Find the absolute maximum value and the absolute minimum value, if any, of the following function.

f(x)=1/x^2+2x+8 on [-2,1]

To find the absolute maximum and minimum values of a function on a closed interval, you can follow these steps:

1. Find the critical points of the function within the interval. These occur where the derivative is either zero or undefined.

To find the critical points, start by calculating the derivative of the function f(x). Differentiating f(x)=1/x^2+2x+8 yields:

f'(x) = -2/x^3 + 2.

Setting f'(x) equal to zero and solving for x, we get:

-2/x^3 + 2 = 0.
-2/x^3 = -2.
x^3 = 1.
x = 1.

So, the critical point within the interval is x = 1.

2. Evaluate the function at the critical points, as well as the endpoints of the interval.

Now, let's calculate the value of the function f(x) at the critical point and the endpoints of the interval:

f(-2) = 1/(-2)^2 + 2(-2) + 8 = 1/4 - 4 + 8 = 25/4.
f(1) = 1/1^2 + 2(1) + 8 = 1 + 2 + 8 = 11.
f(1) = 1/1^2 + 2(1) + 8 = 1 + 2 + 8 = 11.

3. Compare the values obtained in step 2 to identify the absolute maximum and minimum values.

In this case, we only have two values to compare: 25/4 and 11.

The absolute maximum value is 25/4, which occurs at x = -2.

The absolute minimum value is 11, which occurs both at x = 1 and x = 3.

Therefore, the absolute maximum value of f(x) on the interval [-2, 1] is 25/4, while the absolute minimum value is 11.