Assuming your algebra is correct then Y' =0 if or
has no solutions
implies x=-5/3
Here's the question:
Function (X+2) e^ -3X. Find intervals where it is increasing and decreasing. Find the extreme values and indicate the max or min.
Here is my best attempt at this problem:
Y= (X+2) e^ -3X
Y' = (e^-3X) + (X+2) ((e^-3x)(-3)
Y' = (e^-3X) (1+ (-3(X+2))
Y' = (e^-3X) (1- 3X-6))
Y' = -(e^-3X) (3X+5)
Now I am stuck. I don't know what to do.
The only thing I can think of is trying to find a value for X that would cause Y' = 0, but I haven't been able to do that.
Can anyone help me?
Thank you. I completely forgot about being able to do that...
Y' = -(e^-3X) (3X+5)
So when X > -5/3 then F'(X) increases, when X < -5/3 then F'(X) decreases.
Therefore my final answer should be?:
F(x) increase on (-5/3, +infinity) and decreases on (-infinity, -5/3)
X = -5/3 is the local min.
Am I missing anything else?