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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

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