Solving Derivative Problem: e^(-x^2) = 2/e, Find x=-1

[Essnov]

Homework Statement

Hello - I have been messing around with this problem for a while, please help.
I actually know the solution, but cannot reach it on paper:

Find where the slope of the tangent to the curve e^(-x^2) is equal to 2/e

The Attempt at a Solution

d/dx e^(-x^2) = e^(-x^2) * d/dx -x^2 = -2xe^(-x^2)

Set: -2xe^(-x^2) = 2/e

e^(-x^2 + 1) = -1/x

(e^(-x^2 + 1))^-1 = (-1/x)^-1

e^(x^2-1) = -x

At this point I can see that the only possible solution is x = -1, yet cannot actually reach that conclusion on paper. It seems like I'm missing something ridiculously simple.

Any help greatly appreciated!

 

[AKG]

What do you mean you can't reach that conclusion on paper? Plug in x = -1, show that left sides = right side, then all that remains to show is that there are no more solutions. The function mapping x to e^x²-1+x is clearly increasing, so it has at most one root, and you've found it at -1, so you're done.

 

[Essnov]

I mean that I know that the solution is x = -1 because I just happened to notice that it was the solution, not because I solved for x = -1.

Say I'm left with:
e^(x^2-1) = -x

I could easily have no idea what the solution is.
How do I work the equation to show x = -1?

 

[rock.freak667]

sub x=1 into the equation and show the left side is equal to the right side

 

[atqamar]

This is a problem that, as far as I know, can't be solved algebraically. You'd have to use analysis as explained above.

 

[HallsofIvy]

"Happening to notice" that a specific value satisfies an equation is a perfectly good method! (Provided that you check that it does work.) And a perfectly good method of seeing that x= -1 is the ONLY solution is to graph the functions $$y= e^{x^2-1}$$ and $$y= -x$$.