Find the critical values for the function: g(x) = x^3 - 3x^2 - 24x
2 Answers
hala718 | High School Teacher | (Level 1) Educator Emeritus
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
giorgiana1976 | College Teacher | (Level 3) Valedictorian
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}.