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Solving equations with e

  1. May 8, 2008 #1
    For example, let's say that f(x) = e^(-9t)
    Once I've found the second derivative, I equate it to zero in order to determine concavity, but I don't know how to solve it after that :rolleyes:

    I'm just having a hard time solving with the ln or whatever you have to do..:confused:
     
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  3. May 8, 2008 #2

    Hootenanny

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    Are you trying to solve the equation to find t=something? Or are you tring to do something else? Is t a function of x? Could you clarify your problem please because at the moment it makes no sense whatsoever.
     
    Last edited: May 8, 2008
  4. May 8, 2008 #3

    Redbelly98

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    If you're trying to determine concavity, you need to figure out when the 2nd derivative is positive, negative, or zero. Just setting it to zero will tell you where the inflection points are (neither concave up nor down), and if there are no such points then you'll get "no solution" when you do this.
     
  5. May 8, 2008 #4
    sorry, what i meant was set the first derivative to zero, and solve for t. Then use the t value to solve for the second derivative.

    (The function should have been f(t) not f(x))
     
  6. May 8, 2008 #5

    Redbelly98

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    Okay.

    Plug the value of t you have (from setting 1st derivative = 0) into the 2nd derivative.

    Is the result positive, negative, or zero?
     
  7. May 8, 2008 #6

    Hootenanny

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    Err, just one problem with that: et has no roots. A purely exponential function has no stationary points and therefore it's first derviative is never zero. However, functions involving expoentials can have stationary points. For example,

    [tex] y(x) = e^x-x[/tex]

    [tex]y^\prime(x) = e^x-1[/tex]

    Hence, y(x) has a stationary point when,

    [tex]e^x = 1 \Rightarrow x=0[/tex]
     
    Last edited: May 8, 2008
  8. May 8, 2008 #7

    HallsofIvy

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    No, you don't "equate the second derivative to zero in order to determin concavity".

    The concavity of f(t) at a particular t depends upon the sign of f" at that particular t.

    The second derivative of e^(-9t) is 81e^(-9t) and , as pointed out by others, that is never equal to 0. At t= 0, it is 81> 0. Since 81e^(-9t) is continuous and never equal to 0, it is always positive and so f(t)= e^(-9t) is concave upward for all t.
     
  9. May 8, 2008 #8
    Thanks for all your help everyone :smile:
     
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