Question

# The function $$f(x)={ e }^{ -\left| x \right| }$$ is

A
continuous everywhere but not differentiable at x=0
B
continuous and differentiable everywhere
C
not continuous at x=0
D
None of the above

Solution

## The correct option is A continuous everywhere but not differentiable at $$x=0$$Given $$f(x)=\begin{cases} { e }^{ -x },x\ge 0 \\ { e }^{ x },\quad x<0 \end{cases}$$$$LHL=\lim _{ x\rightarrow { 0 }^{ - } }{ f(x) } =\lim _{ x\rightarrow 0 }{ { e }^{ x } } =1$$$$RHL=\lim _{ x\rightarrow { 0 }^{ + } }{ f(x) } =\lim _{ x\rightarrow 0 }{ { e }^{ -x } } =1$$Also, $$f(0)={e}^{0}=1$$$$\because$$ LHL=RHL=f(0)$$\therefore$$ It is continuous for every value of $$x$$.Now $$LHL$$ at $$x=0$$$${ \left( \cfrac { d }{ dz } { e }^{ x } \right) }_{ x=0 }={ \left[ { e }^{ x } \right] }_{ x=0 }={ e }^{ 0 }=1\quad$$ $$RHD$$ at $$x=0$$$${ \left( \cfrac { d }{ dz } { e }^{ -x } \right) }_{ x=0 }={ \left[ { -e }^{ x } \right] }_{ x=0 }=-1$$So, $$f(x)$$ is not differentiable at $$x=0$$Hence, $$f(x)={ e }^{ -\left| x \right| }$$ is continuous every but not differentiable at $$x=0$$Mathematics

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