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

1 Answer
Jul 6, 2018

Answer:

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

absolute minimum: #(0, 0)#

Explanation:

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)** .#