What are the extrema of # f(x)=x/(x^2+9)# on the interval [0,5]?
Check a graph:
graph{x/(x^2+9) [0.02, 5, 0.02, 0.2]}
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To find the extrema of ( f(x) = \frac{x}{x^2 + 9} ) on the interval ([0, 5]), we first need to find the critical points by setting the derivative of ( f(x) ) equal to zero and then checking the endpoints of the interval.

Find the derivative of ( f(x) ): [ f(x) = \frac{x}{x^2 + 9} ] [ f'(x) = \frac{(x^2 + 9)  x(2x)}{(x^2 + 9)^2} ] [ f'(x) = \frac{9  x^2}{(x^2 + 9)^2} ]

Set ( f'(x) = 0 ) to find critical points: [ \frac{9  x^2}{(x^2 + 9)^2} = 0 ] [ 9  x^2 = 0 ] [ x^2 = 9 ] [ x = \pm 3 ]
The critical points are ( x = 3 ) and ( x = 3 ). However, ( x = 3 ) is not in the interval ([0, 5]), so we only consider ( x = 3 ).
 Evaluate ( f(x) ) at the critical point and endpoints:
 At ( x = 0 ): [ f(0) = \frac{0}{0^2 + 9} = 0 ]
 At ( x = 5 ): [ f(5) = \frac{5}{5^2 + 9} = \frac{5}{34} ]
 At ( x = 3 ) (critical point): [ f(3) = \frac{3}{3^2 + 9} = \frac{3}{18} = \frac{1}{6} ]
The function ( f(x) ) has extrema at ( x = 0 ) and ( x = 3 ) on the interval ([0, 5]), with ( x = 3 ) being the minimum and ( x = 0 ) being the maximum.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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