Let g ( x )= ll e^ 2 x , & x<0 e^ -2 x , & x 0 . then g ( x ) doe… - Vidyadip

Question

Let $g ( x )=\left\{\begin{array}{ll}e^{2 x}, & x<0 \\ e^{-2 x}, & x \geq 0\end{array}\right.$ then $g ( x )$ does not satisfy the condition

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Answer

Given $g ( x )=\left\{\begin{array}{ll}e^{2 x}, & x<0 \\ e^{-2 x}, & x \geq 0\end{array}\right.$
$g ^{\prime}( x )=\left\{\begin{aligned} 2 e^{2 x}, & x<0 \\ -2 e^{-2 x}, & x \geq 0\end{aligned}\right.$
$\therefore \ce{LHD}$ at $x =0, g^{\prime}(0)=2 e ^{2 \times 0}=2 e ^0=2$
$\ce{RHD}$ at $x=0, g^{\prime}(0)=-2 e^0=-2 \times 1=-2$
As $\ce{LHD} \neq \ce{RHD}$ at $x=0$
$\therefore g ( x )$ is not differentiable at $x =0$
$\text { Again RHL }=\lim _{x \rightarrow 0^{+}} g(x)$
$=\lim _{x \rightarrow 0^{+}} e^{-2 x}=e^0=1$
$\text { LHL }=\lim _{x \rightarrow 0^{-}} g(x)$
$=\lim _{x \rightarrow 0^{-}} e^{2 x}=e^0=1$
$g(0)=e^0=1$
$\text { As LHL }=\text { RHL }=f(0)$
$\therefore g ( x )$ is continuous $\forall x \in R$

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