# functionFind the critical values for the function: g(x) = x^3 - 3x^2 - 24x

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to use the definition of critical number, such that:

a is critical number `<=> f'(a) = 0`

Hence, you need to solve for x the equation `f'(x) = 0` to find a, such that:

`f'(x) = 0 <=> (x^3 - 3x^2 - 24x)' = 0 => 3x^2 - 6x - 24 = 0`

Factoring out 3, yields:

`3(x^2 - 2x - 8) = 0 => x^2 - 2x - 8 = 0`

Youmay use factorization to solve the equation, such that:

`x^2 - 2x - 4 - 4 = 0`

Forming groups of two members yields:

`(x^2 - 4) - (2x + 4) = 0`

`(x - 2)(x + 2) - 2(x + 2) = 0`

Factoring out `(x + 2)` yields:

`(x + 2)(x - 2 - 2) = 0 => {(x + 2 = 0),(x - 4 = 0):} => {(x = -2),(x = 4):}`

Hence, evaluating the critical numbers of g(x) yields that g(x) reaches its extremes at `x = 4` and `x = -2.`

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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The critical values are the x values that cancels the first derivative of the function.

To calculate the roots of the derivative, we'll determine the expression of the first derivative:

[g(x)]' =  (x^3 - 3x^2 - 24x)'

[g(x)]' = 3x^2 - 6x - 24

We'll put [g(x)]' = 0:

3x^2 - 6x - 24 = 0

We'll divide by 3:

x^2 - 2x - 8 = 0

x1 = [2 + sqrt(4+32)]/2

x1 = (2+6)/2

x1 = 4

x2 = (2-6)/2

x2 = -2

The critical values for the function g(x) = x^3 - 3x^2 - 24x are {-2 ; 4}.