Find the critical points of:
$\displaystyle (3x^2+3x+3)e^{-x}$
I get x=0, x=1 am I correct?
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
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.