# How do you solve x^3-2x>1?

Sep 9, 2017

$\left(- 1 , \frac{1 - \sqrt{5}}{2}\right) \mathmr{and} \left(\frac{1 + \sqrt{5}}{2} , + \in f\right)$

#### Explanation:

$f \left(x\right) = {x}^{3} - 2 x - 1 > 0$
The simplest method is graphing the function f(x), then looking for the parts of the graph that are above the x-axis.
graph{x^3 - 2x - 1 [-2.5, 2.5, -1.25, 1.25]}
On this graph, the x-intercepts are approximately:
x = - 1 , x = - 0.65, and x = 1.60
$\left(- 1 , \frac{1 - \sqrt{5}}{2}\right)$ and $\left(\frac{1 + \sqrt{5}}{2} , + \in f .\right)$