Determine the critical numbers and find the absolute maximum and minimum values of the function y = 8x^3 + x^4 for x e [-8,2]

To determine the critical numbers of a function, we need to find the values of x where the derivative is equal to zero or undefined. The absolute maximum and minimum values can be found by evaluating the function at the critical numbers and the endpoints of the given interval.

Let's start by finding the derivative of the function y = 8x^3 + x^4. The derivative dy/dx will help us identify the critical points.

dy/dx = d/dx (8x^3 + x^4)
= 24x^2 + 4x^3

Now, let's find the critical numbers by solving the equation 24x^2 + 4x^3 = 0:

24x^2 + 4x^3 = 0
4x^2(6 + x) = 0

From this equation, we find two critical numbers:

1. x = 0 (from 4x^2 = 0)
2. x = -6 (from 6 + x = 0)

Now, let's evaluate the function at the critical numbers and at the endpoints of the interval [-8,2].

- When x = -8:
y = 8(-8)^3 + (-8)^4
= 8(-512) + 4096
= -4096 + 4096
= 0

- When x = 2:
y = 8(2)^3 + (2)^4
= 8(8) + 16
= 64 + 16
= 80

- When x = 0 (critical number):
y = 8(0)^3 + (0)^4
= 0 + 0
= 0

- When x = -6 (critical number):
y = 8(-6)^3 + (-6)^4
= 8(-216) + 1296
= -1728 + 1296
= -432

So, we have the following values:

- y = 0 at x = -8, x = 0, and x = 2
- y = -432 at x = -6

Now, we need to compare these values to find the absolute maximum and minimum in the interval [-8,2].

- Absolute Maximum: The maximum value is 80 at x = 2.
- Absolute Minimum: The minimum value is -432 at x = -6.

Therefore, the critical numbers are x = 0 and x = -6. The absolute maximum value of the function is 80, which occurs at x = 2, while the absolute minimum value is -432, which occurs at x = -6.