2x^3+3x^2-12x-7 the absolute maximum. I need help because it seems no one else in my class knows how to.
For an absolute maximum on an interval,
evaluate the function at the end-points, and at local extrema. The maximum value of f(x) among these points is the absolute maximum.
The local extrema are found by equating f'(x)=0 where f"(x)≠0 (to exclude inflexion points).
For f(x)=2x^3+3x^2-12x-7 on the interval (-3,0),
1. calculate the extremum:
f'(x)=6x²+6x-12=0
x=-2 (on interval) or x=1 (not on interval, rejected)
evaluate f(-2)=13
2. calculate f(x) at end-points of the interval. f(-3)=2, f(0)=7.
So the absolute maximum is at f(-2)=13.