# How do you find the local max and min for f(x) = 7x + 9x^(-1)?

Nov 29, 2016

To find the local extrema we find the points where the first derivative is null and study the sign of the second derivative

#### Explanation:

$f \left(x\right) = 7 x + 9 {x}^{- 1}$

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

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

Find the values of $x$ where $f ' \left(x\right) = 0$

$7 - \frac{9}{x} ^ 2 = 0$

$7 {x}^{2} - 9 = 0$

$x = \pm \frac{3}{\sqrt{7}}$

In both points $f ' ' \left(x\right) \ne 0$ so these are local extrema, namely:

1) around $x = - \frac{3}{\sqrt{7}}$ the second derivative is negative and then the point is a local maximum.

2) around $x = + \frac{3}{\sqrt{7}}$ the second derivative is positive and then the point is a local minimum.

graph{7x+9/x [-47.3, 47.2, -73.6, 73.6]}