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

Answer: (A)

$f\left(x\right)=e^{|x|}$
$f(x)=\{\begin{array}{ll}{e}^{-x},& \text{ x<0 }\\ e^x,& \text{ x≥0 }\end{array}$
$\therefore e^x$ is continuous for all $x>0$ and $e^{-x}$ is continuous for all $x<0$.
Now, $\underset{x\to 0^{-}}{\lim }f(x)=$ $\underset{x\to 0^{-}}{\lim }{e}^{-x}=e^0=1$
$\underset{x\to 0^{+}}{\lim }f(x)=$$\underset{x\to 0^{-}}{\lim }{e}^x=e^0=1$
$\underset{x\to 0^{-}}{\lim }f(x)=$$\underset{x\to 0^{+}}{\lim }f(x)=$$\underset{x\to 0}{\lim }f(x)=f(0)$
Hence, $f(x)$ is continuous everywhere.
$L{f}^{\prime }(0)=\underset{h\to 0^{-}}{\lim }$ $\frac{f\left(0-h\right)-f\left(0\right)}{h}$
$=\underset{h\to 0^{-}}{\lim }$$\frac{e^{-h}-1}{h}=$$\underset{h\to 0^{-}}{\lim }$$\frac{\left(1-h+\frac{h^2}{2!}-\frac{h^3}{3!}+...\right)-1}{h}$
$=\underset{h\to 0^{-}}{\lim }$ $\left(-1+\frac{h}{2!}-\frac{h^2}{3!}\right)=-1$
$R{f}^{\prime }(0)=\underset{h\to 0^{+}}{\lim }$$\frac{f\left(0+h\right)-f\left(0\right)}{h}$$=\underset{h\to 0^{+}}{\lim }$$\frac{e^h-1}{h}$
$=\underset{h\to 0^{+}}{\lim }$$\frac{\left(1+h+\frac{h^2}{2!}+\frac{h^3}{3!}+\dots \right)-1}{h}$
$=\underset{h\to 0^{+}}{\lim }$$\left(1+\frac{h}{2!}+\frac{h^2}{3!}+\dots \right)=1$
$L(f^{\prime }(0))\ne R(f^{\prime }(0))$.
So $f(x)$ is not differentiable at $x=0$.