# How do you find the local maximum and minimum values of f(x) = x / (x^2 + 81) using both the First and Second Derivative Tests?

Jan 2, 2018

Minimum $\left(- 9 , - \frac{1}{18}\right)$ , Maximum $\left(9 , \frac{1}{18}\right)$

#### Explanation:

$f \left(x\right) = \frac{x}{{x}^{2} + 81}$

${D}_{f} = \mathbb{R}$

• $f ' \left(x\right) = \frac{{x}^{2} + 81 - x \left(2 x\right)}{{x}^{2} + 81} ^ 2$ $=$

$\frac{- {x}^{2} + 81}{{x}^{2} + 81} ^ 2$ $=$

$- \frac{{x}^{2} - 81}{{x}^{2} + 81} ^ 2$

$f ' \left(x\right) = 0$$\iff {x}^{2} - 81 = 0 \iff {x}^{2} = 81 \iff \left(x = 9 \mathmr{and} x = - 9\right)$

Let's plug in $f '$ a value , ${x}_{1} \in$$\left(- 9 , 9\right)$ . For example, for ${x}_{1} = 0$
we get

$f ' \left(0\right) = - \frac{- 81}{81} ^ 2 = \frac{81}{81} ^ 2 > 0$

Let's plug in $f '$ a value which is $> 9$ . For example, for ${x}_{2} = 10$
we get

$f ' \left(10\right) = - \frac{100 - 81}{100 + 81} ^ 2 = - \frac{19}{181} ^ 2 < 0$

Let's plug in $f '$ a value which is $< - 9$ . For example, for ${x}_{3} = - 10$
we get

$f ' \left(- 10\right) = - \frac{100 - 81}{100 + 81} ^ 2 = - \frac{19}{181} ^ 2 < 0$

As a result we have:

• $f$ continuous in $\left(- \infty , - 9\right]$ and $f ' \left(x\right) < 0$ for $x$$\in$$\left(- \infty , - 9\right)$
so $f$ is strictly decreasing in $\left(- \infty , - 9\right]$

• $f$ continuous in $\left[- 9 , 9\right]$ and $f ' \left(x\right) > 0$ for $x$$\in$$\left(- 9 , - 9\right)$
so $f$ is strictly increasing in $\left[- 9 , 9\right]$

• $f$ continuous in $\left[9 , + \infty\right)$ and $f ' \left(x\right) < 0$ for $x$$\in$$\left(9 , + \infty\right)$
so $f$ is strictly decreasing in $\left[9 , + \infty\right)$

$f$ is decreasing in $\left(- \infty , - 9\right]$ and increasing in $\left[- 9 , 9\right]$
therefore $f$ has a local minimum at $x = - 9$

$f$ is increasing in $\left[- 9 , 9\right]$ and decreasing in $\left[9 , + \infty\right)$
therefore $f$ has a local maximum at $x = 9$