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[#.