Asked by josh
2x^3+3x^2-12x-7 the absolute maximum. I need help because it seems no one else in my class knows how to.
Answers
Answered by
MathMate
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.
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.