# How do you find the Critical Points for #x^3 + 3x^2-24x#?

Critical points are 2, -4

critical point 'c' of a function is where either f'(c)=0 or f'(c) does not exist.

Critical points are 2, -4

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To find the critical points of the function ( f(x) = x^3 + 3x^2 - 24x ), we first need to find its derivative, ( f'(x) ). Then, we set the derivative equal to zero and solve for ( x ). Finally, we identify the ( x )-values obtained as critical points.

The derivative of ( f(x) ) is ( f'(x) = 3x^2 + 6x - 24 ). Setting this derivative equal to zero gives us the equation ( 3x^2 + 6x - 24 = 0 ).

We can solve this quadratic equation using various methods such as factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:

[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} ]

Where ( a = 3 ), ( b = 6 ), and ( c = -24 ).

Substituting these values into the formula and solving for ( x ) yields:

[ x = \frac{{-6 \pm \sqrt{{6^2 - 4(3)(-24)}}}}{{2(3)}} ]

[ x = \frac{{-6 \pm \sqrt{{36 + 288}}}}{{6}} ]

[ x = \frac{{-6 \pm \sqrt{{324}}}}{{6}} ]

[ x = \frac{{-6 \pm 18}}{{6}} ]

This gives us two solutions for ( x ):

[ x_1 = \frac{{-6 + 18}}{{6}} = \frac{{12}}{{6}} = 2 ]

[ x_2 = \frac{{-6 - 18}}{{6}} = \frac{{-24}}{{6}} = -4 ]

So, the critical points of the function ( f(x) = x^3 + 3x^2 - 24x ) are ( x = 2 ) and ( x = -4 ).

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