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

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Answer 1 · 2015-02-19T17:54:31

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 $f'(x)$ :

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

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