If a function $f$ is defined by : $\begin{aligned} f(x) & =0, & \text{ when } x=1, \\ & =x^{3}-1, & \text{…
If a function $f$ is defined by :
$\begin{aligned}
f(x) & =0, & \text{ when } x=1, \\
& =x^{3}-1, & \text{ when } 1 < x < \infty, \\
& =x-1, & \text{ when } -\infty < x < 1,
\end{aligned}$
then at $x=1, f$ is
continuous and differentiable
continuous but not differentiable
discontinuous and differentiable
discontinuous and not differentiable
Solution
We have,
$
f(x)=\begin{cases}
x-1, & -\infty < x < 1 \\
0, & x=1 \\
x^{3}-1, & 1 < x < \infty
\end{cases}
$
Now, (LHL at $x=1$)
$\begin{aligned}
& =\lim _{x \rightarrow 1}(x-1) \\
& =0
\end{aligned}$
(RHL at $x=1$) $=\lim _{x \rightarrow 1} (x^{3}-1) = 0$ and $f(1)=0$
Therefore, LHL = RHL = $f(1)$
So, $f(x)$ is continuous at $x=1$
Now, (LHD at $x=1$)
$=\lim _{x \rightarrow 1} \frac{(x-1)-0}{x-1}=1$
and (RHD at $x=1$)
$\begin{aligned}
& =\lim _{x \rightarrow 1} \frac{(x^{3}-1)-0}{x-1} \\
& =\lim _{x \rightarrow 1}(x^{2}+x+1)=3
\end{aligned}$
Therefore, LHD $\neq$ RHD
So, $f(x)$ is not differentiable at $x=1$