# What are the critical points of #f(x) = -3x^4 + 12x^3 + 6#?

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To find the critical points, we need to find the derivative of the function, set it equal to zero, and solve for x.

The derivative of ( f(x) = -3x^4 + 12x^3 + 6 ) is ( f'(x) = -12x^3 + 36x^2 ).

Setting the derivative equal to zero, we get ( -12x^3 + 36x^2 = 0 ).

Factor out ( -12x^2 ) to get ( -12x^2(x - 3) = 0 ).

This equation is satisfied when ( x = 0 ) or ( x = 3 ).

Therefore, the critical points of ( f(x) ) are ( x = 0 ) and ( x = 3 ).

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