find the extreme values of the function and where they occur

y=x^3+x^2-8x+5

dy/dx = 0 = 3 x^2 + 2 x - 8

(3x-4)(x+2) = 0
x = 4/3 or x = -2
now go back and find y for each x

To find the extreme values of the function y = x^3 + x^2 - 8x + 5 and where they occur, we need to find the critical points. The critical points occur where the derivative of the function is equal to zero or undefined.

1. First, we find the derivative of the function:
y' = 3x^2 + 2x - 8

2. Now, we find the critical points by solving the equation y' = 0:
3x^2 + 2x - 8 = 0

To solve this quadratic equation, we can either factor it or use the quadratic formula.

3. Factoring the quadratic equation:
(3x - 4)(x + 2) = 0

Setting each factor equal to zero and solving for x, we get:
3x - 4 = 0 --> x = 4/3
x + 2 = 0 --> x = -2

So, the critical points are x = 4/3 and x = -2.

4. Now, we need to check if the critical points are maximum or minimum points by analyzing the second derivative of the function.

The second derivative of y is:
y'' = 6x + 2

Evaluating the second derivative at the critical points:
y''(4/3) = 6(4/3) + 2 = 8
y''(-2) = 6(-2) + 2 = -10

Since y''(4/3) > 0, the critical point x = 4/3 is a local minimum.
Since y''(-2) < 0, the critical point x = -2 is a local maximum.

5. Now, we need to evaluate the function at the critical points and at any endpoints of the interval of interest to determine the extreme values of y.

Evaluating y at x = 4/3:
y(4/3) = (4/3)^3 + (4/3)^2 - 8(4/3) + 5 = -67/27

Evaluating y at x = -2:
y(-2) = (-2)^3 + (-2)^2 - 8(-2) + 5 = 25

Since we don't have any other endpoints given, we only have the critical points to consider.

Therefore, the extreme values of the function are:
-67/27 at x = 4/3 (local minimum)
25 at x = -2 (local maximum)