Question

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

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

Answer 1

Answer: (B)

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