However, as the next example shows,

calculus is needed to find the exact values.

68.
EXACT VALUES Example 9

a.Use a graphing device to estimate

the absolute minimum and maximum values

of the function f(x) = x – 2 sin x, 0 ≤ x ≤ 2π.

b.Use calculus to find the exact minimum

and maximum values.

a.Use a graphing device to estimate

the absolute minimum and maximum values

of the function f(x) = x – 2 sin x, 0 ≤ x ≤ 2π.

b.Use calculus to find the exact minimum

and maximum values.

69.
EXACT VALUES Example 9 a

The figure shows a graph of f in

the viewing rectangle [0, 2π] by [-1, 8].

The figure shows a graph of f in

the viewing rectangle [0, 2π] by [-1, 8].

70.
EXACT VALUES Example 9 a

By moving the cursor close to the maximum

point, we see the y-coordinates don’t change

very much in the vicinity of the maximum.

The absolute

maximum value

is about 6.97

It occurs when

x ≈ 5.2

By moving the cursor close to the maximum

point, we see the y-coordinates don’t change

very much in the vicinity of the maximum.

The absolute

maximum value

is about 6.97

It occurs when

x ≈ 5.2

71.
EXACT VALUES Example 9 a

Similarly, by moving the cursor close to

the minimum point, we see the absolute

minimum value is about –0.68 and it occurs

when x ≈ 1.0

Similarly, by moving the cursor close to

the minimum point, we see the absolute

minimum value is about –0.68 and it occurs

when x ≈ 1.0

72.
EXACT VALUES Example 9 a

It is possible to get more accurate

estimates by zooming in toward

the maximum and minimum points.

However, instead, let’s use calculus.

It is possible to get more accurate

estimates by zooming in toward

the maximum and minimum points.

However, instead, let’s use calculus.

73.
EXACT VALUES Example 9 b

The function f(x) = x – 2 sin x is continuous

on [0, 2π].

As f ’(x) = 1 – 2 cos x, we have f ’(x) = 0

when cos x = ½.

This occurs when x = π/3 or 5π/3.

The function f(x) = x – 2 sin x is continuous

on [0, 2π].

As f ’(x) = 1 – 2 cos x, we have f ’(x) = 0

when cos x = ½.

This occurs when x = π/3 or 5π/3.

74.
EXACT VALUES Example 9 b

The values of f at these critical points

f ( / 3) 2sin 3 0.684853

3 3 3

5 5 5

f (5 / 3) 2sin 3 6.968039

3 3 3

The values of f at these critical points

f ( / 3) 2sin 3 0.684853

3 3 3

5 5 5

f (5 / 3) 2sin 3 6.968039

3 3 3

75.
EXACT VALUES Example 9 b

The values of f at the endpoints

f(0) = 0

f(2π) = 2π ≈ 6.28

The values of f at the endpoints

f(0) = 0

f(2π) = 2π ≈ 6.28

76.
EXACT VALUES Example 9 b

Comparing these four numbers and

using the Closed Interval Method, we see

the absolute minimum value is

f(π/3) = π/3 - 3

and the absolute maximum value is

f(5π/3) = 5π/3 + 3

The values from (a) serve as a check on our work.

Comparing these four numbers and

using the Closed Interval Method, we see

the absolute minimum value is

f(π/3) = π/3 - 3

and the absolute maximum value is

f(5π/3) = 5π/3 + 3

The values from (a) serve as a check on our work.

77.
MAXIMUM & MINIMUM VALUES Example 10

The Hubble Space Telescope was

deployed on April 24, 1990, by the space

shuttle Discovery.

The Hubble Space Telescope was

deployed on April 24, 1990, by the space

shuttle Discovery.

78.
MAXIMUM & MINIMUM VALUES Example 10

A model for the velocity of the shuttle

during this mission—from liftoff at t = 0

until the solid rocket boosters were jettisoned

at t = 126 s—is given by:

v(t) = 0.001302t3 – 0.09029t2 + 23.61t – 3.083

(in feet per second)

A model for the velocity of the shuttle

during this mission—from liftoff at t = 0

until the solid rocket boosters were jettisoned

at t = 126 s—is given by:

v(t) = 0.001302t3 – 0.09029t2 + 23.61t – 3.083

(in feet per second)

79.
MAXIMUM & MINIMUM VALUES Example 10

Using this model, estimate the absolute

maximum and minimum values of

the acceleration of the shuttle between

liftoff and the jettisoning of the boosters.

Using this model, estimate the absolute

maximum and minimum values of

the acceleration of the shuttle between

liftoff and the jettisoning of the boosters.

80.
MAXIMUM & MINIMUM VALUES Example 10

We are asked for the extreme values

not of the given velocity function,

but rather of the acceleration function.

We are asked for the extreme values

not of the given velocity function,

but rather of the acceleration function.

81.
MAXIMUM & MINIMUM VALUES Example 10

So, we first need to differentiate to find

the acceleration:

a (t ) v '(t )

d 3 2

(0.001302t 0.09029t

dt

23.61t 3.083)

2

0.003906t 0.18058t 23.61

So, we first need to differentiate to find

the acceleration:

a (t ) v '(t )

d 3 2

(0.001302t 0.09029t

dt

23.61t 3.083)

2

0.003906t 0.18058t 23.61

82.
MAXIMUM & MINIMUM VALUES Example 10

We now apply the Closed Interval Method

to the continuous function a on the interval

0 ≤ t ≤ 126.

Its derivative is:

a’(t) = 0.007812t – 0.18058

We now apply the Closed Interval Method

to the continuous function a on the interval

0 ≤ t ≤ 126.

Its derivative is:

a’(t) = 0.007812t – 0.18058

83.
MAXIMUM & MINIMUM VALUES Example 10

The only critical number occurs

when a’(t) = 0:

0.18058

t1 23.12

0.007812

The only critical number occurs

when a’(t) = 0:

0.18058

t1 23.12

0.007812

84.
MAXIMUM & MINIMUM VALUES Example 10

Evaluating a(t) at the critical number

and at the endpoints, we have:

a(0) = 23.61 a(t1) ≈ 21.52 a(126) ≈ 62.87

The maximum acceleration is about 62.87 ft/s2.

The minimum acceleration is about 21.52 ft/s2.

Evaluating a(t) at the critical number

and at the endpoints, we have:

a(0) = 23.61 a(t1) ≈ 21.52 a(126) ≈ 62.87

The maximum acceleration is about 62.87 ft/s2.

The minimum acceleration is about 21.52 ft/s2.