# Thread: relative extrema and C.P

1. ## relative extrema and C.P

Find the critical points of:

$(3x^2+3x+3)e^{-x}$

I get x=0, x=1 am I correct?

2. ## Re: relative extrema and C.P

Originally Posted by Dobby
Find the critical points of:

$(3x^2+3x+3)e^{-x}$

I get x=0, x=1 am I correct?
And how did you get this answer?

3. ## Re: relative extrema and C.P

found the derivative and set it=0

Let f(x) = (3x^2 + 3x + 3) * e^(-x)

Using the product rule, we have:
f'(x) = (6x + 3) * e^(-x) + (3x^2 + 3x + 3) * (-e^(-x))
f'(x) = (6x + 3) * e^(-x) - (3x^2 + 3x + 3) * e^(-x)
f'(x) = ((6x + 3) - (3x^2 + 3x + 3)) * e^(-x)
f'(x) = (6x + 3 - 3x^2 - 3x - 3) * e^(-x)
f'(x) = (-3x^2 + 3x) * e^(-x)

Setting f'(x) to zero, we have:
f'(x) = 0
(-3x^2 + 3x) * e^(-x) = 0
(-3x)(x - 1) * e^(-x) = 0
-3x = 0 or x - 1 = 0 or e^(-x) = 0
x = 0 or x = 1

Looks good.

5. ## Re: relative extrema and C.P

Prove It could have just said "yes" but it would still be possible that you got the right answer by making two mistakes that happened to cancel- and so you would have thought you had done it right when, perhaps, you hadn't. It is always better, when turning in homework or posting on a forum, to show all of your work.