The equation of the normal to the curve `y=x^(-x) ` at the point of its maximum is

Options

A. `x=e`
B. `x=e^(-1)`
C. `y=e`
D. `y=e^(-1)`

Answer

B

Explanation

We have,
` y=x^(-x)=e^(-x log_(e)x), x gt 0 `
` rArr (dy)/(dx) =x^(-x)(-1-log_(e)x)`
` rArr (dy)/(dx) = -x^(-x)(1+log_(e)x ) `
For the point of local maximum, we must have
` (dy)/(dx)=0 rArr 1+log_(e)x =0 rArr x=(1)/(e). `
Clearly, `1+log_(e)x lt 0 " for " 0 lt x lt (1)/(e) " and " 1+ log_(e)x gt 0 " for " x gt (1)/(e) `
Thus,
` (dy)/(dx) gt 0 " for " 0 lt x lt (1)/(e) " and " (dy)/(dx) lt 0 " for " x gt (1)/(e). `
` rArr x=(1)/(e) ` is the point of local maximum.
Clearly, ` (dy)/(dx)=0 " at " x=(1)/(e). " So, the normal at " x=(1)/(e) ` is parallel to y-axis and its equation is given by ` x=(1)/(e) `