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 $(a,b)$ .

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 $(a,b)$ .
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.

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 $f'$ .
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 $f'(b). \qquad \blacksquare$