Posted by Anonymous on Tuesday, April 20, 2010 at 4:58pm.
Find the local max/min points for the function y=x^2 * e^x

calculus  Reiny, Tuesday, April 20, 2010 at 5:05pm
by 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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