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 ]
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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 ]
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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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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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