Question

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

Answer 1

Hello,

1) You have to calculate the derivative :

`f'(x) = 6x^2 + 6x - 180 = 6(x^2 + x - 30)`.

2) Solve `x^2+x-30=0` : the discriminant is `\Delta = 1^2-4\times 1\times (-30) = 121`. So, roots are

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

Therefore, `f'(x) = 6(x+6)(x-5)`.

3) You can now study the sign of `f'(x)` :

• `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 `[-6,5]`.
- `f` is creasing on `]-oo,-6]` and on `[5,+oo[`.