cbrt(2x-1)<2 inequation

Given the inequality:
$$\sqrt[3]{2 x - 1} < 2$$
To solve this inequality, we must first solve the corresponding equation:
$$\sqrt[3]{2 x - 1} = 2$$
Solve:
Given the equation
$$\sqrt[3]{2 x - 1} = 2$$
Because equation degree is equal to = 1/3 - does not contain even numbers in the numerator, then
the equation has single real root.
We raise the equation sides to 3-th degree:
We get:
$$\left(\sqrt[3]{2 x - 1}\right)^{3} = 2^{3}$$
or
$$2 x - 1 = 8$$
Move free summands (without x)
from left part to right part, we given:
$$2 x = 9$$
Divide both parts of the equation by 2

x = 9 / (2)

We get the answer: x = 9/2

$$x_{1} = \frac{9}{2}$$
$$x_{1} = \frac{9}{2}$$
This roots
$$x_{1} = \frac{9}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{9}{2}$$
=
$$\frac{22}{5}$$
substitute to the expression
$$\sqrt[3]{2 x - 1} < 2$$
$$\sqrt[3]{-1 + \frac{2 \cdot 22}{5}} < 2$$

 2/3 3 ____    
5   *\/ 39     
----------- < 2
     5         
    

the solution of our inequality is:
$$x < \frac{9}{2}$$

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