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

Question

Evelyn Agard · Accepted Answer

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	imes 1	imes (-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[#.

Ezra Adams · Answer

undefined To find the intervals on which $ f(x) $ is increasing or decreasing: 1. Calculate the derivative of $ f(x) $, denoted as $ f'(x) $. 2. Set $ f'(x) $ equal to zero and solve for $ x $. These are the critical points. 3. Test the intervals between the critical points by evaluating $ f'(x) $ at a point within each interval. 4. If $ f'(x) > 0 $ in an interval, then $ f(x) $ is increasing on that interval. If $ f'(x) < 0 $, then $ f(x) $ is decreasing on that interval. So, for $ f(x) = 2x^3 + 3x^2 - 180x $: 1. $ f'(x) = 6x^2 + 6x - 180 $. 2. Set $ f'(x) = 0 $: $ 6x^2 + 6x - 180 = 0 $. Solve for $ x $ to find the critical points. 3. After finding the critical points, evaluate $ f'(x) $ at a point within each interval. 4. Determine the intervals where $ f'(x) > 0 $ (increasing) and $ f'(x) < 0 $ (decreasing).