# Thread: Find the intervals and extreme values

1. ## Find the intervals and extreme values

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?

2. Assuming your algebra is correct then Y' =0 if $\displaystyle e^{-3x} =0$ or $\displaystyle 3x + 5 =0$

$\displaystyle e^{-3x} =0$ has no solutions

$\displaystyle 3x + 5 =0$ implies x=-5/3

3. Originally Posted by SpringFan25
Assuming your algebra is correct then Y' =0 if $\displaystyle e^{-3x} =0$ or $\displaystyle 3x + 5 =0$

$\displaystyle e^{-3x} =0$ has no solutions

$\displaystyle 3x + 5 =0$ implies x=-5/3
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?