Question

What is the equation of the normal line of `f(x)=xe^-x-x` at `x=2`?

Answer 1

`0.7124x-y-3.1536`, nearly. See the normal-inclusive Socratic graph.

Explanation:

`f=x(e^(-x)-1)`

At x = 2, f =-2(1-e^(-2))=-1.72933, nearly.

The foot of the normal is P(2, -1.729), nearly

`f'=x(e^(-x)-1)'+(e^(-x)-1)(x)'=-xe^(-x)-e^(-x)-1=-3e^(-2)-1`

`=-1.406`, nearly. at P.

The slope of the normal at P is -1/f'= 0.7124, nearly.

So, the equation to the normal at P is

#y+1.729=0.7124(x-2), giving

`0.7124x-y-3.1538`, nearly.

graph{(x(e^(-x)-1)-y)((x-2)^2+(y+1.729)^2-.01)(0.7124x-y-3.1538)=0 [-10, 10, -5, 5]}