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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- Jun 3rd 2007, 07:11 AMconfusedagainexponential 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? - Jun 3rd 2007, 07:42 AMearboth
Hello,

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

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

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

f'(x) = 0 if x = 0

Use product rule to get the second derivation:

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

f''(x) = 0 if 4x² - 2 = 0 that means $\displaystyle x=\sqrt{\frac{1}{2}} \ \vee \ x= -\sqrt{\frac{1}{2}}$ - Jun 3rd 2007, 08:46 AMcurvaturethe curv
the figure show the inflection points

- Jun 3rd 2007, 09:19 AMJhevon