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

Asked on by abigaile

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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).

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