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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Originally Posted by confusedagain 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... . Use chain rule to get the first derivation: f'(x) = 0 if x = 0 Use product rule to get the second derivation: f''(x) = 0 if 4x² - 2 = 0 that means
the figure show the inflection points
Last edited by curvature; June 3rd 2007 at 02:35 PM.
Originally Posted by curvature the figure show the inflection points umm...there's no figure
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