# What are the local extrema, if any, of #f (x) =80+108x-x^3 #?

We have a local minima at

graph{80+108x-x^3 [-20, 20, -700, 700]}

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To find the local extrema of the function ( f(x) = 80 + 108x - x^3 ), first, find the derivative of the function, then set it equal to zero and solve for ( x ). The critical points obtained will be checked for local extrema using the second derivative test.

First derivative: ( f'(x) = 108 - 3x^2 )

Setting ( f'(x) = 0 ) to find critical points: ( 108 - 3x^2 = 0 ) ( 3x^2 = 108 ) ( x^2 = 36 ) ( x = \pm 6 )

Second derivative: ( f''(x) = -6x )

Test the critical points: For ( x = 6 ), ( f''(6) = -6(6) = -36 ) (negative), so ( x = 6 ) is a local maximum. For ( x = -6 ), ( f''(-6) = -6(-6) = 36 ) (positive), so ( x = -6 ) is a local minimum.

Therefore, the local maximum occurs at ( x = 6 ), and the local minimum occurs at ( x = -6 ).

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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.

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