ISI2016-MMA-27

Question

Consider the function $f(x) = \dfrac{e^{- \mid x \mid}}{\text{max}\{e^x, e^{-x}\}}, \: \: x \in \mathbb{R}$. Then

• A. $f$ is not continuous at some points
• B. $f$ is continuous everywhere, but not differentiable anywhere
• C. $f$ is continuous everywhere, but not differentiable at exactly one point
• D. $f$ is differentiable everywhere

Answer 1

$f(x) = \frac{e^{-|x|}}{max{(e^x,e^{-x})}} \\ \implies f(x) = \frac{e^{-|x|}}{e^{|x|}} \\ \implies f(x) = e^{-2|x|}$

Now we need to find the critical points

$f'(x) = \frac{\mathrm{d} }{\mathrm{d} x} (e^{-2|x|}) \\ \implies f'(x) = \frac{-2xe^{-2|x|}}{|x|}$

Now we need to set $f'(x) = 0$ so

$\therefore \frac{-2xe^{-2|x|}}{|x|} = 0 \\ \implies -2xe^{-2|x|} = 0 \\ \implies e^{-2|x|} = 0$

$\mathrm{Critical\:points\:are\:points\:where\:the\:function\:is\:defined\:and\:its\:derivative\:is\:zero\:or\:undefined}$

x = 0 is the critical point

the left hand limit

$\lim_{x\rightarrow0^-}(f(x)) = \lim_{x\rightarrow0^-}(e^{-2|x|}) \\ = \lim_{x\rightarrow0^-}(e^{-2(-x)}) = \lim_{x\rightarrow0^-}(e^{2x}) = 1$

and the right hand limit

$\lim_{x\rightarrow0^+}(f(x)) = \lim_{x\rightarrow0^+}(e^{-2|x|}) \\ = \lim_{x\rightarrow0^+}(e^{-2(x)}) = 1$

we can see that both right hand and left hand limit is equal which means function is continuous.

For Differentiability we need to check for x = 0

$\lim_{x\rightarrow0^+}\frac{\partial }{\partial x}(e^{-2|x|}) = \lim_{x\rightarrow0^-}\frac{\partial }{\partial x}(e^{-2|x|}) \\ \implies \lim_{x\rightarrow0^+}\frac{-2xe^{-2|x|}}{|x|}= \lim_{x\rightarrow0^-}\frac{-2xe^{-2|x|}}{|x|} \\ \implies \lim_{x\rightarrow0^+}\frac{-2xe^{-2x}}{x}= \lim_{x\rightarrow0^-}\frac{-2xe^{-2(-x)}}{-x} \\ \implies -2 \neq 2$

$\therefore$ f(x) is not diffrentiable at x =0

C is the correct Answer

Comments

First of all i love this question and second your explanation is awesome. Thanks for the answer @nirban.das .

Answer 2

Break the mod and then try to solve.

Answer 3

LETS DISCUSS THIS QUESTION

TO FIND CONTINUITY AND DIFFERENTIABLITY OF A FUNCTION,WE NEED TO UNDERSTAND THAT WE NEED A VALUE WHERE WE HAVE TO FIND CONTI & DIFF OKAY.......SO WHERE IS THE VALUE ???

SO WE ALL KNOW THE GRAPH OF e^x & e^-x.....if you dont know then try to remember graph of important function but also about their domain please it will help you

now observe in this graph how the value of e^x become 1 at x=0 so in this question there is no restriction on domain so we can take domain as all real values isn't it .......

so just check at zero 

and also you can find critical point because critical point is a point of function where function derivative becomes zero

so when you find critical point it will be zero (0)

check every thing on zero