Rolle's theorem is not applicable to the func | vedclass

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(B) The function is defined as $f(x) = \begin{cases} -x, & 	ext{if } -1 \le x < 0 \ x, & 	ext{if } 0 \le x \le 1 \end{cases}$.First,we check the co

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

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(B) The function is defined as $f(x) = \begin{cases} -x, & 	ext{if } -1 \le x First,we check the conditions for Rolle's theorem:$1$. $f(x)$ is continuous on $[-1, 1]$.$2$. $f(-1) = |-1| = 1$ and $f(1) = |1| = 1$,so $f(-1) = f(1)$.$3$. We check for differentiability at $x = 0$:Right-hand derivative: $Rf'(0) = \lim_{h 	o 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^+} \frac{|h|}{h} = \lim_{h 	o 0^+} \frac{h}{h} = 1$.Left-hand derivative: $Lf'(0) = \lim_{h 	o 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^-} \frac{|h|}{h} = \lim_{h 	o 0^-} \frac{-h}{h} = -1$.Since $Rf'(0) 
eq Lf'(0)$,the function is not differentiable at $x = 0$,which lies in the interval $(-1, 1)$.Therefore,Rolle's theorem is not applicable because the function is not differentiable on $(-1, 1)$.

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

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