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

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.

To find the absolute maximum value for the function f(x) = x^2 - 3 on the interval [-4,0) U (0,3], we need to evaluate the function at its critical points and endpoints.

Step 1: Find the critical points by taking the derivative of the function and setting it equal to zero:
f'(x) = 2x
Setting f'(x) = 0, we have:
2x = 0
x = 0

Step 2: Evaluate the function at the critical point, as well as at the endpoints of the interval.
For x = 0:
f(0) = (0)^2 - 3 = -3

For x = -4:
f(-4) = (-4)^2 - 3 = 16 - 3 = 13

For x = 3:
f(3) = (3)^2 - 3 = 9 - 3 = 6

Step 3: Compare the values of the function at the critical point and endpoints to determine the absolute maximum value.

The maximum value of f(x) = x^2 - 3 on the interval [-4,0) U (0,3] is 13, which occurs at x = -4.

To find the absolute maximum value for the function f(x) = x^2 - 3 on the interval [-4,0) U (0,3], we first need to find the critical points and the endpoints of the interval.

1. Find the critical points:
The critical points are the values of x where the derivative f'(x) is equal to zero or does not exist.
In this case, the derivative of f(x) = x^2 - 3 is f'(x) = 2x.

Setting f'(x) = 0, we get:
2x = 0
x = 0

So x = 0 is a critical point.

2. Find the endpoints:
The given interval is [-4,0) U (0,3]. The endpoints of the interval are x = -4, 0, 3.

3. Evaluate the function at the critical points and endpoints:
Evaluate f(x) = x^2 - 3 at the critical points and endpoints.

f(-4) = (-4)^2 - 3 = 16 - 3 = 13
f(0) = (0)^2 - 3 = 0 - 3 = -3
f(3) = (3)^2 - 3 = 9 - 3 = 6

4. Compare the values obtained:
We have the values f(-4) = 13, f(0) = -3, and f(3) = 6.

The absolute maximum value of f(x) on the interval [-4,0) U (0,3] is 13, which occurs at x = -4.

Therefore, the absolute maximum value for the function f(x) = x^2 - 3 on the interval [-4,0) U (0,3] is 13.