Question

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

• A. continuous and differentiable
• B. continuous and not differentiable
• C. discontinuous and differentiable
• D. discontinuous and not differentiable

Answer 1

Answer: (B)

$LHL= \displaystyle \lim _{x \rightarrow 1^{-}} f(x)$
$= \displaystyle \lim _{x \rightarrow 1}(x-1)=0$
$RHL= \displaystyle \lim _{x \rightarrow 1^{+}} f(x)$
$= \displaystyle \lim _{x \rightarrow 1}\left(x^{3}-1\right)=0$
Also, $f(1)=1- 1= 0$
$\therefore f$ is continuous at $x=1$
Now, $L f'(1) = \displaystyle \lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}$
$=\displaystyle \lim _{h \rightarrow 0} \frac{(1-h)-1-0}{-h} $
$=\displaystyle \lim _{h \rightarrow 0} \frac{-h}{-h}=1$
and $R f'(1)= \displaystyle \lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$
$= \displaystyle \lim _{h \rightarrow 0} \frac{(1+h)^{3}-1-0}{h}$
$= \displaystyle \lim _{h \rightarrow 0} \frac{1+h^{3}+3 h+3 h^{2}-1}{h}$
$= \displaystyle \lim _{h \rightarrow 0} h^{2}+3+3 h=3$
Clearly, $L f'(1) \neq R f'(1)$
$\therefore f(x)$ is not differentiable at $x=1$