Question

If f(x) = \begin{cases} xe^{-\left(\frac{1}{\left|x\right|}+\frac{1}{x}\right)}, & \text{x \ne0} \\\[2ex] 0, & \text{x=0} \end{cases} then $f (x)$ is

• A. continuous as well as differentiable for all $x$
• B. continuous for all x but not differentiable at $x = 0$
• C. neither differentiable nor continuous at $x = 0$
• D. discontinuous everywhere

Answer 1

Answer: (B)

continuous for all x but not differentiable at $x = 0$

Answer 2

$f' (0)$ $f' (0 - h) = 1$ $f' (0 + h) = 0$ $LHD ? RHD.$