The function $f(x)=e^{|x|}$ is
(a) continuous every where but not differentiable at x = 0
(b) continuous and differentiable everywhere
(c) not continuous at x = 0
(d) none of these
The given function is $f(x)=e^{|x|}$.
We know
If $f$ is continuous on its domain $D$, then $|f|$ is also continuous on $D$.
Now, the identity function $p(x)=x$ is continuous everywhere.
So, $g(x)=|p(x)|=|x|$ is also continuous everywhere.
Also, the exponential function $a^{x}, a>0$ is continuous everywhere.
So, $h(x)=e^{x}$ is continuous everywhere.
The composition of two continuous functions is continuous everywhere.
$\therefore f(x)=(h o g)(x)=e^{|x|}$ is continuous everywhere.
Now,
$g(x)=|x|= \begin{cases}x, & x \geq 0 \\ -x, & x<0\end{cases}$
$L g^{\prime}(0)=\lim _{h \rightarrow 0} \frac{g(0-h)-g(0)}{-h}$
$\Rightarrow L g^{\prime}(0)=\lim _{h \rightarrow 0} \frac{-(-h)-0}{-h}$
$\Rightarrow L g^{\prime}(0)=\lim _{h \rightarrow 0} \frac{h}{-h}$
$\Rightarrow L g^{\prime}(0)=-1$
And
$R g^{\prime}(0)=\lim _{h \rightarrow 0} \frac{g(0+h)-g(0)}{h}$
$\Rightarrow R g^{\prime}(0)=\lim _{h \rightarrow 0} \frac{h-0}{h}$
$\Rightarrow R g^{\prime}(0)=\lim _{h \rightarrow 0} \frac{h}{h}$
$\Rightarrow R g^{\prime}(0)=1$
$\therefore L g^{\prime}(0) \neq R g^{\prime}(0)$
So, $g(x)=|x|$ is not differentiable at $x=0$.
We know
The exponential function $a^{x}, a>0$ is differentiable everywhere.
So, $h(x)=e^{x}$ is differentiable everywhere.
We know that, the composition of differentiable functions is differentiable.
Now, $e^{x}$ is differentiable everywhere, but $|x|$ is not differentiable at $x=0$.
$\therefore f(x)=(h o g)(x)=e^{|x|}$ is differentiable everywhere except at $x=0$
Thus, the function $f(x)=e^{|x|}$ is continuous every where but not differentiable at $x=0$.
Hence, the correct answer is option (a).
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