The function `f(x)=x^(-x),(x epsilonR)` attains a maximum value at `x` is

Options

A. 2
B. 3
C. `1/e`
D. `1`

Answer

C

Explanation

Given curve is `f(x)=y=x^(-x)`
On takinglog both sides we get
`logy=-x logx`
On differentiating w.r.t `x` we get
`1/y(dy)/(dx)=-[x1/x+logx]` ltbRgt `=1/y(dy)/(dx)=-[1+logx]`
`=(dy)/(dx)=-x^(-x)[1+logx]`
`implies(dy)/(dx)=x^(-x)["log"1/x-1]`
For maximum put `(dy)/(dx)=0`
`implies"log"1/x=log_(e)e`
`implies 1/x=eimpliesx=1/e`