# If f(x) = 2x^3 + 3x^2 - 180x, how do I find the intervals on which f is increasing and decreasing?

Feb 19, 2015

Hello,

1) You have to calculate the derivative :

$f ' \left(x\right) = 6 {x}^{2} + 6 x - 180 = 6 \left({x}^{2} + x - 30\right)$.

2) Solve ${x}^{2} + x - 30 = 0$ : the discriminant is $\setminus \Delta = {1}^{2} - 4 \setminus \times 1 \setminus \times \left(- 30\right) = 121$. So, roots are

$\setminus \frac{- 1 - 11}{2} = - 6$ and $\setminus \frac{- 1 + 11}{2} = 5$

Therefore, $f ' \left(x\right) = 6 \left(x + 6\right) \left(x - 5\right)$.

3) You can now study the sign of $f ' \left(x\right)$ :

• f'(x) <0 iff x \in ]-6,5[.
• f'(x) >0 iff x\in ]-oo , -6[\cup ]5,+oo[.

4) Conclusion :
- $f$ is decreasing on $\left[- 6 , 5\right]$.
- $f$ is creasing on ]-oo,-6] and on [5,+oo[.
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