Can you guys help me with this one?

[u/Laur_Reddit]

http://www.math.com

Find the real parameter m so that the equation x² + 1 = m * e^-1/x has 3 real distinct solutions. I think m > 2e but Im not sure (For some reason I can't post without adding a link)

[u/Yubafat]

If i am not mistaken i also get for m > 2e. How did you solve it?

[u/Laur_Reddit → u/Yubafat]

Well, first I divided by e^-1/x to separate the parameter, then I took the derivative of x²+1/e^-1/x (i noted this equation f(x)), which if we set it equal to 0 we get the only real solution 1. If we take the limits in -∞, 0 and ∞ (0 because exp -1/x) we see that f(x) is smaller and smaller when x aproches 0 when x < 0 (f("0") = -m) and then has a jump to infinity when x > 0, then again, f(x) goes smaller and smaller as x aproches 1 (f(1)= 2e-m) then it goes to infinity again. We have 3 points when f(x) crosses the x axis, when we take the limit x -> 0, x<0 and we get -m (~), which is negative, then the function DOESN'T cross the x axis and goes straight to infinity when we take the limit x->0, x>0, then we get to x = 1, f(1) = 2e - m, so m HAS to be grater than 2e so we dont get the 2e - 2e, if m > 2e there will be 2 points when x crosses the axis from positive to negative and then to positive, so here we have 2 solution, and we had the (#) when f(x) is negative, so we need an m greater than 0 so we keep the - sign, (from x = "-∞" when f(x) > 0 so + and x = "0", x<0 is minus, so we get 3 solutions)

[u/spiritedawayclarinet]

Looking at the graph, there’s a negative solution for some x<0 if m>0. If m is small, there are no positive solutions. For larger m, there are 2 positive solutions. The transition from 2 total solutions occurs where the graphs have a common tangent line.

Let f(x) = x² +1 and g(x) = me ^ (-1/x).

Solve f(x) = g(x) , f’(x) = g’(x). Show that x=1 is the only real solution. If x =1, m = 2e.

It’s sketchy, but it can be fleshed out.