Question

At the point $x=2,$ the function

$f\left(x\right)=\{\begin{array}{cc}{x}^3-8& 2<x<\infty \\ x-2& -\infty <x\le 2\end{array}$ 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
$\underset{x\to 2-}{lim}f\left(x\right)=\underset{x\to 2^{-}}{lim}(x-2)=0$
$\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}(x^3-8)=0$

Also $f\left(2\right)=0$
Thus $\underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}f\left(x\right)=f\left(2\right)$

$\therefore$ f is continuous at $x=2$

and $L{f}^{\prime }\left(2\right)=1andR{f}^{\prime }\left(2\right)=12$

$\therefore f$ is not differentiable at $x=2.$