Question

Calculus II Online Homework (**Due at midnight**):

Find the absolute maximum and absolute minimum values of f on the given interval.

f(x)= x^(2) e^(-x/2) [-1,8]

Answer 1

max{f(x) = e^(-x/2) x^2} = 16/e^2 at x = 4

Answer 2

min{f(x) = e^(-x/2) x^2} = 0 at x = 0

Answer 3

not sure about these though, been a while since calculus

Answer 4

anyone can check?

Answer 5

I typed them in, and they're right. :)
Thank you so much!

Answer 6

fan me!

Answer 7

Will do! :)

Answer 8

That is correct