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]
To find the absolute extrema of the function g(x) = (x^2 − 4)^2/3 on the closed interval [−5, 3], we can follow these steps:
1. First, we find the critical points by taking the derivative of g(x) and set it equal to 0:
g'(x) = 0
To do this, we need to use the chain rule since there is a function raised to the power of 2/3. The chain rule states that if we have a function f(g(x)) raised to the power n, then the derivative is n * f'(g(x)) * g'(x).
For g(x) = (x^2 − 4)^2/3, we have f(u) = u^2/3 and g(x) = x^2 - 4.
So, using the chain rule, the derivative becomes:
g'(x) = (2/3) * (x^2 - 4)^(-1/3) * 2x = 4x / (3 * (x^2 - 4)^(1/3))
2. Next, we check the endpoints of the interval, x = -5 and x = 3, since they are part of the closed interval.
3. To find the absolute extrema, we evaluate g(x) at the critical points and endpoints and compare the values.
Let's calculate the critical points and endpoints:
1. Critical Points:
Set g'(x) = 0:
4x / (3 * (x^2 - 4)^(1/3)) = 0
The numerator can be 0 when x = 0
The denominator can never be 0 since (x^2 - 4)^(1/3) is always positive except at x = -2 and x = 2 (which are not in the interval)
So, we have one critical point: x = 0
2. Endpoints:
Evaluate g(-5):
g(-5) = ((-5)^2 - 4)^2/3 = (25 - 4)^2/3 = 21^(2/3) ≈ 7.17
Evaluate g(3):
g(3) = ((3)^2 - 4)^2/3 = (9 - 4)^2/3 = 5^(2/3) ≈ 2.92
Now, we compare the values:
- The critical point at x = 0 gives us g(0) = (0^2 - 4)^2/3 = 16^(1/3) = 2.52
- The endpoint at x = -5 gives us g(-5) ≈ 7.17
- The endpoint at x = 3 gives us g(3) ≈ 2.92
By comparing these values, we can see that the absolute minimum is g(3) ≈ 2.92 and the absolute maximum is g(-5) ≈ 7.17.
To verify these results, you can plot the function g(x) = (x^2 − 4)^2/3 on a graphing utility and visually confirm the locations and values of the absolute extrema.