Question
Discuss the applicability of the Rolle's theorem for the following function on the indicated interval
$\text{f}(\text{x})=\text{x}^{\frac{2}{3}}\text{ on }[-1,1]$

Answer

Here, $\text{f}(\text{x})=\text{x}^{\frac{2}{3}}\text{ on }[-1,1]$
$\text{f}'(\text{x})=\frac{2}{3\text{x}^{\frac{1}{3}}}$
$\text{f}'(0)=\frac{2}{3(0)^{\frac{1}{3}}}$
$\text{f}'(0)=\infty$
So, f'(x) does not exist at $\text{x}=0\in(-1,1)$
⇒ f(x) is not differentialble in $\text{x}\in(-1,1)$
So, Rolle's theorem is not applicable on f(x) in [-1, 1].

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