Example 1. Using the method of iterations, find the smaller positive root of the equation $x^{3}-5 x+1=0$.
Solution. Graphically separating the roots of the given equation, we conclude that the equation has three real roots lying in the intervals $[-3 ;-2],[0 ; 1],[2 ; 3]$. Let's find the smallest positive root belonging to the interval $[0 ; 1]$. Specify a segment of lesser length on which the root lies. Since $f(x)=x^{3}-5 x+1, f(0)=1>0$, $f\left(\frac{1}{2}\right)=\frac{1}{8}-\frac{5}{2}+1=-\frac{11}{8} $=\left(x_{n}^{3}+1\right) / 5$ |
| :---: | :---: | :---: | :---: | :---: |
| 0 | 0.25 | 0.01563 | 1.01563 | 0.20313 |
| 1 | 0.20313 | 0.00838 | 1.00838 | 0.20168 |
| 2 | 0.20168 | 0.00821 | 1.00821 | 0.20164 |
| 3 | 0.20164 | 0.00820 | 1.00820 | 0.20164 |

Remark. When finding the other two roots of the original equation using the method of successive approximations, one cannot use the formula $x=\frac{\left(x^{3}+1\right)}{5}$, since

$$
\max _{24 \mid1
$$

i.e., condition (12.9) is not satisfied. In this case, the equation should be represented in another form, for example, $x=\sqrt[3]{5 x-1}$; for the function $\varphi(x)=\sqrt[3]{5 x-1}$, condition (12.9) on the intervals $[-3 ;-2],[2 ; 3]$ will be satisfied.