If f(x)=begin{cases} frac{2 x e^{frac{1}{2 x} | vedclass

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(C) Given the function $f(x) = \begin{cases} \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}}} & x 
eq 0 \

Vedclass · Accepted Answer

$f$ is not differentiable at $x=0$

Vedclass · Answer

(C) Given the function $f(x) = \begin{cases} \frac{2 x e^{\frac{1}{2 x}}-3 x e^{\frac{-1}{2 x}}}{e^{\frac{1}{2 x}}+4 e^{-\frac{1}{2 x}}} & x 
eq 0 \ 0 & x=0 \end{cases}$.First,calculate the Left Hand Derivative ($L$.$H$.$D$.) at $x=0$:$f^{\prime}(0^{-}) = \lim_{h ightarrow 0^{+}} \frac{f(0-h)-f(0)}{-h} = \lim_{h ightarrow 0^{+}} \frac{f(-h)-0}{-h}$$= \lim_{h ightarrow 0^{+}} \frac{-2h e^{\frac{-1}{2h}} + 3h e^{\frac{1}{2h}}}{-h(e^{\frac{-1}{2h}} + 4e^{\frac{1}{2h}})} = \lim_{h ightarrow 0^{+}} \frac{2e^{\frac{-1}{2h}} - 3e^{\frac{1}{2h}}}{e^{\frac{-1}{2h}} + 4e^{\frac{1}{2h}}}$Divide numerator and denominator by $e^{\frac{1}{2h}}$:$= \lim_{h ightarrow 0^{+}} \frac{2e^{\frac{-1}{h}} - 3}{e^{\frac{-1}{h}} + 4} = \frac{0 - 3}{0 + 4} = -\frac{3}{4}$.Next,calculate the Right Hand Derivative ($R$.$H$.$D$.) at $x=0$:$f^{\prime}(0^{+}) = \lim_{h ightarrow 0^{+}} \frac{f(0+h)-f(0)}{h} = \lim_{h ightarrow 0^{+}} \frac{f(h)-0}{h}$$= \lim_{h ightarrow 0^{+}} \frac{2h e^{\frac{1}{2h}} - 3h e^{\frac{-1}{2h}}}{h(e^{\frac{1}{2h}} + 4e^{\frac{-1}{2h}})} = \lim_{h ightarrow 0^{+}} \frac{2e^{\frac{1}{2h}} - 3e^{\frac{-1}{2h}}}{e^{\frac{1}{2h}} + 4e^{\frac{-1}{2h}}}$Divide numerator and denominator by $e^{\frac{1}{2h}}$:$= \lim_{h ightarrow 0^{+}} \frac{2 - 3e^{\frac{-1}{h}}}{1 + 4e^{\frac{-1}{h}}} = \frac{2 - 0}{1 + 0} = 2$.Since $f^{\prime}(0^{-}) 
eq f^{\prime}(0^{+})$,the function $f(x)$ is not differentiable at $x=0$.

If $f(x)=\begin{cases} \frac{2 x e^{\frac{1}{2 x}}-3 x e^{\frac{-1}{2 x}}}{e^{\frac{1}{2 x}}+4 e^{\frac{-1}{2 x}}} & \text{if } x \neq 0 \\ 0 & \text{if } x=0 \end{cases}$ is a real valued function,then:

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