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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?

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1 Answer 1

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

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  • $\begingroup$ Correct, but is a function "not differentiable" at a point if the derivative fails to exist due to a division by zero? $\endgroup$
    – alxmke
    Mar 26, 2015 at 7:05
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    $\begingroup$ Look at the definition of differentiable. Derivatives must be finite. $\endgroup$
    – Karl
    Mar 26, 2015 at 7:13

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