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March 5, 2015

March 5, 2015

Posted by **Shan** on Friday, July 12, 2013 at 3:12pm.

f(x) = xe-x2 on [0,2]

Absolute maximum value:

at x =

Absolute minimum value:

at x =

- Calculus -
**Steve**, Friday, July 12, 2013 at 3:58pmf = xe^-x^2

f' = (1-2x^2) e^-x^2

so, where is f' = 0?

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