How do you find the relative extrema for #f(x)=(x^4)+(4x^3)12#?
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To find the relative extrema of ( f(x) = x^4 + 4x^3  12 ), follow these steps:
 Find the derivative of the function, ( f'(x) ).
 Set ( f'(x) = 0 ) and solve for ( x ) to find critical points.
 Determine the nature of each critical point by considering the sign of the second derivative, ( f''(x) ), or by using the first derivative test.
 The critical points where the derivative changes sign correspond to relative extrema.
Let's proceed with these steps:

Find the derivative: [ f'(x) = 4x^3 + 12x^2 ]

Set ( f'(x) = 0 ) and solve for ( x ): [ 4x^3 + 12x^2 = 0 ] [ 4x^2(x + 3) = 0 ] [ x = 0 \quad \text{or} \quad x = 3 ]

Determine the nature of each critical point:
 For ( x = 0 ): ( f''(x) = 12x + 24 = 24 > 0 ), so ( x = 0 ) is a relative minimum.
 For ( x = 3 ): ( f''(x) = 12x + 24 = 0 < 0 ), so ( x = 3 ) is a relative maximum.
Therefore, the relative minimum occurs at ( x = 0 ) and the relative maximum occurs at ( x = 3 ).
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