# Mean Value Theorem for a function $f(x) = \sqrt[n]{x}$ on the interval $[a, b]$ where $a < 0 < b$

The Mean Value Theorem is as follows:

Let $f$ be a function that satisfies the following hypotheses:

1. $f$ is continuous on the closed interval $[a, b]$.
2. $f$ is differentiable on the open interval $(a, b)$.

Then there is a number $c$ in $(a, b)$ such that:

$\hspace{.6 in} f'(c) = \dfrac{f(b) - f(a))}{b - a}$

Say we have a function $f$ that is any $n^{th}$ degree root function $\sqrt[n]{x}$.

This function will not satisfy the MVT if it is restricted to any closed interval $[a, b]$ such that $a < 0 < b$.

Now, a function $f$ is not differentiable on the open interval $(a, b)$ if it is not continuous on the closed interval $[a, b]$, or if it comes to a sharp point at any value $c$ on the closed interval $[a, b]$.

What I find odd is that the graph of $\sqrt[5]{x}$ comes to no such points, is both continuous and differentiable over this interval, but still fails to satisfy the MVT.

Why is this?