The Mean Value Theorem

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Sharp Tutor Tue, Jan 18, 2022 11:46 PM UTC

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

1. 4
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.

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. MEAN VALUE THEOREM
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.

6. ROLLE’S THEOREM
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

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.

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)

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).

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].

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.

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.

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.

15. ROLLE’S THEOREM Example 2
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.

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.

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.

19. ROLLE’S THEOREM Example 2
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.

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 )

22. MEAN VALUE THEOREM
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

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.

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.

26. MEAN VALUE THEOREM
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.

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

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

30. MEAN VALUE THEOREM
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.

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.

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).

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

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.

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)

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 .

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.

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).

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.

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.

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.

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?

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)

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.

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.

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).

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].

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)

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).

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.

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.

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.

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).

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.

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.

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.

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