Question

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

• A. Continuous and differentiable
• B. Continuous and not differentiable
• C. Discontinuous and differentiable
• D. Discontinuous and not differentiable

Answer 1

The correct option is B Continuous and not differentiable

LHL $=\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1}{lim}(x-1)=0$

RHL $=\underset{x\to 1^{+}}{lim}=\underset{x\to 1}{lim}(x^3-1)=0$

Also, $f\left(1\right)=1-1=0$

$\therefore f$ is continuous at $x=1$

Now, $L{f}^{\prime }\left(1\right)=\underset{h\to 0}{lim}\frac{f(1-h)-f\left(1\right)}{-h}=\underset{h\to 0}{lim}\frac{(1-h)-1-0}{-h}=1$

and $R{f}^{\prime }\left(1\right)=\underset{h\to 0}{lim}\frac{f(1+h)-f\left(1\right)}{h}=\underset{h\to 0}{lim}\frac{(1+h{)}^3-1-0}{h}=\underset{h\to 0}{lim}{h}^2+3+3h=3$

Clearly, $L{f}^{\prime }\left(1\right)\ne R{f}^{\prime }\left(1\right)$

$\therefore f\left(x\right)$ is not differentiable at $x=1$