Find the absolute maximum value and the absolute minimum value, if any, of the function. (If an answer does not exist, enter DNE.)

f(x)=x^3+3x^2-1 on [-3,1]

To find the absolute maximum and minimum values of a function on a closed interval, we need to follow these steps:

1. Find the critical points of the function on the interval.
2. Evaluate the function at these critical points as well as the endpoints of the interval.
3. Compare the function values to identify the absolute maximum and minimum values.

Step 1: Find the critical points
To find the critical points, we need to take the derivative of the function and set it equal to zero:

f'(x) = 3x^2 + 6x

Setting this equal to zero and factoring:

3x(x + 2) = 0

This gives us two critical points: x = 0 and x = -2.

Step 2: Evaluate the function
Now we need to evaluate the function at the critical points and the endpoints of the interval.

- For x = 0: f(0) = 0^3 + 3(0)^2 - 1 = -1
- For x = -2: f(-2) = (-2)^3 + 3(-2)^2 - 1 = -7
- For x = -3: f(-3) = (-3)^3 + 3(-3)^2 - 1 = -29
- For x = 1: f(1) = 1^3 + 3(1)^2 - 1 = 3

Step 3: Identify the maximum and minimum values
Comparing the function values:

- The maximum value is 3 when x = 1.
- The minimum value is -29 when x = -3.

Therefore, the absolute maximum value of the function f(x) = x^3 + 3x^2 - 1 on the interval [-3,1] is 3, and the absolute minimum value is -29.