# Show that the Mean Value Theorem does not apply to $f(x)=x^{-2}$ on $(-1,1)$

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

$$f'(c)=-2c^{-3}=0.$$

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.

• 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. Dec 9, 2017 at 13:04

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.
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.