Question

Let $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}0,& x<0\end{array}\\ \begin{array}{cc}{x}^2,& x\ge 0\end{array}\end{array}$. Then for all $x$

• A. $f^{\prime \prime }$ is differentiable
• B. $f$ is differentiable
• C. $f^{\prime }$ is not continuous
• D. $f$ is not continuous

Answer 1

Answer: (B)

$f$ is differentiable

For all $x,f\left(x\right)$ is differentiable, but the only point where differentiability is to be examined is $x=0.$

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

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

$\Rightarrow f$ is differentiable.