Find all relative extrema of the function. (Enter NONE in any unused answer blanks.)

f(x) = x4 – 8x3

To find the relative extrema of a function, we need to find the points where the derivative changes sign.

Let's start by finding the derivative of the function f(x) = x^4 - 8x^3.

Step 1: Calculate the derivative
f'(x) = 4x^3 - 24x^2

Step 2: Set the derivative equal to zero and solve for x to find critical points
4x^3 - 24x^2 = 0

Factor out common term:
4x^2(x - 6) = 0

Setting each factor equal to zero gives us two critical points:
4x^2 = 0, which implies x = 0
x - 6 = 0, which implies x = 6

Step 3: Determine the values of f(x) at the critical points
Substitute the critical points into the original function to find the corresponding y-values:

For x = 0:
f(0) = (0^4) - 8(0^3) = 0

For x = 6:
f(6) = (6^4) - 8(6^3) = 1296 - 1728 = -432

Step 4: Analyze the results
Since we have two critical points, we need to analyze the values of f(x) at these points to determine if they are relative extrema.
- At x = 0, f(x) = 0, which indicates that this point is a local minimum.
- At x = 6, f(x) = -432, which indicates that this point is a local maximum.

Therefore, the relative extrema of the function f(x) = x^4 - 8x^3 are:
- Local minimum at (0, 0)
- Local maximum at (6, -432)