What is the absolute maximum for f(x) = 2 - x^2 on [-2, 3] ??

Question

I believe the answer is -3

Answer 1

solve f'(x)=0
f'(x)=-2x=0
thus the max is at x=0 (which is on the interval [-2, 3])
f(0) = 2
so the absolute max is 2

Answer 2

derivative of 2 - x^2 is -2x.
This is 0 at x = 0, and x = 0 is in the interval.

So 0 is a minimum or a maximum. This is a parabola with negative argument of the square, so it opens down and the point is a max. x = 0, y = 2.

Answer 3

Answer is 2
The equasion is a parabola along the Y=0 axis, opening downward. So the absolute max will occur at x=0, which lies within range [-2,3]. Solving for f(0) gives 2, which is the absolute max.