1549901369-Elements_of_Real_Analysis__Denlinger_

324 Chapter 6 • Differentiable Functions

Thus, h'(c) = 0::::} f'(c) - f(b) - f(a) = 0, and hence,

b-a

f'(c) = f(b~ =~(a). •

APPLICATIONS OF THE MEAN VALUE THEOREM

In Corollary 6.1.3 we proved that if f is a constant, then f'(x) = 0. Surely

you believe intuitively that the converse must also be true: if f'(x) = 0 on an

interval, then f is constant on that interval. The proof of that converse had to

wait until this point, because it is based on the mean value theorem.

Theorem 6.4.4 Suppose f is differentiable on an internal I , and 'v'x E I,

f'(x) = 0. Then f is constant on I.

Proof. Suppose f is differentiable on an interval I, and 'v'x EI, f'(x) = 0.

Let x1,x2 EI. Without loss of generality, x 1 < x 2. By the mean value theorem

applied to f on the interval [x 1 ,x2J, :Jc E (x1,x2) 3

f'(c) = f(x2) - f(x1).

X2 - X1

But f'(c) = 0. That is , f(x^2 ) - f(xi) = 0. But then f(x2) - f(x1) = 0, or
X2 - X1
f(x1) = f(x2).
We have proved that 'v'x1, x2 E I , f(x1) = f(x2). That is, f takes on the
same value at any two points of I. But that means that f is constant on I. •

Corollary 6.4.5 Suppose f and g are differentiable on an internal I , and 'v'x E

I , f'(x) = g'(x). Then 3 constant CE JR 3 'v'x EI, f(x) = g(x) + C.

Proof. Suppose f and g are differentiable on an interval I, and Vx E I,
f'(x) = g'(x). Define the function hon I by h(x) = f(x)-g(x). By the algebra
of derivatives, h is differentiable on I, and 'v'x E I,

h'(x) = f'(x) - g'(x) = 0.

Thus, by Theorem 6.4.4, h is constant on I. That is, 3 C E JR 3 'v'x E I ,
h(x) = C. Thus, 'v'x EI, f(x) - g(x) = C , or

f(x) = g(x) + C. •