Tuesday

March 31, 2015

March 31, 2015

Posted by **Anonymous** on Tuesday, April 20, 2010 at 4:58pm.

- calculus -
**Reiny**, Tuesday, April 20, 2010 at 5:05pmby the product rule

y' = x^2(e^x) + 2x(e^x)

= e^x(x^2 + 2x) = 0 for max/min

e^x = 0 ---> no solution or

x^2 + 2x = 0

x(x+2) = 0

x = 0 or x = -2

if x=0 then y 0

if x = -2 then y = 4(e^2)

so (0,0) and ( -2,4e^2) are max/min points.

I will let you decide which is which.

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