Question

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

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

Answer 1

Answer: (B)

continuous but not differentiable

$f\left(1\right)=0,f\left(x\right)$ is continuous.

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

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

$\Rightarrow f\left(x\right)$ is not differentiable.