Question

What is the equation of the normal line of `f(x)= x^e/e^x` at `x=e`?

Answer 1

`y=1.`

Explanation:

`f(x)=x^e/e^x=x^e*e^-x rArr f(e)=e^e/e^e=1.`

Thus, we require the eqn. of Normal to the Curve at the point

`(e,1)," say "P(e,1).`

`:. f'(x)=x^e*d/dx(e^-x)+e^-x*d/dx(x^e).`

`=x^e*(-e^-x)+e^-x*(ex^(e-1)).`

`:. f'(e)=e^e*(-e^-e)+e^-e*(ee^(e-1)).`

`=-e^e*1/e^e+e^(-e+1+e-1)`

`:. f'(e)=-1+e^0=-1+1=0.`

This means that, the Slope of Tangent at the point `x=e` to

the given curve is `0.`

In other words, the tgt. line at `P(e,1)` to the curve is Horizontal.

Since the Normal at `P(e,1)" is "bot` to the tgt. at that point, we

find that the reqd. normal is a Vertical line through P.

Therefore, the eqn. of the normal is `y=1.`