y=x^2e^(-x) has a maximum turning point at A(2,4/e^2). The equation x^2e^(-x)-2=0 has 3 real, distinct roots. What are the possible values of k?

The answers say 0<k<2, but I have no idea how to get to this.

Thanks!

1 Answer
Nov 27, 2017

See below.

Explanation:

The equation x^2e^ -x -2 = 0 can be solved with the contribution of the so called Lambert function W(cdot)

https://en.wikipedia.org/wiki/Lambert_W_function

We have

x^2e^ -x = 2 rArr x e^(-x/2) = pm sqrt2 rArr (-x/2)e^(-x/2) = pm(-sqrt2/2)

Now using the property

Xe^X = Y hArr X = W(Y)

we have

-x/2 = W(pm(-sqrt2/2))rArr x = -2 W(pm(-sqrt2/2))

with the only real solution

x = -2W(sqrt2/2) = -0.901201031729666

NOTE

Attached the representation of how to obtain a graphical solution as the intersection of e^-x in blue and 2/x^2 in red.

enter image source here