Rolle's theorem is not applicable to the | vedclass

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Question

(b) $f(x) = \left\{ \begin{array}{l} - x,\,{m{when\,\, -1}} \le x < 0\{m{ }}x,\;{m{when\,\,}}\;{m{0}} \le x \le {m{1}}\end{array} ight

Vedclass · Accepted Answer

$f$  is not differentiable on $ (-1,1)$

Vedclass · Answer

(b) $f(x) = \left\{ \begin{array}{l} - x,\,{m{when\,\, -1}} \le x < 0\{m{ }}x,\;{m{when\,\,}}\;{m{0}} \le x \le {m{1}}\end{array} ight.$

Clearly $f( - 1) = | - 1| = 1 = f(1)$

But $Rf'(0) = \mathop {\lim }\limits_{h 	o 0}\frac{{f(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{h 	o 0} \frac{{|h|}}{h}$

$ = \mathop {\lim }\limits_{h 	o 0} \frac{h}{h} = 1$

$Lf'(0) = \mathop {\lim }\limits_{h 	o 0} \frac{{f(0 - h) - f(0)}}{h} = \mathop {\lim }\limits_{h 	o 0} \frac{{| - h|}}{{ - h}}$

$ = \mathop {\lim }\limits_{h 	o 0} \frac{h}{{ - h}} = - 1$

$	herefore Rf'(0) 
e Lf'(0)$

Hence it is notdifferentiable on $( - 1,\,\,1)$.

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

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