Question

$f\left(x\right)=\left|x\right|$ in the interval [-1, 1] Is Rolle's Theorem applicable?

Answer 1

$=f\left(x\right)=\left|x\right|$ in interal [-1, 1]
We show that f(x) is continuous in [-1, 1]
$=f\left(1\right)=\left|1\right|=1,f(-1)=|-1|=1,f\left(1\right)=f(-1)=1.$
Also f(x) is continous in [-1,1]. for all values of x Now, $R{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}$
$=\underset{h\to 0}{lim}\frac{[0+h]-\left|0\right|}{h}\underset{h\to 0}{lim}=\frac{h-0}{h}=1$ and
$L{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}$
$=\underset{h\to 0}{lim}\frac{|0+h|-\left|0\right|}{-h}=\underset{h\to 0}{lim}\frac{h-0}{-h}=-1$
$R{f}^{\prime }\left(0\right)\ne L{f}^{\prime }\left(0\right).$
Therefore f'(0) does not exist.

Hence, the function f(x) is not differentiable in open interval (0, 2) and so Rolle's theorem is not applicable to given fucntion f(x) in [-1, 1].