Given: f(x) = x/(x^2 + 25) " on interval "[0, 9]f(x)=xx2+25 on interval [0,9]
Absolute extrema can be found by evaluating the endpoints and finding any relative maximums or minimums and comparing their yy-values.
Evaluate end points:
f(0) = 0/25 = 0 => (0, 0)f(0)=025=0⇒(0,0)
f(9) = 9/(9^2 + 25) = 9/(81 + 25) = 9/106 => (9, 9/106) ~~(9, .085)f(9)=992+25=981+25=9106⇒(9,9106)≈(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)** .
