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Math Help - exponential differentiation

  1. #1
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    Post exponential differentiation

    f(x) = e^-x^2

    f'(x) = e^-x^2 + (-2x)
    f'(x) = -2xe^-x^2

    f"(x) = -2x3^-x^2 +e^-x^2 + (-2x)
    f"(x) =-4x(2e^-x^2)

    how do you solve these types (e^-x^2) of problems to find the inflection point and concavity?
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  2. #2
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    Quote Originally Posted by confusedagain View Post
    f(x) = e^-x^2

    f'(x) = e^-x^2 + (-2x)
    f'(x) = -2xe^-x^2

    f"(x) = -2x3^-x^2 +e^-x^2 + (-2x)
    f"(x) =-4x(2e^-x^2)

    how do you solve these types (e^-x^2) of problems to find the inflection point and concavity?
    Hello,

    I'm not quite certain if I read your problem correctly...

    f(x)=e^{-x^2} . Use chain rule to get the first derivation:

    f'(x) = e^{-x^2} \cdot (-2x) = -2x\cdot e^{-x^2}

    f'(x) = 0 if x = 0

    Use product rule to get the second derivation:

    f''(x) = e^{-x^2} \cdot (-2) + (-2x) \cdot e^{-x^2} \cdot (-2x) = e^{-x^2} ( 4x^2 -2)

    f''(x) = 0 if 4x - 2 = 0 that means x=\sqrt{\frac{1}{2}} \ \vee \ x= -\sqrt{\frac{1}{2}}
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  3. #3
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    the curv

    the figure show the inflection points
    Attached Thumbnails Attached Thumbnails exponential differentiation-mathhelp.jpg  
    Last edited by curvature; June 3rd 2007 at 02:35 PM.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by curvature View Post
    the figure show the inflection points
    umm...there's no figure
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