# How do you find the local maximum and minimum values of  f(x) = 7x + 9x^(-1)?

Jun 27, 2017

$\text{local min at } \left(\frac{3 \sqrt{7}}{7} , 6 \sqrt{7}\right)$

$\text{local max at } \left(- \frac{3 \sqrt{7}}{7} , - 6 \sqrt{7}\right)$

#### Explanation:

$\text{to find the x-coordinates of the stationary points, differentiate}$
$f \left(x\right) \text{ and equate to zero}$

$f ' \left(x\right) = 7 - 9 {x}^{-} 2 = 0$

$\Rightarrow \frac{9}{x} ^ 2 = 7$

$\Rightarrow {x}^{2} = \frac{9}{7} \Rightarrow x = \pm \frac{3}{\sqrt{7}} = \pm \frac{3 \sqrt{7}}{7}$

$f \left(\frac{3 \sqrt{7}}{7}\right) = 7 \left(\frac{3 \sqrt{7}}{7}\right) + \frac{9}{\frac{3 \sqrt{7}}{7}} = 6 \sqrt{7}$

$f \left(- \frac{3 \sqrt{7}}{7}\right) = - 6 \sqrt{7}$

$\Rightarrow \text{stationary points at " ((3sqrt7)/7,6sqrt7)" and}$

$\left(- \frac{3 \sqrt{7}}{7} , - 6 \sqrt{7}\right)$

$\text{to find the nature of the stationary points}$

$\text{use the "color(red)"second derivative test}$

$f ' \left(x\right) = 7 - 9 {x}^{-} 2$

$\Rightarrow f ' ' \left(x\right) = 18 {x}^{-} 3 = \frac{18}{x} ^ 3$

$f ' ' \left(\frac{3 \sqrt{7}}{7}\right) > 0 \Rightarrow \textcolor{red}{\text{ local minimum}}$

$f ' ' \left(- \frac{3 \sqrt{7}}{7}\right) < 0 \Rightarrow \textcolor{red}{\text{ local maximum}}$

$\Rightarrow \left(\frac{3 \sqrt{7}}{7} , 6 \sqrt{7}\right) \text{ is a local minimum}$

$\text{and " (-(3sqrt7)/7,-6sqrt7)" is a local maximum}$ graph{7x+9x^-1 [-38.9, 38.88, -19.45, 19.45]}