Find the absolute max and min values of f(x)=(x^2)(e^x)....I know f'(x) is e^x(x^2+2x)...just can't get the critical points

Question

zarkon · Answer

set equal to zero and solve...what is the problem?

anonymous · Answer

it said its 0 and e...I got 0 but not e

zarkon · Answer

x=0,-2

myininaya · Answer

$$e^x \neq 0$$

myininaya · Answer

so all you have to do is look at x^2+2x=0
and thats what zarkon did

anonymous · Answer

thats what I got but according to the answer  e and 0 are the max & min

myininaya · Answer

e is crazy talk

anonymous · Answer

it was on my exam and I got it wrong and that was the correct answer

myininaya · Answer

is there an interval?

anonymous · Answer

[-1,1]

zarkon · Answer

look at the graph http://www.wolframalpha.com/input/?i=plot [x^2*e^x%2C+{x%2C-3%2C1}]

zarkon · Answer

AH....IS THE THE RESTRICTED DOMAIN?

zarkon · Answer

sorry for the caps  ;)

zarkon · Answer

on [-1,1]   e is the max value of the function

myininaya · Answer

so the critical numbers are 0 and -2

we don't care about -2 its not the interval [-1,1]

$$f(-1)=(-1)^2e^{-1}=\frac{1}{e} ; f(1)=(1)^2e^{1}=e; f(0)=0$$

myininaya · Answer

largest is e so e is the abs max value
0 is smallest so 0 is the abs min value

anonymous · Answer

ooooo ok I get it

zarkon · Answer

you probably should have included the [-1,1] in the original problem ;)

myininaya · Answer

yep yep
thats important info

anonymous · Answer

sorry