1549901369-Elements_of_Real_Analysis__Denlinger_

6.4 Mean-Value Type Theorems 325

The result we have just proved should be quite familiar to you. Where

have you seen it before? Do you remember indefinite integration in Calculus I?

Corollary 6.4.5 tells us that any two antiderivatives of the same function must

have a constant difference. One antiderivative must differ from any other by a

constant. This is the reason for the "C" in indefinite integration formulas such

as

J

x^2 dx --3 ~x^3 + C.

The mean value theorem is also used as the basis for proving results about

monotone functions-spec ifically, a converse of Theorem 6.3.5.

Theorem 6.4.6 Suppose f is differentiable on an interval I.

(a) If f'(x);::: 0, \:/x EI, then f is monotone increasing on I.

(b) If f'(x) s:; 0, \:/x EI, then f is monotone decreasing on I.

(c) If f'(x) > 0, \:/x EI, then f is strictly increasing on I.

(d) If f'(x) < 0, \:/x EI, then f is strictly decreasing on I.

Beware: The converses of (c) and (d) are false! (See Exercise 15.)

Proof. Suppose f is differentiable on an interval I.
(a) Suppose f'(x) ;::: 0, \:/x EI. Let x 1 < x2 in I. Applying the mean value
f (x2) - f (x1)
theorem to f on the interval [x 1 , x2], :Jc E (xi, X2) 3 f' ( c) =.
X2 - Xi
Since f'(c) 2': 0,

f(x2) - f(x1) > O.

X2 - X1 -

But x 2 - xi > 0 since x1 < x2. Hence, f(x2) - f(x1) 2': O; that is,

We have proved that \:/xi < x 2 in I , f (xi) s:; f (x2). That is, f is monotone
increasing on I.

(b) Exercise 13.

(c) Exercise 14.

(d) Exercise 14. •

Example 6.4. 7 (Using the MVT to Prove General Inequalities) Prove that
\:/x, y E IR, I sin x - sin YI s:; Ix -YI· Consequently, \:/x E IR, I sin xi s:; lxl.