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

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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 -...

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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

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