Show that the Mean Value Theorem does not apply to $f(x)=x^{-2}$ on $(-1,1)$. Is this a contradiction to the Mean Value Theorem?

My solution so far:

$$f(1)-f(-1)=f'(c)(1-(-1)) \Leftrightarrow 1-1=2f'(c) \Leftrightarrow f'(c)=0$$

Since $ f'(x)=-2x^{-3} \Rightarrow f'(c)=-2c^{-3}$


This is not satisfied by any value $c$ so I think I've now got the first part of the task. However, I'm not quite sure how do I figure if this contradicts with the Mean Value Theorem. I know that the MVT tells us that if $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, there is a point $c\in (a,b )$ such that $f(b)-f(a)=f'(c)(b-a)$ but I don't know how to apply it here.

  • $\begingroup$ You already proved, indeed you already know that you cannot apply the Mean Value thorem since the function is not defined on a closed bounded interval. Then if you assume by contradiction that the theorem holds anyway, you find that $f'(c) = -2c^{3}=0$ which is impossible. Therefore, you conclude that the fact that the function has to be defined (and continuous) on a bounded closed interval is a necessary condition to apply the theorem. $\endgroup$ Dec 9, 2017 at 13:04

2 Answers 2


The mean-value theorem only applies for continous functions. But $x^{-2}=\frac{1}{x^2}$ is not defined at $x=0$ , the singularity is not even removeable.

  • 1
    $\begingroup$ It would be better to show directly that it fails for some endpoints, indeed this is not a clear evidence that the theorem could not be extended to cover this function (even if it can't) $\endgroup$ Dec 9, 2017 at 12:59

I think your idea is correct, maybe I would phrase it differently:

If MVT did apply to $f$, it would be that

$$\frac{f(-1) - f(1)}{2} = 0 = f'(c)$$

would hold for some $c$, but, as you say, $f'(c) = \frac{1}{-2c^3}$, so no such $c$ can be found, contradicting MVT.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.