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

Jul 6, 2018

absolute maximum: $\left(5 , \frac{1}{10}\right)$

absolute minimum: $\left(0 , 0\right)$

Explanation:

Given: $f \left(x\right) = \frac{x}{{x}^{2} + 25} \text{ on interval } \left[0 , 9\right]$

Absolute extrema can be found by evaluating the endpoints and finding any relative maximums or minimums and comparing their $y$-values.

Evaluate end points:

$f \left(0\right) = \frac{0}{25} = 0 \implies \left(0 , 0\right)$

$f \left(9\right) = \frac{9}{{9}^{2} + 25} = \frac{9}{81 + 25} = \frac{9}{106} \implies \left(9 , \frac{9}{106}\right) \approx \left(9 , .085\right)$

Find any relative minimums or maximums by setting $f ' \left(x\right) = 0$.

Use the quotient rule: $\left(\frac{u}{v}\right) ' = \frac{v u ' - u v '}{v} ^ 2$

Let u = x; " "u' = 1; " "v = x^2 + 25; " "v' = 2x

$f ' \left(x\right) = \frac{\left({x}^{2} + 25\right) \left(1\right) - x \left(2 x\right)}{{x}^{2} + 25} ^ 2$

$f ' \left(x\right) = \frac{- {x}^{2} + 25}{{x}^{2} + 25} ^ 2 = 0$

Since ${\left({x}^{2} + 25\right)}^{2} \cdot 0 = 0$, we only need to set the numerator = 0

$- {x}^{2} + 25 = 0$

${x}^{2} = 25$

critical values: $x = \pm 5$

Since our interval is $\left[0 , 9\right]$, we only need to look at $x = 5$

$f \left(5\right) = \frac{5}{{5}^{2} + 25} = \frac{5}{50} = \frac{1}{10} \implies \left(5 , \frac{1}{10}\right)$

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

intervals: $\text{ "(0, 5)," } \left(5 , 9\right)$
test values: $\text{ "x = 1, " } x = 6$
$f ' \left(x\right) : \text{ } f ' \left(1\right) > 0 , f ' \left(6\right) < 0$

This means at $f \left(5\right)$ we have a relative maximum . This becomes the absolute maximum in the interval $\left[0 , 9\right]$, since the $y$-value of the point $\left(5 , \frac{1}{10}\right) = \left(5 , 0.1\right)$ is the highest $y$-value in the interval.

**The absolute minimum occurs at the lowest $y$-value at the endpoint $\left(0 , 0\right) \ast .$