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Graph + number of solutions problem

I've been working through a problem and once again seem to be stuck and in need of all your erudite support, if anyone's available!

I first had to sketch y = x^2*e^-x, which was easy enough investigating the intercept, considering limits and looking at first and second derivatives. I've checked my answer against a graphical calculator so know that I'm not overcomplicating the follow-up by getting this wrong!

What's stumping me is the next part:

A > 0
x^2*e^-x = A(1+x) has TWO solutions
What is the larger solution?

I've tried a few scenarios graphically, but the problem is it seems to me that there are multiple values of m that will work (in fact surely any value below a certain critical value somewhere about 0.4 by my very crude estimations). So I assume that I need to express it in terms of A. I've thought long and hard about it but I really can't seem to derive any equations to work with so any help would be great!

Thanks very much in advance.
Reply 1
Original post by Moustachian
I've been working through a problem and once again seem to be stuck and in need of all your erudite support, if anyone's available!

I first had to sketch y = x^2*e^-x, which was easy enough investigating the intercept, considering limits and looking at first and second derivatives. I've checked my answer against a graphical calculator so know that I'm not overcomplicating the follow-up by getting this wrong!

What's stumping me is the next part:

A > 0
x^2*e^-x = A(1+x) has TWO solutions
What is the larger solution?

I've tried a few scenarios graphically, but the problem is it seems to me that there are multiple values of m that will work (in fact surely any value below a certain critical value somewhere about 0.4 by my very crude estimations). So I assume that I need to express it in terms of A. I've thought long and hard about it but I really can't seem to derive any equations to work with so any help would be great!

Thanks very much in advance.


I haven't spent too much time looking at this but...if that equation has two solutions, doesn't the line y = A(x + 1) have to be a tangent to the curve? As the graphs will always intercept once in the second quadrant and otherwise you are either crossing the graph twice in the first quadrant and thus three times overall or missing it altogether in the first quadrant giving one solution overall.
(edited 8 years ago)

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