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Interview question regarding graph and intercepts

I've been doing a question which starts by asking the reader to sketch

y = x^2*e^-x

which is easy enough using standard methods. I've checked my result on a graphics calculator to confirm that errors here aren't overcomplicating the part I'm stuck on.

You're then told that
x^2*e^-x = A(x+1) where A > 0
has two real solutions, and are asked to find the larger solution.

Firstly it seems to me that below a critical value of A, there will always be two solutions, so I think that our answer will have to be in terms of A. But I really can't seem to derive any equations to work with. Any ideas?
Has exactly 2 solutions, so the curves touch. Maybe take derivative?
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Moustachian
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Original post by EricPiphany
Has exactly 2 solutions, so the curves touch. Maybe take derivative?


Well I suppose it's clear that the gradient of our A graph reach a certain critical value - the turning point of the e curve's y-co-ordinate - AFTER the x-co-ordinate that this turning point has (which turns out to be 1). Maybe this gives me something to work with...
My point is that one solution will be negative, and since we are given that there are exactly two solutions, the positive solution has to touch the curve.
So you have TWO conditions: Curves touch and gradients are the same at x.
This gives you a unique solution for A and x (if A > 0, x > 0).
(edited 8 years ago)
Reply 4
Avatar for Moustachian
Moustachian
OP
Original post by EricPiphany
My point is that one solution will be negative, and since we are given that there are exactly two solutions, the positive solution has to touch the curve.
So you have TWO conditions: Curves touch and gradients are the same at x.
This gives you a unique solution for A and x (if A > 0, x > 0).


Oh, how silly of me! Thank you very much.

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