#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#?

Question

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

Thanks!

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Answer 1 · 2017-11-27T10:24:45

See below.

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

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.

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