Find the absolute maximum value for the function f(x)=x^2 - 3 , on the interval [-4,0) U (0,3]

Answer 1

First, check for any local maxima within the interval.
f'(x) = 2x, which is zero at x=0.
But x=0 is not in the domain.
So, except for the hole there, f(x) = x^2-3, so the limit as x->0 is -3
f(-4) = 16-3 = 13
f(3) = 9-3 = 6
So, the absolute maximum of f(x) over the given domain is f(-4) = 13

Of course, knowing that the graph is a parabola with its vertex at (0,-3) we could have easily discerned that the maximum would be at the point farthest from x=0.