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

Answer 1

Paisley Johnson

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

#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]}

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Answer 2

Evelyn Agard

The equation of the normal line of f(x)=xe^-x-x at x=2 is y = -3x + 4.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.