Question

Rolle's theorem is not applicable to the function f(x)=|x| defined on [-1,1] because

• A. f is not continuous on [ –1, 1]
• B. f is not differentiable on (–1,1)
• C. $f(-1)\ne f\left(1\right)$
• D. $f(-1)=f\left(1\right)\ne 0$

Answer 1

The correct option is B f is not differentiable on (–1,1)

$f\left(x\right)=\{\begin{array}{cc}-x,& When-1\le x<0\\ x,& when0\le x\le 1\end{array}$
Clearly f(-1)=|-1|=1 =f(1)
But $R{f}^{\prime }\left(0\right)={lim}_{h\to 0}\frac{f(0+h)-f\left(0\right)}{h}={lim}_{h\to 0}\frac{|h|}{h}={lim}_{h\to 0}\frac{h}{h}=1$
$L{f}^{\prime }\left(0\right)={lim}_{h\to 0}\frac{f(0-h)-f\left(0\right)}{h}={lim}_{h\to 0}\frac{|-h|}{-h}={lim}_{h\to 0}\frac{h}{-h}=-1\therefore R{f}^{\prime }\left(0\right)\ne L{f}^{\prime }\left(0\right)$
Hence it is not differentiable on (-1.1).