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

Feb 4, 2017

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

#### Explanation:

$f = x \left({e}^{- x} - 1\right)$

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

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

$f ' = x \left({e}^{- x} - 1\right) ' + \left({e}^{- x} - 1\right) \left(x\right) ' = - x {e}^{- x} - {e}^{- x} - 1 = - 3 {e}^{- 2} - 1$

$= - 1.406$, 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

$0.7124 x - y - 3.1538$, 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]}