# How do you find all local maximum and minimum points given y=(x^2-1)/x?

Mar 15, 2018

None

#### Explanation:

The lokal minimum and maximum can be found by $f ' \left(x\right) = 0$

$y = \frac{{x}^{2} - 1}{x}$
$y ' = \frac{2 x}{x} - \frac{{x}^{2} - 1}{x} ^ 2 = 1 + \frac{1}{x} ^ 2$

$0 = 1 + \frac{1}{x} ^ 2 | - 1 | \cdot {x}^{2}$

$- {x}^{2} = 1 | \cdot \left(- 1\right) | \sqrt{}$

$x = \pm \sqrt{- 1}$

Can't take the square root of negative numbers->No roots->No maximum or minimum