# What are the absolute extrema of # f(x)= 6x^3 − 9x^2 − 36x + 3 in [-4,8]#?

In order to find the extrema, you need to take the derivative of the function and find the roots of the derivative.

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To find the absolute extrema of ( f(x) = 6x^3 - 9x^2 - 36x + 3 ) in the interval ([-4, 8]), you need to evaluate the function at the critical points and endpoints within that interval.

- Find the critical points by taking the derivative of ( f(x) ), setting it equal to zero, and solving for ( x ).
- Evaluate ( f(x) ) at the critical points and at the endpoints of the interval.
- The largest and smallest values obtained in step 2 will be the absolute maximum and minimum, respectively.

Derivative of ( f(x) ): [ f'(x) = 18x^2 - 18x - 36 ]

Setting ( f'(x) = 0 ) and solving for ( x ): [ 18x^2 - 18x - 36 = 0 ] [ x^2 - x - 2 = 0 ] [ (x - 2)(x + 1) = 0 ]

So, critical points are ( x = 2 ) and ( x = -1 ).

Evaluate ( f(x) ) at critical points and endpoints: [ f(-4) = 6(-4)^3 - 9(-4)^2 - 36(-4) + 3 = 339 ] [ f(2) = 6(2)^3 - 9(2)^2 - 36(2) + 3 = -123 ] [ f(8) = 6(8)^3 - 9(8)^2 - 36(8) + 3 = 1479 ] [ f(-1) = 6(-1)^3 - 9(-1)^2 - 36(-1) + 3 = 48 ]

The absolute maximum is ( f(8) = 1479 ) and the absolute minimum is ( f(2) = -123 ).

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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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