# What are the extrema of #f(x) = 64-x^2# on the interval #[-8,0]#?

graph{64-x^2 [-8, 0, -2, 66]}

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The function ( f(x) = 64 - x^2 ) has extrema on the interval ([-8, 0]). To find these extrema, we first need to find the critical points by taking the derivative of the function and setting it equal to zero.

( f'(x) = -2x )

Setting ( f'(x) ) equal to zero, we get:

( -2x = 0 )

Solving for ( x ), we find the critical point at ( x = 0 ).

To determine if this critical point is a maximum or minimum, we can use the second derivative test.

( f''(x) = -2 )

Since the second derivative is negative, the critical point ( x = 0 ) corresponds to a maximum.

Now we need to check the endpoints of the interval, ( x = -8 ) and ( x = 0 ), to see if they are also extrema.

( f(-8) = 64 - (-8)^2 = 64 - 64 = 0 )

( f(0) = 64 - 0^2 = 64 )

So, the function has an extremum at ( x = -8 ) where it reaches a minimum of ( 0 ), and at ( x = 0 ) where it reaches a maximum of ( 64 ).

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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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