Find the absolute extrema of the function on the closed interval. Use a graphing utility to verify your results. (Round your answers to two decimal places. If an answer does not exist, enter DNE.)

g(x) = (x^2 − 4)^2/3, [−5, 3]
find
absolute maximum (x, y) =
absolute minimum (x, y) = (smaller x-value)
(x, y) = (larger x-value)

To find the absolute extrema of a function on a closed interval, we need to locate the highest and lowest points on that interval.

To find the absolute extrema of the function g(x) = (x^2 - 4)^2/3 on the closed interval [-5, 3], we'll follow these steps:

1. Find the critical points: These are the points where the derivative is either zero or undefined. We'll find the derivative of g(x) first and solve for x when g'(x) = 0 or g'(x) is undefined.

2. Evaluate the function at the critical points: We'll plug the critical points we found in step 1 into the original function g(x) to get the corresponding y-values.

3. Evaluate the function at the endpoints: We'll substitute the endpoints of the interval (x = -5 and x = 3) into g(x) to find the y-values at these points.

4. Compare the y-values: We'll compare all the y-values we found in steps 2 and 3 to determine the absolute maximum and minimum.

Now let's follow these steps to find the absolute extrema:

Step 1: Find the derivative of g(x):
g'(x) = 2/3 * (x^2 - 4)^(2/3 - 1) * 2x
= 4x(x^2 - 4)^(-1/3)

To find the critical points, we set g'(x) equal to zero and solve for x:
4x(x^2 - 4)^(-1/3) = 0

Either x = 0 or (x^2 - 4)^(-1/3) = 0

Simplifying the second equation, we have:
x^2 - 4 = 0
(x - 2)(x + 2) = 0

Therefore, the critical points are x = 0, x = 2, and x = -2.

Step 2: Evaluate the function at the critical points:
g(0) = (0^2 - 4)^2/3 = (-4)^2/3 = 16/3 ≈ 5.33
g(2) = (2^2 - 4)^2/3 = 0^2/3 = 0
g(-2) = ((-2)^2 - 4)^2/3 = 0^2/3 = 0

So the function evaluated at the critical points is:
g(0) ≈ 5.33, g(2) = 0, and g(-2) = 0.

Step 3: Evaluate the function at the endpoints:
g(-5) = ((-5)^2 - 4)^2/3 = (25 - 4)^2/3 = 21^2/3 ≈ 14.17
g(3) = ((3)^2 - 4)^2/3 = (9 - 4)^2/3 = 5^2/3 ≈ 3.08

So the function evaluated at the endpoints is:
g(-5) ≈ 14.17 and g(3) ≈ 3.08.

Step 4: Compare the y-values:
We have the following y-values:
g(0) ≈ 5.33, g(2) = 0, g(-2) = 0, g(-5) ≈ 14.17, and g(3) ≈ 3.08.

The absolute maximum is the highest y-value, which is g(-5) ≈ 14.17, and it occurs at x = -5.
The absolute minimum is the lowest y-value, which is g(2) = 0, and it occurs at x = 2.

To verify these results, you can use a graphing utility to plot the function and identify the highest and lowest points visually.

g ' (x) = (2/3)(x^2 - 4)^(-1/3) ( 2x)

= 0 for a max/min

so 2x = 0
x = 0
when x = 0, g(0) = (-4)^(2/3)
or - cuberoot(16)

value at endpoints:
g(-5) = (21)^(2/3)
g(3) = 5^(2/3)

I am sure you can make use of the above to input your answers