If a function f is defined by:begin{cases} f( | vedclass

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(B) We have the function defined as:$f(x) = \begin{cases} x-1, & -\infty < x < 1 \ 0, & x=1 \ x^3-1, & 1 < x < \infty \end{cases}$First,we check for

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continuous but not differentiable

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(B) We have the function defined as:$f(x) = \begin{cases} x-1, & -\infty First,we check for continuity at $x=1$:Left Hand Limit $(LHL)$ = $\lim_{x ightarrow 1^-} (x-1) = 1-1 = 0$.Right Hand Limit $(RHL)$ = $\lim_{x ightarrow 1^+} (x^3-1) = 1^3-1 = 0$.Value of the function $f(1) = 0$.Since $LHL$ = $RHL$ = $f(1)$,the function is continuous at $x=1$.Next,we check for differentiability at $x=1$:Left Hand Derivative $(LHD)$ = $\lim_{x ightarrow 1^-} \frac{f(x)-f(1)}{x-1} = \lim_{x ightarrow 1^-} \frac{(x-1)-0}{x-1} = 1$.Right Hand Derivative $(RHD)$ = $\lim_{x ightarrow 1^+} \frac{f(x)-f(1)}{x-1} = \lim_{x ightarrow 1^+} \frac{(x^3-1)-0}{x-1} = \lim_{x ightarrow 1^+} \frac{(x-1)(x^2+x+1)}{x-1} = \lim_{x ightarrow 1^+} (x^2+x+1) = 1^2+1+1 = 3$.Since $LHD$ $
eq$ $RHD$,the function is not differentiable at $x=1$.

If a function $f$ is defined by:
$\begin{cases} f(x) = x-1, & \text{when } -\infty < x < 1 \\ f(x) = 0, & \text{when } x=1 \\ f(x) = x^3-1, & \text{when } 1 < x < \infty \end{cases}$
then at $x=1$,$f$ is:

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If a function $f$ is defined by:$\begin{cases} f(x) = x-1, & \text{when } -\infty < x < 1 \\ f(x) = 0, & \text{when } x=1 \\ f(x) = x^3-1, & \text{when } 1 < x < \infty \end{cases}$then at $x=1$,$f$ is: