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

Options

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

Answer

C

Explanation

Given `f(x)=x^(-x)`
Taking log on both sides, we get
`implies1/(f(x).f'(x))=-logx-1`
`impliesf'(x)=-f(x)(1+logx)`
Put `f'(x)=0`
`implieslogx=-1impliesx=e^(-1)`
Now `f''(x)=-f'(x)(1+logx)-f(x)1/x`
`=f'(x)(1+logx)^(2)-(f(x))/x`
At `x=1/e, f''(x)=-ef(1/e)lt0`, maxima
Hence at `x=1/3,f(x)` is maximum.