If $f(x)=\left\{\begin{array}{cc}\frac{2 x e^{\frac{1}{2 x}}-3 x e^{\frac{-1}{2 x}}}{e^{\frac{1}{2 x}}+4…
If $f(x)=\left\{\begin{array}{cc}\frac{2 x e^{\frac{1}{2 x}}-3 x e^{\frac{-1}{2 x}}}{e^{\frac{1}{2 x}}+4 e^{\frac{-1}{2 x}}} & \text { if } x \neq 0 \\ 0 & \text { if } x=0\end{array}\right.$
is a real valued function then
$f^{\prime}\left(0^{+}\right)=\frac{-3}{4}$
$f^{\prime}\left(0^{-}\right)=2$
$f$ is not differentiable at $x=0$
$f$ is differentiable at $x=0$
Solution
Given the function
$f(x)=\left\{\begin{array}{cc}\frac{2 x e^{\frac{1}{2 x}}-3 x e^{\frac{-1}{2 x}}}{e^{\frac{1}{2 x}}+4 e^{-\frac{1}{2 x}}} & \text { if } x \neq 0 \\ 0 & \text { if } x=0\end{array}\right.$
$\begin{aligned} & \text { L.H.D. }=f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h} \\ & =\lim _{h \rightarrow 0} \frac{f(-h)-f(0)}{-h}\end{aligned}$
$=\lim _{h \rightarrow 0} \frac{-2 h e^{\frac{-1}{2 h}}+3 h e^{\frac{1}{2 h}}-0}{-h\left(e^{\frac{-1}{2 h}}+4 e^{\frac{1}{2 h}}\right)}$
$\left.=\lim _{h \rightarrow 0} \frac{2 e^{\frac{-1}{2 h}}-3 e^{\frac{1}{2 h}}}{\left(e^{\frac{-1}{2 h}}+4 e^{\frac{1}{2 h}}\right.}\right)=\lim _{h \rightarrow 0}\left(\frac{2 e^{\frac{-2}{2 h}}-3}{e^{\frac{-2}{2 h}}+4}\right)$
$\begin{aligned} & =\frac{0-3}{0+4}=\frac{-3}{4} \text { and, R.H.D. }=\mathrm{f}^{\prime}\left(0^{+}\right) \\ & =\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{f(h)-0}{h}\end{aligned}$
$\begin{aligned} & =\frac{0-3}{0+4}=\frac{-3}{4} \text { and, R.H.D. }=\mathrm{f}^{\prime}\left(0^{+}\right) \\ & =\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{f(h)-0}{h}\end{aligned}$
$\left.=\lim _{h \rightarrow 0} \frac{2 h e^{\frac{1}{2 h}}-3 h e^{\frac{-1}{2 h}}}{h\left(e^{\frac{1}{2 h}}+4 e^{\frac{-1}{2 h}}\right.}\right)=\lim _{h \rightarrow 0}\left(\frac{2-3 e^{-\frac{1}{h}}}{1+4 e^{\frac{-1}{h}}}\right)=\frac{2-0}{1+0}=2$
Since $\left(f^{\prime}\left(0^{-}\right) \neq f^{\prime}\left(0^{+}\right)\right.$
So, $f(x)$ is not differentiable at $x=0$