calculus - 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?

Answer

No, look at zero! The deriavtive there does not exist. So the conditions of the MVT are not fulfilled.