# What are the local extrema of f(x)= x^3-x+3/x?

Dec 6, 2016

${x}_{1} = - 1$ is a maximum
${x}_{2} = 1$ is a minimum

#### Explanation:

First find the critical points by equating the first derivative to zero:

$f ' \left(x\right) = 3 {x}^{2} - 1 - \frac{3}{x} ^ 2$

$3 {x}^{2} - 1 - \frac{3}{x} ^ 2 = 0$

As $x \ne 0$ we can multiply by ${x}^{2}$

$3 {x}^{4} - {x}^{2} - 3 = 0$

${x}^{2} = \frac{1 \pm \sqrt{1 + 24}}{6}$

so ${x}^{2} = 1$ as the other root is negative, and $x = \pm 1$

Then we look at the sign of the second derivative:

$f ' ' \left(x\right) = 6 x + \frac{6}{x} ^ 3$

$f ' ' \left(- 1\right) = - 12 < 0$

$f ' ' \left(1\right) = 12 > 0$

so that:

${x}_{1} = - 1$ is a maximum
${x}_{2} = 1$ is a minimum

graph{x^3-x+3/x [-20, 20, -10, 10]}