# What are the extrema of #f(x)=x^2-192x+8 # on #x in[-4,9]#?

The minimum is

Therefore, The minimum and maximum occur at the endpoints.

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To find the extrema of ( f(x) = x^2 - 192x + 8 ) on ( x ) in the interval ( [-4, 9] ):

- Find the critical points by taking the derivative of ( f(x) ) and setting it equal to zero.
- Evaluate ( f(x) ) at the critical points and at the endpoints of the interval.
- Identify the maximum and minimum values among these values.

First, find the derivative of ( f(x) ):

[ f'(x) = 2x - 192 ]

Set ( f'(x) ) equal to zero to find the critical points:

[ 2x - 192 = 0 ] [ x = 96 ]

Evaluate ( f(x) ) at the critical point and the endpoints of the interval ( [-4, 9] ):

At ( x = -4 ): ( f(-4) = (-4)^2 - 192(-4) + 8 = 360 )

At ( x = 9 ): ( f(9) = (9)^2 - 192(9) + 8 = -1459 )

At ( x = 96 ): ( f(96) = (96)^2 - 192(96) + 8 = -9016 )

The critical point ( x = 96 ) lies outside the interval ( [-4, 9] ), so we only consider ( x = -4 ) and ( x = 9 ) within the interval.

The minimum value occurs at ( x = 9 ), where ( f(x) = -1459 ).

The maximum value occurs at ( x = -4 ), where ( f(x) = 360 ).

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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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