Question

$f\left(x\right)=\left|x\right|$ in $[-1,1]$ verify Rolle's theorem.

Answer 1

$f\left(x\right)=\left|x\right|$
Then, $f\left(x\right)=\{\begin{array}{c}-x,-1\le x<0\\ x,0\le x\le 1\end{array}$
$\because$ Modulus function is continuous but it is not differentiable.
At $x=0$, Right hand derivative
$=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}$
$=\underset{h\to 0}{lim}\frac{|h|-0}{h}$
$=\underset{h\to 0}{lim}\frac{h}{h}=1$
and at $x=0$, Left hand derivative
$=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}$
$=\underset{h\to 0}{lim}\frac{|-h|}{-h}$
$=\underset{h\to 0}{lim}\frac{h}{-h}=-1$
At $x=0,$ R.H.D $\ne$ L.H.D
$R{f}^{{}^{\prime }}\left(0\right)\ne L{f}^{{}^{\prime }}\left(0\right)$
Function is not differentiable at $x=0$. So, Rolle's theorem does not satisfied.