g(x) = x^3 - 3x^2 - 24x

To find the critical values, first we will determine the first derivative of g(x):

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

Now we will determine the derivative's zeros:

==> 3x^2 - 6x - 24 = 0

Divide by 3:

==> x^2 - 2x - 8 = 0

Now let us factor the equation:

==> (x-4)(x+2) = 0

==> x1= 4

==> x2= -2

**Then the critical values for g(x) are x = { -2, 4}**

**Or, the function changes directiosn when x= -2 and x= 4**

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

We'll apply the quadratic formula:

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}.**