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 -...
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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