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

Question

Discuss the continuity and differentiability of $\text{f(x)}=\text{e}^{|\text{x}|}.$

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Answer

Given:
$\text{f(x)}=\text{e}^{|\text{x}|}$
$\Rightarrow\text{f(x)}=\begin{cases}\text{e}^\text{x},&\text{x}\geq0\\\text{e}^{-\text{x}},&\text{x}<0\end{cases}$
f is Continuity:
(LHL at x = 0)
$\lim_\limits{\text{x}\rightarrow0{^-}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{e}^{-(0-\text{h})}$
$=\lim_\limits{\text{h}\rightarrow0}\text{e}^{-\text{h}}$
$=1$
(RHL at x = 0)
$\lim_\limits{\text{x}\rightarrow0{^{+}}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{e}^{(0+\text{h})}$
$=1$
and f(0)
$=\text{e}^0=1$
Thus, $\lim_\limits{\text{x}\rightarrow0^{-}}-\text{f(x)}=\lim_\limits{\text{h}\rightarrow0^{+}}-\text{f(x)}=\text{f(0)}$
Hence, function is continuous at x = 0.
Differentiability at x = 0.
(LHL at x = 0)
$=\lim_\limits{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0-\text{h})-\text{f}(0)}{0-\text{h}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{-(0-\text{h})}-1}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^\text{h}-1}{-\text{h}}=-1\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^\text{h}-1}{\text{h}}=1\Big]$
(RHL at x = 0)
$=\lim_\limits{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0-\text{h})-\text{f}(0)}{0+\text{h}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{-(0-\text{h})}-1}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^\text{h}-1}{\text{h}}=1\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^\text{h}-1}{\text{h}}=1\Big]$
LHL at (x = 0) $\neq$ RHL at (x = 0)
Hence the function is not differentiable at x = 0.

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