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Rolle's theorem is not applicable to the function $f(x) \neq x$ defined on $[-1,1]$ because
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Verified Answer
The correct answer is:
$f$ is not differentiable on $(-1,1)$
$f(x)=\left\{\begin{array}{c}
-x, \text { when }-1 \leq x \lt 0 \\
x, \text { when } 0 \leq x \leq 1\end{array}\right.$
Clearly $f(-1)=|-1|=1=f(1)$
$\begin{aligned}& \text { But } R f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{|h|}{h}=\lim _{h \rightarrow 0} \frac{h}{h}=1 \\
& L f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{|-h|}{-h}=\lim _{h \rightarrow 0} \frac{h}{-h}=-1 \\
& \therefore R f^{\prime}(0) \neq L f^{\prime}(0)\end{aligned}$
But
Hence it is not differentiable on $(-1,1)$.
-x, \text { when }-1 \leq x \lt 0 \\
x, \text { when } 0 \leq x \leq 1\end{array}\right.$
Clearly $f(-1)=|-1|=1=f(1)$
$\begin{aligned}& \text { But } R f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{|h|}{h}=\lim _{h \rightarrow 0} \frac{h}{h}=1 \\
& L f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{|-h|}{-h}=\lim _{h \rightarrow 0} \frac{h}{-h}=-1 \\
& \therefore R f^{\prime}(0) \neq L f^{\prime}(0)\end{aligned}$
But
Hence it is not differentiable on $(-1,1)$.
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