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).