Question

The function f (x) = e^−|x| is
(a) continuous everywhere but not differentiable at x = 0
(b) continuous and differentiable everywhere
(c) not continuous at x = 0
(d) none of these

Answer 1

(a) continuous everywhere but not differentiable at x = 0



Given: $$\begin{gathered} f\left(x\right)=e^{-\left|x\right|}=\left\{\begin{array}{l}{e}^x,x\ge 0\\ e^{-x},x<0\end{array}\right. \\ Continuity: \\ \underset{x\to 0^{-}}{\lim }f\left(x\right) \\ =\underset{h\to 0}{\lim }f(0-h) \\ =\underset{h\to 0}{\lim }{e}^{-(0-h)} \\ =\underset{h\to 0}{\lim }{e}^h \\ =1 \end{gathered}$$

RHL at x = 0

$$\begin{gathered} \underset{x\to 0^{+}}{\lim }f\left(x\right) \\ =\underset{h\to 0}{\lim }f(0+h) \\ =\underset{h\to 0}{\lim }{e}^{(0+h)} \\ =1 \end{gathered}$$

and f(0) = $f\left(0\right)=e^0=1$
Thus, $\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)=f$

Hence, function is continuous at x = 0

Differentiability at x = 0

(LHD at x = 0)

$$\begin{gathered} \underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0} \\ =\underset{h\to 0}{\lim }\frac{f(0-h)-f\left(0\right)}{0-h-0} \\ =\underset{h\to 0}{\lim }\frac{e^{-(0-h)}-1}{-h} \\ =\underset{h\to 0}{\lim }\frac{e^h}{h} \\ =\infty \end{gathered}$$

Therefore, left hand derivative does not exist.
Hence, the function is not differentiable at x = 0.