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

Answer 1

absolute maximum: #(5, 1/10)#

absolute minimum: #(0, 0)#

Given: #f(x) = x/(x^2 + 25) " on interval "[0, 9]#
Absolute extrema can be found by evaluating the endpoints and finding any relative maximums or minimums and comparing their #y#-values.

Evaluate end points:

#f(0) = 0/25 = 0 => (0, 0)#
#f(9) = 9/(9^2 + 25) = 9/(81 + 25) = 9/106 => (9, 9/106) ~~(9, .085)#
Find any relative minimums or maximums by setting #f'(x) = 0#.
Use the quotient rule: #(u/v)' = (vu' - uv')/v^2#
Let #u = x; " "u' = 1; " "v = x^2 + 25; " "v' = 2x#
#f'(x) = ((x^2+25)(1) - x(2x))/(x^2 + 25)^2#
#f'(x) = (-x^2 + 25)/(x^2 + 25)^2 = 0#
Since #(x^2 + 25)^2 * 0 = 0#, we only need to set the numerator = 0
#-x^2 + 25 = 0#
#x^2 = 25#
critical values: # x = +- 5#
Since our interval is #[0, 9]#, we only need to look at #x = 5#
#f(5) = 5/(5^2 + 25) = 5/50 = 1/10 => (5, 1/10)#

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

intervals: #" "(0, 5)," " (5, 9)# test values: #" "x = 1, " "x = 6# #f'(x): " "f'(1) > 0, f'(6) < 0#
This means at #f(5)# we have a relative maximum . This becomes the absolute maximum in the interval #[0, 9]#, since the #y#-value of the point #(5, 1/10) = (5, 0.1)# is the highest #y#-value in the interval.
**The absolute minimum occurs at the lowest #y#-value at the endpoint #(0,0)** .#
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Answer 2

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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Answer from HIX Tutor

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.

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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