Question

Which among the following function is continuous everywhere in its domain but has at least one point where it is not differentiable

• A. $f\left(x\right)=x^{\frac{1}{3}}$
• B. $f\left(x\right)=\frac{|x|}{x}$
• C. $f\left(x\right)=e^{-x}$
• D. $f\left(x\right)=\tan x$

Answer 1

The correct option is A $f\left(x\right)=x^{\frac{1}{3}}$

For $f\left(x\right)=x^{\frac{1}{3}}$
It is continuous everywhere in its own domain, $x\in [0,\infty )$
$f^{\prime }\left(x\right)=\frac{1}{3}\frac{1}{x^{\frac{2}{3}}}$
$\Rightarrow f\left(x\right)$ is not differentiable but it has vertical tangent at $x=0.$

For $f\left(x\right)=\frac{|x|}{x}$
Since for $x<0,f\left(x\right)=-1$ and for $x>0,f\left(x\right)=+1$
It is continuous and differentiable $\forall x\in R-\left\{0\right\}$

For $f\left(x\right)=e^{-x},$$f^{\prime }\left(x\right)=-e^{-x}$
So, it is continuous and differentiable $x\in R$

For $f\left(x\right)=\tan x$
By using graph we can say that it is continuous and differentiable $\forall x\in R-\left\{(2k-1)\frac{\pi }{2}\right\}\text{where k}\in I$