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.
Answer
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.
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