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Prove the intermediate value theorem for derivatives

Consider the following statement of the intermediate value theorem for derivatives:

Assume f is differentiable on an open interval I. Let a < b be two points in I. Then, the derivative f' takes every value between f'(a) and f'(b) somewhere in (a,b).

  1. Define a function

        \[ g(x) = \frac{f(x) - f(a)}{x-a} \qquad \text{if } x \neq a, \qquad g(a) = f'(a). \]

    Prove that g takes every value between f'(a) and f'(b) in the interval (a,b). Then, use the mean-value theorem for derivatives to show f' takes all values between f'(a) and g(b) somewhere in the interval (a,b).

  2. Define a function

        \[ h(x) = \frac{f(x) - f(b)}{x-b}, \qquad \text{if } x \neq b, \qquad h(b) = f'(b). \]

    Show that the derivative f' takes on all values between f'(b) and h(a) in the interval (a,b). Conclude that the statement of the intermediate-value theorem is true.


  1. Proof. First, since f is differentiable everywhere on the interval I, we know f is continuous on [a,b] and differentiable on (a,b). Thus, if x \in [a,b] and x \neq a then g is continuous at x since it is the quotient of continuous functions and the denominator is nonzero. If x = a then

        \[ \lim_{x \to a} g(x) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a} = f'(a) = g(a); \]

    hence, g is continuous at x = a as well. Therefore, g is continuous on the closed interval [a,b]. So, by the intermediate value theorem for continuous functions we know g takes on every value between g(a) and g(b) somewhere on the interval (a,b). Since g(a) = f'(a) this means g takes on every value between f'(a) and g(b) somewhere on the interval (a,b).
    By the mean-value theorem for derivatives, we then know there exists some c \in (a,b) such that

        \begin{align*}  &&f(x) - f(a) &= f'(c)(x-a) \\ \implies && f'(c) &= \frac{f(x) - f(a)}{x-a} \\ \implies && f'(c) &= g(x) \end{align*}

    for some c \in (a,x). Since x < b, we then conclude there is some c \in (a,b) such that f'(c) = g(x) for any x. Since g takes on every value between f'(a) and g(b), so does f'.

  2. Proof. This is very similar to part (a). By the same argument we have the function h is continuous on [a,b]; thus, h takes on every value between h(a) and f'(b) by the intermediate value theorem for continuous functions. Then, by the mean value theorem, we know there exists a c \in (x, b) such that

        \[ f'(c) = \frac{f(x) - f(b)}{x-b} = h(x) \qquad \text{for some } c \in (x,b). \]

    Thus, f' takes on every value between f'(b) and h(a). Since h(a) = g(b); f' takes on every value between f'(a) and f'(b). \qquad \blacksquare

One comment

  1. Anonymous says:

    The proof shows that f’ takes all values between f'(a) and g(b) , and also between f'(b) and g(b). But don’t we also need to prove that it actually takes on the value g(b) before concluding that it takes all values between f'(a) and f'(b)? Am I missing something?

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