# Theorem 5.16 in Apostol's MATHEMATICAL ANALYSIS, 2nd ed: Intermediate-Value Theorem for Derivatives

Here is Theorem 5.16 (intermediate-value theorem for derivatives) in the book Mathematical Analysis by Tom M. Apostol, 2nd edition:

Assume that $f$ is defined on a compact interval $[a, b]$ and that $f$ has a derivative (finite or infinite) at each interior point. Assume also that $f$ has finite one-sided derivatives $f^\prime_+(a)$ and $f^\prime_-(b)$ at the endpoints, with $f^\prime_+(a) \neq f^\prime_-(b)$. Then, if $c$ is a real number between $f^\prime_+(a)$ and $f^\prime_-(b)$, there exists at least one interior point $x$ such that $f^\prime(x) = c$.

And, here is Apostol's proof:

Define a new function $g$ as follows: $$g(x) = \frac{ f(x) - f(a) }{ x-a } \ \mbox{ if } x \neq a, \qquad g(a) = f^\prime_+(a).$$ Then $g$ is continuous on the closed interval $[a, b]$. By the intermediate-value theorem for continuous functions, $g$ takes on every value between $f^\prime_+(a)$ and $[ f(b) - f(a) ]/(b-a)$ in the interior $(a, b)$. By the Mean-Value Theorem, we have $g(x) = f^\prime (k)$ for some $k$ in $(a, x)$ whenever $x \in (a, b)$. Therefore $f^\prime$ takes on every value between $f^\prime_+(a)$ and $[ f(b) - f(a) ]/( b-a )$ in the interior $(a, b)$. A similar argument applies to the function $h$, defined by $$h(x) = \frac{ f(x) - f(b) }{ x-b } \ \mbox{ if } x \neq b, \qquad h(b) = f^\prime_-(b),$$ shows that $f^\prime$ takes on every value between $[ f(b) - f(a) ]/(b-a)$ and $f^\prime_-(b)$ in the interior $(a, b)$. Combining these results, we see that $f^\prime$ takes on every value between $f^\prime_+(a)$ and $f^\prime_-(b)$ in the interior $(a, b)$, and this proves the theorem.

What Apostol has not mentioned is the fact that $f^\prime$ does actually take on the value $[ f(b) - f(a) ]/(b-a)$ somewhere in $(a, b)$, by virtue of the Mean-Value Theorem. Am I right?

Immediately following the above proof is this Note:

Theorem 5.16 is still valid if one or both of the one-sided derivatives $f^\prime_+(a)$, $f^\prime_-(b)$, is infinite. The proof in this case can be given by considering the auxiliary function $g$ defined by the equation $g(x) = f(x) - cx$, if $x \in [a, b]$. Details are left to the reader.

Although I think I fully understand Apostol's proof of Theorem 5.16 as given above, I'm unable to figure out how to prove the result if either one or both of the one-sided derivatives $f^\prime_+(a)$ and $f^\prime_-(b)$ is infinite; I have no idea of how the new auxiliary function $g$ is going to be helpful in this regard.

• The proof is the same. I don't know why they wrote that hint of using $g(x)=f(x)-cx$. That is not necessary. In the proof, if say $g(x)\to-\infty$ when $x\to a^+$, then take $a'>a$ such that $g(a')<c$ and apply the argument to $g$ on $[a',b]$. Similarly, do the cases $g\to+\infty$, $h\to\pm\infty$. – user561777 May 22 '18 at 19:22