At the point \(x=1\), the function \(f(x)=\left\{\begin{array}{l}x^{3}-1,1 < x < \infty \ x-1,-\infty < x \leq 1\end{array}\right.\) is

LHL \(=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}(x-1)=0\)
\(\mathrm{RHL} =\lim _{x \rightarrow 1^{+}} f(x) \)
\( =\lim _{x \rightarrow 1}\left(x^{3}-1\right)=0 \)
\( \text { Also, } f(1) =1-1=0\)
\(\therefore f\) is continuous at \(x=1\) Now, \(L f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}\)
\(=\lim _{h \rightarrow 0} \frac{(1-h)-1-0}{-h} \)
\( =\lim _{h \rightarrow 0} \frac{-h}{-h}=1 \)
\( \text { and } R f^{\prime}(1) =\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}\)
\(
\begin{array}{l}
=\lim _{h \rightarrow 0} \frac{(1+h)^{3}-1-0}{h} \\
=\lim _{h \rightarrow 0} \frac{1+h^{3}+3 h+3 h^{2}-1}{h} \\
=\lim _{h \rightarrow 0} h^{2}+3+3 h=3
\end{array}
\)
\(\therefore f(x)\) is not differentiable at \(x=1\)