At x=1,the function f(x)=begin{cases} x^{3}-1 | vedclass

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(B) Given $f(x)=\begin{cases} x^{3}-1, & 1 < x < \infty \ x-1, & -\infty < x \leq 1 \end{cases}$First,we check the continuity at $x=1$.$RHL = \lim_{x

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continuous and non-differentiable.

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(B) Given $f(x)=\begin{cases} x^{3}-1, & 1 First,we check the continuity at $x=1$.$RHL = \lim_{x ightarrow 1^{+}} f(x) = \lim_{x ightarrow 1^{+}} (x^{3}-1) = 1^{3}-1 = 0$$LHL = \lim_{x ightarrow 1^{-}} f(x) = \lim_{x ightarrow 1^{-}} (x-1) = 1-1 = 0$$f(1) = 1-1 = 0$Since $LHL = RHL = f(1)$,the function is continuous at $x=1$.Next,we check the differentiability at $x=1$.$f'(x) = \begin{cases} 3x^{2}, & 1 $LHD = \lim_{x ightarrow 1^{-}} f'(x) = 1$$RHD = \lim_{x ightarrow 1^{+}} f'(x) = 3(1)^{2} = 3$Since $LHD 
eq RHD$,the function is non-differentiable at $x=1$.Therefore,the function is continuous and non-differentiable.

At $x=1$,the function $f(x)=\begin{cases} x^{3}-1, & 1 < x < \infty \\ x-1, & -\infty < x \leq 1 \end{cases}$ is

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