Question

Discuss the continuity and differentiability of f (x) = e^|x| .

Answer 1

Given: $f\left(x\right)=e^{\left|x\right|}$
$\Rightarrow f\left(x\right)=\left\{\begin{array}{l}{e}^x,x\ge 0\\ e^{-x},x<0\end{array}\right.$

Continuity:

(LHL at x = 0)

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

(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 $f\left(0\right)=e^0=1$

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

Hence,function is continuous at x = 0 .

Differentiability at x = 0.

(LHD at x = 0)

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

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

LHD at (x = 0)$\ne$RHD at (x = 0)

Hence the function is not differentiable at x = 0.