# What are the absolute extrema of # f(x)= x/(x^2 + 25)# on the interval [0,9]?

absolute maximum:

absolute minimum:

Evaluate end points:

Using the first derivative test, set up intervals to find out if this point is a relative maximum or a relative minimum:

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To find the absolute extrema of ( f(x) = \frac{x}{x^2 + 25} ) on the interval ([0, 9]), we first find the critical points of ( f(x) ) on the interval by setting its derivative equal to zero and solving for ( x ). Then, we evaluate ( f(x) ) at the critical points and the endpoints of the interval to determine the absolute extrema.

The derivative of ( f(x) ) is ( f'(x) = \frac{25 - x^2}{(x^2 + 25)^2} ).

Setting ( f'(x) = 0 ), we get ( 25 - x^2 = 0 ), which simplifies to ( x^2 = 25 ). So, ( x = \pm 5 ).

The critical points on the interval ([0, 9]) are ( x = 0 ), ( x = 5 ), and ( x = 9 ).

Now, we evaluate ( f(x) ) at these points and the endpoints of the interval:

- At ( x = 0 ): ( f(0) = \frac{0}{0^2 + 25} = 0 )
- At ( x = 5 ): ( f(5) = \frac{5}{5^2 + 25} = \frac{5}{50} = \frac{1}{10} )
- At ( x = 9 ): ( f(9) = \frac{9}{9^2 + 25} = \frac{9}{106} )
- At ( x = 9 ): ( f(9) = \frac{9}{9^2 + 25} = \frac{9}{106} )

Therefore, the absolute maximum of ( f(x) ) on ([0, 9]) is ( \frac{1}{10} ) at ( x = 5 ), and the absolute minimum is ( 0 ) at ( x = 0 ).

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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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