What are the extremes of the function f(x)=x^3-3x^2+6?=

To determine the extremes of the function, we'll have to
verify if the function has critical values. The critical values of a function are the
zeroes of the derivative of that function.

We'll
differentiate the function with respect to x:

f'(x) = `3x^2
- 6x`

Now, we'll cancel the derivative's
equation:

`3x^2 - 6x =
0`

We'll factor 3x:

3x(x - 2)
= 0

We'll cancel each
factor:

3x = 0 => x =
0

x - 2 = 0

x =
2

The critical values of the function are x = 0 and x =
2.

Between these values, the function is decreasing,
therefore, the function will have a maximum point at x = 0 and a minimum point at x =
2.

We'll calculate the maximum and minimum
points:

x = 0 => f(0) =
6

x = 2 => f(2) = 8 - 12 + 6 =
2

Therefore, the extreme points of the
function are: minimum:(2 ; 2) and maximum:(0 ;
6).