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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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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