KindiakMath

The Intermediate Value Derivative

Problem 1. Let f : [a, b] \to \mathbb R differentiable. Assume f'(a) < f'(b). Prove that for any m \in (f'(a), f'(b)), there exists c \in [a, b] such that f'(c) = m.

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Solution. We observe that f'(c) = m \iff (f(x)-mx)'(c) = 0. Define the differentiable function g : [a, b] \to \mathbb R by g(x) = f(x) - mx. Then it suffices to find c \in [a, b] such that g'(c) = 0.

Since g is differentiable on [a, b], it is continuous on [a, b]. By the extreme value theorem, there exists c \in [a, b] such that g(x) \geq g(c) for any x \in [a, b]. We claim that g'(c) = 0.

Firstly, we note that g'(a) = f'(a) - m < 0, so that there exists \delta_1 > 0 such that g(c) \leq g(a+\delta_1) < g(a). Similarly, g'(b) = f'(b) - m > 0, so that there exists \delta_2 > 0 such that g(c) \leq g(b-\delta_2) < g(b). Hence, g(c) \neq g(a), g(b). Therefore, c \in (a, b), which implies g'(c) = 0, as required.

Remark 1. Any function that satisfies the conclusion of the intermediate value theorem is called a Darboux function. The intermediate value theorem demonstrates that continuous functions are Darboux functions. This problem demonstrates that derivatives are automatically Darboux functions, even if they are not continuous.

—Joel Kindiak, 22 Jan 25, 1234H

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