The Mean Value Theorem - Pennsylvania State University

[Pages:5]The Mean Value Theorem

Rolle's Theorem: Let f be a function such that:

1. It is continuous on the closed interval [a, b]. 2. It is differentiable on the open interval (a, b). 3. f (a) = f (b).

Then there is (at least) a number c in (a, b) such that f (c) = 0.

ex. Show that the function f (x) = x4 + 32x + k has at most 2 real roots (where k is any constant).

Solution: Suppose there are 3 (or more) real roots, denote them x1, x2, and x3. Without the loss of generality, assume that x1 < x2 < x3. Since f (x1) = f (x2) = 0, by the Rolle's Theorem there is a number c1 in (x1, x2) such that f ( c1) = 0. And since f (x2) = f (x3) = 0, there is a second number c2 in (x2, x3), which is necessarily different than c1, such that f ( c2) = 0. But f (x) = 4x3 + 32 = 4(x + 2)(x2 - 2x + 4) equals zero at only one real number, when x = -2. (It also has two complex zeroes.) Therefore, c2 cannot possibly exist ? a violation of the Rolle's Theorem unless there are at most 2 roots, x1 and x2.

Note: The above only says that f (x) can have at most 2 real roots. It doesn't say that f (x) will actually have 2 real roots. In fact, it can be easily verified that for any constant k > 48, f (x) will not have any real root at all.

The "slanted" version of the Rolle's Theorem is the Mean Value Theorem.

The Mean Value Theorem: Let f be a function such that:

1. It is continuous on the closed interval [a, b]. 2. It is differentiable on the open interval (a, b).

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

f (c) =

f (b) - f (a) b-a

Comment: Notice that the left side is the instantaneous rate of change / slope of the tangent line at c, while the right side is the average rate of change / slope of the secant line from a to b. Thus the theorem guarantees that there is at least one place in (a, b) where the tangent line there is parallel to the secant line connecting the points (a, f (a)) and (b, f (b)) on the curve y = f (x).

ex. Find the value(s) of c such that the Mean Value Theorem is satisfied for the function f (x) = x2/3 on the interval [0, 1].

First, let's note that f is indeed continuous on [0, 1], and is differentiable on (0, 1), so the prerequisites for the M.V.T. are met. (The function is not differentiable at x = 0, but it nevertheless meets the condition of being differentiable on the open interval (0, 1).)

The average rate of change from x = 0 to x = 1 is

f

(1) - f 1- 0

(0)

=

1 1

=1

The Mean Value Theorem says that there is at least one number c on

(0, 1) such that f (c) = 1. So set f (x) = 1 and solve:

f (x) = 2 x -1/ 3 = 1

x -1/ 3 = 3

3

2

1 =3 3x 2

3 x=2 3

x = 2 3 = 8 3 27

Therefore, c = 8/27, which is indeed within the interval (0, 1).

ex. Suppose it took 14 seconds for a thermometer to rise from -19?C to 100?C. Show that sometime between t = 0 and t = 14 (sec) the mercury is rising at the exact rate of 8.5?C/sec.

The average rate of change of the temperature function f , between t = 0 and t = 14 is

f

(14) 14

- -

f 0

(0)

=

100

- (-19) 14

=

119 14

=

8.5

Since the temperature, as a function, is continuous and differentiable everywhere, the Mean Value Theorem applies. Therefore, by the M.V.T., there is, at least, one moment within the given time interval when the mercury is rising at the exact rate of 8.5?C/sec. (Notice that we don't even need to know what the function was to prove the above.)

ex. Consider the function

f

(

x)

=

x

,

0 ,

0 x ................
................

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