Question

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

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Solution

## Given: $f\left(x\right)={e}^{\left|x\right|}$ $⇒f\left(x\right)=\left\{\begin{array}{l}{e}^{x},x\ge 0\\ {e}^{-x},x<0\end{array}\right\\phantom{\rule{0ex}{0ex}}$ Continuity: (LHL at x = 0) $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}{e}^{-\left(0-h\right)}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}{e}^{h}\phantom{\rule{0ex}{0ex}}=1$ (RHL at x = 0) $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}{e}^{\left(0+h\right)}\phantom{\rule{0ex}{0ex}}=1$ and $f\left(0\right)={e}^{0}=1$ Thus, $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$ Hence,function is continuous at x = 0 . Differentiability at x = 0. (LHD at x = 0) $\underset{x\to {0}^{-}}{\mathrm{lim}}\frac{f\left(x\right)-f\left(0\right)}{x-0}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(0-h\right)-f\left(0\right)}{0-h-0}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{{e}^{-\left(0-h\right)}-1}{-h}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{{e}^{h}-1}{-h}\phantom{\rule{0ex}{0ex}}=-1\left[\because \underset{h\to 0}{\mathrm{lim}}\frac{{e}^{h}-1}{h}=1\right]$ (RHD at x = 0) $\underset{x\to {0}^{+}}{\mathrm{lim}}\frac{f\left(x\right)-f\left(0\right)}{x-0}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(0+h\right)-f\left(0\right)}{0+h-0}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{{e}^{\left(0+h\right)}-1}{h}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{{e}^{h}-1}{h}\phantom{\rule{0ex}{0ex}}=1\left[\because \underset{h\to 0}{\mathrm{lim}}\frac{{e}^{h}-1}{h}=1\right]$ LHD at (x = 0)$\ne$RHD at (x = 0) Hence the function is not differentiable at x = 0.

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