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**1. Homework Statement**

Find the 1st and 2nd derivative:

[tex]f(x) = x^{1/3}* e^{-x^2}[/tex]

**3. The Attempt at a Solution**

[tex]f'(x) = x^{1/3} * -2xe^{-x^2} + e^{-x^2} * \frac {1}{3}x^{-2/3}[/tex]

I simplified this to:

[tex][e^{-x^2}]*[-2x^{4/3} + \frac {1}{3}x^{-2/3}][/tex]

Also to find the x values is

[tex]-2x^{4/3} + \frac {1}{3}x^{-2/3}[/tex]

[tex]x^{-2/3}*[-2x^{2} + \frac {1}{3}][/tex]

[tex]x^{2} = 1/6[/tex]

[tex]x = +- \sqrt{1/6}[/tex]

Okay now I'm trying to make the 2nd derivative be simplified so I can solve for the inflection points.

[tex]f''(x) = [e^{-x^{2}}]*[\frac {-8}{3}x^{1/3} - \frac {2}{9}x^{-5/3}] + [-2x^{4/3} + \frac {1}{3}x^{-2/3}]*[-2xe^{-x^{2}}] [/tex]

Im kind of stuck after this. I think it would involve taking out an [tex]e^{-x{2}}[/tex] but it's very confusing.

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