Question

Discuss the applicapibility of Rolle's Theorem to the function $f\left(x\right)=x^{2/3}$ in (-1,1).

• A. Yes Rolle's theorem is applicable and the stationary point $x=0$
• B. Yes Rolle's theorem is applicable and the stationary point $x=\frac{1}{2}$
• C. No Rolle's theorem is not applicable in the given interval
• D. none of these

Answer 1

The correct option is C No Rolle's theorem is not applicable in the given interval

$f\left(x\right)=x^{\frac{2}{3}}\Rightarrow f^{\prime }\left(x\right)=\frac{2}{x}{x}^{-\frac{1}{3}}$
$\therefore \underset{x\to 0}{lim}{f}^{\prime }\left(x\right)=\underset{x\to 0}{lim}\frac{2}{3}{(0+h)}^{-\frac{1}{3}}=\infty$
$R{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\left\{\frac{f(0+h)-f\left(0\right)}{h}\right\}=\underset{h\to 0}{lim}\left\{\frac{h^{\frac{2}{3}}-1}{h}\right\}=+\infty$
$L{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\left\{\frac{f(0+h)-f\left(0\right)}{-h}\right\}=f^{\prime }\left(0\right)=\underset{h\to 0}{lim}\left\{\frac{{(-h)}^{\frac{2}{3}}}{h}\right\}=-\infty$
$\therefore L{f}^{\prime }\left(0\right)\ne R{f}^{\prime }\left(0\right)$
$\therefore f^{\prime }\left(0\right)$ does not exist showing that $f^{\prime }\left(x\right)$ does not exists in the open interval $(-1,1)$
Hence, Rolle's Theorem is not applicable
although $f(-1)=f\left(1\right)=1$ and $f\left(x\right)$ is continuous in the closed interval $(-1,1)$