f'(x) = 12x^2 - 36x - 480
=0 for max/mins
divide by 12
x^2 - 3x - 40=0
(x-8)(x+5) = 0
x = 8 or x = -5
You should know the general behaviour of cubic curve with a positive x^3 term and critical values at -5 and 8
I will give you the decreasing interval, it is
between -5 and 8 or
-5 < x < 8
f(x)=4x3−18x2−480x−2
is decreasing on what interval?
It is increasing on what interval(s) ?
The function has a local maximum at ?
1 answer