Question

Assume that f is continuous on [a,b] and differentiable on (a,b). Also assume that f(a) and f(b) have opposite signs and f' DOES NOT equal zero between a and b. Show that f(x)=0 exactly once between a and b.

Answer 1

Since f(a) and f(b) are of opposite signs and f(x) is continuous, f(x) MUST cross the x-axis at least once. Therefore, f(x) must be zero at least once.

Since f'(x) does not equal zero in the interval (a,b), f'(x) must be EITHER positive or negative in the interval [a,b].

If f'(x) is positive in the interval [a,b], then f(x) is increasing in the interval [a,b]. Therefore, it can cross the x-axis just once. To cross the x-axis a second time the function will have to "turn around" or decrease in the interval at some point. But f'(x) is positive in the interval and so f(x) does not decrease at all and so it cannot cross the x-axis a second time.

The same argument can be made when f'(x) is negative in the interval [a,b].

Answer 2

Thanks!

Answer 3

Welcome!