Assume that ƒ is continuous on 3a, b4 and differentiable on (a, b). Also assume that ƒ(a) and ƒ(b) have opposite signs and that ƒ′ ≠ 0 between a and b. Show that ƒ(x) = 0 exactly once between a and b.

Answer 1

If f ' ≠ 0 on (a, b) then the sign of f ' stays the same. So (1) it must either be positive, meaning the function is increasing over (a, b), or (2) it must be negative, meaning the function is decreasing over (a, b).

For Case (1), if the function is increasing and f(a) and f(b) are different signs, then f(a) must be negative and f(b) must be positive. Since the function is continuous and increasing, there must be a value c between a nd b where f(c) = 0.

For Case (2), if the function is decreasing and f(a) and f(b) are different signs, then f(a) must be positive and f(b) must be negative. Since the function is continuous and decreasing, there must be a value c between a nd b where f(c) = 0.

Here is a sketch for Case 1: