SOLUTION: Trying to find the intervals where this function increases and decreases but running into some problems. So I know to find the critical values first: f(x) = x^4 + 2x^3 +1 f'(

Question 999407: Trying to find the intervals where this function increases and decreases but running into some problems.
So I know to find the critical values first:
f(x) = x^4 + 2x^3 +1
f'(x) = 4x^3 +6x^2
f'(x) = 2x^2(2x+3)
x = 0 and x = -3/2 are the critical values.
Now, if I understand correctly I use the first derivative test to determine in between these two critical values where it's increasing/decreasing.
Therefore, I make a number line and I plug in values around them into the derivative.
<----(-2)----(-3/2)----(-1)----(0)----(1)---->
f'(-2) = 4(-2)^3 + 6(-2)^2 = -8 therefore (-) on the left of (-3/2)
f'(-1) = 4(-1)^3 + 6(-1)^2 = 2 therefore (+) on the right side (-3/2) and left of (0)
f'(1) = 4(1)^3 + 6(1) = 10 therefore (+) on the right of (0)
bit confused on why on the left of (0) there would be another (+) I thought it worked by -,+,-,+ what do I get when I have two (+)'s or two (-)'s in a row like this?
All I can deduce now is that I have a minimum between -2 and -3/2.
Maybe I am getting things confused here with other methods.
But since I have proven I have a (-) the left of (-3/2) does that imply that the function is decreasing from (-inf,-3/2) then increasing from (-3/2, +inf) and thus increasing (0,inf)
Feel like I am really misunderstanding something here
Please clarify
Thank yu

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
It is possible to have the increasing/decreasing intervals not alternate. There is no rule that says they have to alternate. You can easily have "decreasing, increasing, increasing" which is the case here

Answer:

Increasing Interval: $\left(-\frac{3}{2}, 0\right) \cup (0,\infty)$

Decreasing Interval: $\left(-\infty, -\frac{3}{2}\right)$

Side Note: the local min occurs at x = -3/2. There is NO local min at x = 0 because we do NOT have a sign change in f ' (x) when we pass through x = 0.
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