Question

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

We have $R{f}^{\prime }\left(1\right)={lim}_{h\to 0}\frac{f(1+h)-f\left(1\right)}{h}$
$={lim}_{h\to 0}\frac{\left\{\right(1+h{)}^3-1\}}{h}=3$
$L{f}^{\prime }\left(1\right)={lim}_{h\to 0}\frac{f(1-h)-f\left(1\right)}{-h}={lim}_{h\to 0}\frac{\left\{\right(1-h)-1\}-0}{-h}=1$
$\therefore R{f}^{\prime }\left(1\right)\ne L{f}^{\prime }\left(1\right)\Rightarrow f\left(x\right)$ is not differentiable at x = 1.
Now, f(1+0) = ${lim}_{h\to 0}$ f (1 + h) = 0
and f(1 - 0) = ${lim}_{h\to 0}$ f (1 - h) = 0
$\therefore f(1+0)=f(1-0)=f\left(0\right)\Rightarrow f\left(x\right)$ is continuous at x = 1. Hence at x = 1, f (x) is continuous and not differentiable.