Let $a,b \in \Bbb R$, $a<b$, and let $f$ be a differentiable real-valued function on an open subset of $\Bbb R$ that contains $[a,b]$. Show that if $\gamma$ is any real number between $f'(a)$ and $f'(b)$ then there exists a number $c \in (a,b)$ such that $\gamma = f'(c)$ .

So I was trying to use the Mean Value Theorem and the Intermediate Value Theorem for the function $\frac {f(x_1) -f(x_2)}{x_1 - x_2}$ on this set: $\{(x_1,x_2) \in E^2: a \le x_1 < x_2 \le b \}$ but I am stuck how do you go from here (if my thinking is correct).

  • $\begingroup$ This problem is subtler than it may appear. I first thought of applying the intermediate value theorem to $f'$, but we have no guarantee that $f'$ is continuous everywhere. The original function $f$ must be continuous everywhere since $f$ is differentiable, but there do exist derivatives that are everywhere defined but not everywhere continuous. But I think the answer posted by Reveillark solves the problem. ${}\qquad{}$ $\endgroup$ Apr 25, 2015 at 3:26
  • $\begingroup$ I have been thinking about what you are saying, could you give me an example of a continuous function on an interval but it's derivative is not continuous within that interval? I am having trouble imagining a case for this, especially in $R$. $\endgroup$
    – Meecolm
    Apr 25, 2015 at 18:48
  • $\begingroup$ $\displaystyle f(x) = \left.\begin{cases} x^2 \sin(1/x) & \text{if }x\ne 0, \\ 0 & \text{if }x=0.\end{cases}\right\}$ This function is differentiable everywhere on $\mathbb R$, but $\displaystyle\lim_{x\to 0} f'(x)$ does not exist because it oscillates between $\pm1$ as $x\to0$. Thus $f'$ is not continuous at $0$. Consequently the intermediate value theorem cannot be applied on any interval that contains $0$. But the function nonetheless satisfies the conclusion of the intermediate value theorem. And Reveillark's answer shows you how to prove that it satisfies that conclusion.${}\qquad{}$ $\endgroup$ Apr 25, 2015 at 21:49

1 Answer 1


Suppose $f'(a)<\gamma <f'(b)$ and consider $g(x)=f(x)-\gamma x$, then $g'(x)=f'(x)-\gamma$ and therefore $g'(a)<0<g'(b)$.

Use the last inequality to deduce that there exist $c,d \in (a,b)$ such that $g(c)<g(a)$ and $g(d)<g(b)$. It then follows that the minimum of $g$ (which exists, since $g$ is differentiable and hence continuous in $[a,b]$, which is compact) occurs in $(a,b)$. If $g$ achieves its minimum at $x_0$, it follows that that $g'(x_0)=0$

  • $\begingroup$ Should it be $g(x) = f(x) + \gamma x$? in order to cancel the $\gamma$ in the substitution in the first inequality? $\endgroup$
    – Meecolm
    Apr 25, 2015 at 2:31
  • 1
    $\begingroup$ Not unless I'm missing something. You want to define a function $g$ such that its derivative has different signs at $a$ and $b$. The way to achieve this is to put $g(x)=f(x)-\gamma x$ $\endgroup$
    – Reveillark
    Apr 25, 2015 at 2:34

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