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In this section, we will learn about: The significance of the mean value theorem.

1.
4

APPLICATIONS OF DIFFERENTIATION

APPLICATIONS OF DIFFERENTIATION

2.
APPLICATIONS OF DIFFERENTIATION

We will see that many of the results

of this chapter depend on one central

fact—the Mean Value Theorem.

We will see that many of the results

of this chapter depend on one central

fact—the Mean Value Theorem.

3.
APPLICATIONS OF DIFFERENTIATION

4.2

The Mean Value Theorem

In this section, we will learn about:

The significance of the mean value theorem.

4.2

The Mean Value Theorem

In this section, we will learn about:

The significance of the mean value theorem.

4.
MEAN VALUE THEOREM

To arrive at the theorem, we

first need the following result.

To arrive at the theorem, we

first need the following result.

5.
ROLLE’S THEOREM

Let f be a function that satisfies the following

three hypotheses:

1. f is continuous on the closed interval [a, b]

2. f is differentiable on the open interval (a, b)

3. f(a) = f(b)

Then, there is a number c in (a, b) such

that f’(c) = 0.

Let f be a function that satisfies the following

three hypotheses:

1. f is continuous on the closed interval [a, b]

2. f is differentiable on the open interval (a, b)

3. f(a) = f(b)

Then, there is a number c in (a, b) such

that f’(c) = 0.

6.
ROLLE’S THEOREM

Before giving the proof, let’s look

at the graphs of some typical functions

that satisfy the three hypotheses.

Before giving the proof, let’s look

at the graphs of some typical functions

that satisfy the three hypotheses.

7.
ROLLE’S THEOREM

The figures show the graphs of

four such

The figures show the graphs of

four such

8.
ROLLE’S THEOREM

In each case, it appears there is at least one

point (c, f(c)) on the graph where the tangent

and thus

f’(c) = 0.

So, Rolle’s

Theorem is

plausible.

In each case, it appears there is at least one

point (c, f(c)) on the graph where the tangent

and thus

f’(c) = 0.

So, Rolle’s

Theorem is

plausible.

9.
ROLLE’S THEOREM Proof

There are three cases:

1. f(x) = k, a constant

2. f(x) > f(a) for some x in (a, b)

3. f(x) < f(a) for some x in (a, b)

There are three cases:

1. f(x) = k, a constant

2. f(x) > f(a) for some x in (a, b)

3. f(x) < f(a) for some x in (a, b)

10.
ROLLE’S THEOREM Proof—Case 1

f(x) = k, a constant

Then, f ’(x) = 0.

So, the number c can be

taken to be any number

in (a, b).

f(x) = k, a constant

Then, f ’(x) = 0.

So, the number c can be

taken to be any number

in (a, b).

11.
ROLLE’S THEOREM Proof—Case 2

f(x) > f(a) for some x

in (a, b)

By the Extreme Value Theorem

(which we can apply by

hypothesis 1), f

has a maximum value

somewhere in [a, b].

f(x) > f(a) for some x

in (a, b)

By the Extreme Value Theorem

(which we can apply by

hypothesis 1), f

has a maximum value

somewhere in [a, b].

12.
ROLLE’S THEOREM Proof—Case 2

As f(a) = f(b), it must attain this

maximum value at a number c

in the open interval (a, b).

Then, f has a local maximum

at c and, by hypothesis 2, f

is differentiable at c.

Thus, f ’(c) = 0 by Fermat’s

Theorem.

As f(a) = f(b), it must attain this

maximum value at a number c

in the open interval (a, b).

Then, f has a local maximum

at c and, by hypothesis 2, f

is differentiable at c.

Thus, f ’(c) = 0 by Fermat’s

Theorem.

13.
ROLLE’S THEOREM Proof—Case 3

f(x) < f(a) for some x

in (a, b)

By the Extreme Value

Theorem, f has a minimum

value in [a, b] and, since

f(a) = f(b), it attains this

minimum value at a number

c in (a, b).

Again, f ’(c) = 0 by

Fermat’s Theorem.

f(x) < f(a) for some x

in (a, b)

By the Extreme Value

Theorem, f has a minimum

value in [a, b] and, since

f(a) = f(b), it attains this

minimum value at a number

c in (a, b).

Again, f ’(c) = 0 by

Fermat’s Theorem.

14.
ROLLE’S THEOREM Example 1

Let’s apply the theorem to the position

function s = f(t) of a moving object.

If the object is in the same place at two different

instants t = a and t = b, then f(a) = f(b).

The theorem states that there is some instant of

time t = c between a and b when f ’(c) = 0; that is,

the velocity is 0.

In particular, you can see that this is true when

a ball is thrown directly upward.

Let’s apply the theorem to the position

function s = f(t) of a moving object.

If the object is in the same place at two different

instants t = a and t = b, then f(a) = f(b).

The theorem states that there is some instant of

time t = c between a and b when f ’(c) = 0; that is,

the velocity is 0.

In particular, you can see that this is true when

a ball is thrown directly upward.

15.
ROLLE’S THEOREM Example 2

Prove that the equation

x3 + x – 1 = 0

has exactly one real root.

Prove that the equation

x3 + x – 1 = 0

has exactly one real root.

16.
ROLLE’S THEOREM Example 2

First, we use the Intermediate Value

Theorem (Equation 10 in Section 2.5)

to show that a root exists.

Let f(x) = x3 + x – 1.

Then, f(0) = – 1 < 0 and f(1) = 1 > 0.

Since f is a polynomial, it is continuous.

So, the theorem states that there is a number c

between 0 and 1 such that f(c) = 0.

Thus, the given equation has a root.

First, we use the Intermediate Value

Theorem (Equation 10 in Section 2.5)

to show that a root exists.

Let f(x) = x3 + x – 1.

Then, f(0) = – 1 < 0 and f(1) = 1 > 0.

Since f is a polynomial, it is continuous.

So, the theorem states that there is a number c

between 0 and 1 such that f(c) = 0.

Thus, the given equation has a root.

17.
ROLLE’S THEOREM Example 2

To show that the equation has no

other real root, we use Rolle’s Theorem

and argue by contradiction.

To show that the equation has no

other real root, we use Rolle’s Theorem

and argue by contradiction.

18.
ROLLE’S THEOREM Example 2

Suppose that it had two roots a and b.

Then, f(a) = 0 = f(b).

As f is a polynomial, it is differentiable on (a, b)

and continuous on [a, b].

Thus, by Rolle’s Theorem, there is a number c

between a and b such that f ’(c) = 0.

However, f ’(x) = 3x2 + 1 ≥ 1 for all x

(since x2 ≥ 0), so f ’(x) can never be 0.

Suppose that it had two roots a and b.

Then, f(a) = 0 = f(b).

As f is a polynomial, it is differentiable on (a, b)

and continuous on [a, b].

Thus, by Rolle’s Theorem, there is a number c

between a and b such that f ’(c) = 0.

However, f ’(x) = 3x2 + 1 ≥ 1 for all x

(since x2 ≥ 0), so f ’(x) can never be 0.

19.
ROLLE’S THEOREM Example 2

This gives a contradiction.

So, the equation can’t have two real roots.

This gives a contradiction.

So, the equation can’t have two real roots.

20.
ROLLE’S THEOREM

Our main use of Rolle’s Theorem is in proving

the following important theorem—which was

first stated by another French mathematician,

Joseph-Louis Lagrange.

Our main use of Rolle’s Theorem is in proving

the following important theorem—which was

first stated by another French mathematician,

Joseph-Louis Lagrange.

21.
MEAN VALUE THEOREM Equations 1 and 2

Let f be a function that fulfills two hypotheses:

1. f is continuous on the closed interval [a, b].

2. f is differentiable on the open interval (a, b).

Then, there is a number c in (a, b) such that

f (b) f (a)

f '(c)

b a

or, equivalently,

f (b) f (a) f '(c)(b a )

Let f be a function that fulfills two hypotheses:

1. f is continuous on the closed interval [a, b].

2. f is differentiable on the open interval (a, b).

Then, there is a number c in (a, b) such that

f (b) f (a)

f '(c)

b a

or, equivalently,

f (b) f (a) f '(c)(b a )

22.
MEAN VALUE THEOREM

Before proving this theorem,

we can see that it is reasonable by

interpreting it geometrically.

Before proving this theorem,

we can see that it is reasonable by

interpreting it geometrically.

23.
MEAN VALUE THEOREM

The figures show the points A(a, f(a)) and

B(b, f(b)) on the graphs of two differentiable

The figures show the points A(a, f(a)) and

B(b, f(b)) on the graphs of two differentiable

24.
MEAN VALUE THEOREM Equation 3

The slope of the secant line AB is:

f (b) f (a )

mAB

b a

This is the same expression as on the right side

of Equation 1.

The slope of the secant line AB is:

f (b) f (a )

mAB

b a

This is the same expression as on the right side

of Equation 1.

25.
MEAN VALUE THEOREM

f ’(c) is the slope of the tangent line at (c, f(c)).

So, the Mean Value Theorem—in the form given by

Equation 1—states that there is at least one point

P(c, f(c)) on the graph where the slope of the tangent

line is the same as the slope of the secant line AB.

f ’(c) is the slope of the tangent line at (c, f(c)).

So, the Mean Value Theorem—in the form given by

Equation 1—states that there is at least one point

P(c, f(c)) on the graph where the slope of the tangent

line is the same as the slope of the secant line AB.

26.
MEAN VALUE THEOREM

In other words, there is a point P

where the tangent line is parallel to

the secant line AB.

In other words, there is a point P

where the tangent line is parallel to

the secant line AB.

27.
We apply Rolle’s Theorem to a new

function h defined as the difference

between f and the function whose graph

is the secant line AB.

function h defined as the difference

between f and the function whose graph

is the secant line AB.

28.
Using Equation 3, we see that the equation

of the line AB can be written as:

f (b) f (a )

y f (a ) ( x a)

b a

or as:

f (b) f (a)

y f (a) ( x a)

b a

of the line AB can be written as:

f (b) f (a )

y f (a ) ( x a)

b a

or as:

f (b) f (a)

y f (a) ( x a)

b a

29.
MEAN VALUE THEOREM Equation 4

So, as shown in the figure,

f (b) f (a)

h( x ) f ( x ) f ( a ) ( x a)

b a

So, as shown in the figure,

f (b) f (a)

h( x ) f ( x ) f ( a ) ( x a)

b a

30.
MEAN VALUE THEOREM

First, we must verify that h satisfies the

three hypotheses of Rolle’s Theorem—

as follows.

First, we must verify that h satisfies the

three hypotheses of Rolle’s Theorem—

as follows.

31.
HYPOTHESIS 1

The function h is continuous on [a, b]

because it is the sum of f and a first-degree

polynomial, both of which are continuous.

The function h is continuous on [a, b]

because it is the sum of f and a first-degree

polynomial, both of which are continuous.

32.
HYPOTHESIS 2

The function h is differentiable on (a, b)

because both f and the first-degree polynomial

are differentiable.

In fact, we can compute h’ directly from Equation 4:

f (b) f (a )

h '( x) f '( x)

b a

Note that f(a) and [f(b) – f(a)]/(b – a) are constants.

The function h is differentiable on (a, b)

because both f and the first-degree polynomial

are differentiable.

In fact, we can compute h’ directly from Equation 4:

f (b) f (a )

h '( x) f '( x)

b a

Note that f(a) and [f(b) – f(a)]/(b – a) are constants.

33.
HYPOTHESIS 3

f (b) f (a)

h( a ) f ( a ) f ( a ) (a a)

b a

0

f (b) f (a)

h(b) f (b) f (a) (b a )

b a

f (b) f (a) [ f (b) f (a)]

0

Therefore, h(a) = h(b).

f (b) f (a)

h( a ) f ( a ) f ( a ) (a a)

b a

0

f (b) f (a)

h(b) f (b) f (a) (b a )

b a

f (b) f (a) [ f (b) f (a)]

0

Therefore, h(a) = h(b).

34.
MEAN VALUE THEOREM

As h satisfies the hypotheses of Rolle’s

Theorem, that theorem states there is

a number c in (a, b) such that h’(c) = 0.

f (b) f (a )

Therefore, 0 h '(c) f '(c)

b a

f (b) f (a)

So, f '(c)

b a

As h satisfies the hypotheses of Rolle’s

Theorem, that theorem states there is

a number c in (a, b) such that h’(c) = 0.

f (b) f (a )

Therefore, 0 h '(c) f '(c)

b a

f (b) f (a)

So, f '(c)

b a

35.
MEAN VALUE THEOREM Example 3

To illustrate the Mean Value Theorem

with a specific function, let’s consider

f(x) = x3 – x, a = 0, b = 2.

To illustrate the Mean Value Theorem

with a specific function, let’s consider

f(x) = x3 – x, a = 0, b = 2.

36.
MEAN VALUE THEOREM Example 3

Since f is a polynomial, it is continuous

and differentiable for all x.

So, it is certainly continuous on [0, 2]

and differentiable on (0, 2).

Therefore, by the Mean Value Theorem,

there is a number c in (0,2) such that:

f(2) – f(0) = f ’(c)(2 –

0)

Since f is a polynomial, it is continuous

and differentiable for all x.

So, it is certainly continuous on [0, 2]

and differentiable on (0, 2).

Therefore, by the Mean Value Theorem,

there is a number c in (0,2) such that:

f(2) – f(0) = f ’(c)(2 –

0)

37.
MEAN VALUE THEOREM Example 3

Now, f(2) = 6, f(0) = 0, and f ’(x) = 3x2 – 1.

So, this equation becomes

6 = (3c2 – 1)2 = 6c2 – 2

4

This gives c2 = , that is, c = 2 / 3 .

3

However, c must lie in (0, 2), so c = 2 / 3 .

Now, f(2) = 6, f(0) = 0, and f ’(x) = 3x2 – 1.

So, this equation becomes

6 = (3c2 – 1)2 = 6c2 – 2

4

This gives c2 = , that is, c = 2 / 3 .

3

However, c must lie in (0, 2), so c = 2 / 3 .

38.
MEAN VALUE THEOREM Example 3

The figure illustrates this calculation.

The tangent line at this value of c is parallel

to the secant line OB.

The figure illustrates this calculation.

The tangent line at this value of c is parallel

to the secant line OB.

39.
MEAN VALUE THEOREM Example 4

If an object moves in a straight line with

position function s = f(t), then the average

velocity between t = a and t = b is

f (b) f (a)

b a

and the velocity at t = c is f ’(c).

If an object moves in a straight line with

position function s = f(t), then the average

velocity between t = a and t = b is

f (b) f (a)

b a

and the velocity at t = c is f ’(c).

40.
MEAN VALUE THEOREM Example 4

Thus, the Mean Value Theorem—in the form

of Equation 1—tells us that, at some time

t = c between a and b, the instantaneous

velocity f ’(c) is equal to that average velocity.

For instance, if a car traveled 180 km in 2 hours, the

speedometer must have read 90 km/h at least once.

Thus, the Mean Value Theorem—in the form

of Equation 1—tells us that, at some time

t = c between a and b, the instantaneous

velocity f ’(c) is equal to that average velocity.

For instance, if a car traveled 180 km in 2 hours, the

speedometer must have read 90 km/h at least once.

41.
MEAN VALUE THEOREM Example 4

In general, the Mean Value Theorem can be

interpreted as saying that there is a number

at which the instantaneous rate of change

is equal to the average rate of change over

an interval.

In general, the Mean Value Theorem can be

interpreted as saying that there is a number

at which the instantaneous rate of change

is equal to the average rate of change over

an interval.

42.
MEAN VALUE THEOREM

The main significance of the Mean Value

Theorem is that it enables us to obtain

information about a function from information

about its derivative.

The next example provides an instance

of this principle.

The main significance of the Mean Value

Theorem is that it enables us to obtain

information about a function from information

about its derivative.

The next example provides an instance

of this principle.

43.
MEAN VALUE THEOREM Example 5

Suppose that f(0) = -3 and f ’(x) ≤ 5

for all values of x.

How large can f(2) possibly be?

Suppose that f(0) = -3 and f ’(x) ≤ 5

for all values of x.

How large can f(2) possibly be?

44.
MEAN VALUE THEOREM Example 5

We are given that f is differentiable—and

therefore continuous—everywhere.

In particular, we can apply the Mean Value

Theorem on the interval [0, 2].

There exists a number c such that

f(2) – f(0) = f ’(c)(2 –

0)

So, f(2) = f(0) + 2 f ’(c) = – 3 + 2 f ’(c)

We are given that f is differentiable—and

therefore continuous—everywhere.

In particular, we can apply the Mean Value

Theorem on the interval [0, 2].

There exists a number c such that

f(2) – f(0) = f ’(c)(2 –

0)

So, f(2) = f(0) + 2 f ’(c) = – 3 + 2 f ’(c)

45.
MEAN VALUE THEOREM Example 5

We are given that f ’(x) ≤ 5 for all x.

So, in particular, we know that f ’(c) ≤ 5.

Multiplying both sides of this inequality by 2,

we have 2 f ’(c) ≤ 10.

So, f(2) = – 3 + 2 f ’(c) ≤ – 3 + 10 = 7

The largest possible value for f(2) is 7.

We are given that f ’(x) ≤ 5 for all x.

So, in particular, we know that f ’(c) ≤ 5.

Multiplying both sides of this inequality by 2,

we have 2 f ’(c) ≤ 10.

So, f(2) = – 3 + 2 f ’(c) ≤ – 3 + 10 = 7

The largest possible value for f(2) is 7.

46.
MEAN VALUE THEOREM

The Mean Value Theorem can be used

to establish some of the basic facts of

differential calculus.

One of these basic facts is the following theorem.

Others will be found in the following sections.

The Mean Value Theorem can be used

to establish some of the basic facts of

differential calculus.

One of these basic facts is the following theorem.

Others will be found in the following sections.

47.
MEAN VALUE THEOREM Theorem 5

If f ’(x) = 0 for all x in an

interval (a, b), then f is constant

on (a, b).

If f ’(x) = 0 for all x in an

interval (a, b), then f is constant

on (a, b).

48.
MEAN VALUE THEOREM Theorem 5—Proof

Let x1 and x2 be any two numbers

in (a, b) with x1 < x2.

Since f is differentiable on (a, b), it must be

differentiable on (x1, x2) and continuous on [x1, x2].

Let x1 and x2 be any two numbers

in (a, b) with x1 < x2.

Since f is differentiable on (a, b), it must be

differentiable on (x1, x2) and continuous on [x1, x2].

49.
MEAN VALUE THEOREM Th. 5—Proof (Eqn. 6)

By applying the Mean Value Theorem to f

on the interval [x1, x2], we get a number c

such that x1 < c < x2 and

f(x2) – f(x1) = f ’(c)(x2 – x1)

By applying the Mean Value Theorem to f

on the interval [x1, x2], we get a number c

such that x1 < c < x2 and

f(x2) – f(x1) = f ’(c)(x2 – x1)

50.
MEAN VALUE THEOREM Theorem 5—Proof

Since f ’(x) = 0 for all x, we have f ’(c) = 0.

So, Equation 6 becomes

f(x2) – f(x1) = 0 or f(x2) = f(x1)

Therefore, f has the same value at any two numbers

x1 and x2 in (a, b).

This means that f is constant on (a, b).

Since f ’(x) = 0 for all x, we have f ’(c) = 0.

So, Equation 6 becomes

f(x2) – f(x1) = 0 or f(x2) = f(x1)

Therefore, f has the same value at any two numbers

x1 and x2 in (a, b).

This means that f is constant on (a, b).

51.
MEAN VALUE THEOREM Corollary 7

If f ’(x) = g ’(x) for all x in an interval (a, b),

then f – g is constant on (a, b).

That is, f(x) = g(x) + c where c is

a constant.

If f ’(x) = g ’(x) for all x in an interval (a, b),

then f – g is constant on (a, b).

That is, f(x) = g(x) + c where c is

a constant.

52.
MEAN VALUE THEOREM Corollary 7—Proof

Let F(x) = f(x) – g(x).

F’(x) = f ’(x) – g ’(x) = 0

for all x in (a, b).

Thus, by Theorem 5, F is constant.

That is, f – g is constant.

Let F(x) = f(x) – g(x).

F’(x) = f ’(x) – g ’(x) = 0

for all x in (a, b).

Thus, by Theorem 5, F is constant.

That is, f – g is constant.

53.
Care must be taken in applying

Theorem 5.

Let f ( x ) x 1 if x 0

| x | 1 if x 0

The domain of f is D = {x | x ≠ 0} and f ’(x) = 0

for all x in D.

Theorem 5.

Let f ( x ) x 1 if x 0

| x | 1 if x 0

The domain of f is D = {x | x ≠ 0} and f ’(x) = 0

for all x in D.

54.
However, f is obviously not a constant

This does not contradict Theorem 5

because D is not an interval.

Notice that f is constant on the interval (0, ∞)

and also on the interval (-∞, 0).

This does not contradict Theorem 5

because D is not an interval.

Notice that f is constant on the interval (0, ∞)

and also on the interval (-∞, 0).

55.
MEAN VALUE THEOREM Example 6

Prove the identity

tan-1 x + cot -1 x = π/2.

Although calculus isn’t needed to prove this

identity, the proof using calculus is quite simple.

Prove the identity

tan-1 x + cot -1 x = π/2.

Although calculus isn’t needed to prove this

identity, the proof using calculus is quite simple.

56.
MEAN VALUE THEOREM Example 6

If f(x) = tan-1 x + cot -1 x ,

then 1 1

f '( x) 2

2

0

1 x 1 x

for all values of x.

Therefore, f(x) = C, a constant.

If f(x) = tan-1 x + cot -1 x ,

then 1 1

f '( x) 2

2

0

1 x 1 x

for all values of x.

Therefore, f(x) = C, a constant.

57.
MEAN VALUE THEOREM Example 6

To determine the value of C, we put x = 1

(because we can evaluate f(1) exactly).

1

1

C f (1) tan 1 cot 1

4 4 2

Thus, tan-1 x + cot-1 x = π/2.

To determine the value of C, we put x = 1

(because we can evaluate f(1) exactly).

1

1

C f (1) tan 1 cot 1

4 4 2

Thus, tan-1 x + cot-1 x = π/2.