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Derivative problem

  1. Oct 17, 2007 #1
    1. The problem statement, all variables and given/known data
    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


    3. 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!
     
  2. jcsd
  3. Oct 17, 2007 #2

    AKG

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    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 ex²-1+x is clearly increasing, so it has at most one root, and you've found it at -1, so you're done.
     
  4. Oct 17, 2007 #3
    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?
     
  5. Oct 17, 2007 #4

    rock.freak667

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    sub x=1 into the equation and show the left side is equal to the right side
     
  6. Oct 18, 2007 #5
    This is a problem that, as far as I know, can't be solved algebraically. You'd have to use analysis as explained above.
     
  7. Oct 18, 2007 #6

    HallsofIvy

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    "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 [tex]y= e^{x^2-1}[/tex] and [tex] y= -x[/tex].
     
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