Question

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)`

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)** .`