Example 1. Using the method of tangents, find the real root of the equation $x^{3}+x-3=0$.
Solution. Writing the given equation in the form $x^{3}=-x+3$ and plotting the graphs of the functions $f_{1}(x)=x^{3}, f_{2}(x)=-x+3$, we find that the only root of the equation lies in the interval $[1,2]$. Let's specify an interval of lesser length in which the root lies. Since $f(x)=x^{3}+x-3$, $f(1.2)=(1.2)^{3}+1.2-3=-0.0720$, the root lies in the interval $[1.2 ; 1.3]$. The midpoint of this interval is the point $x=1.25$. Since $f(1.25)=(1.25)^{3}+1.25-3=0.203125>0$ and $f(1.2) $=x_{n}^{3}+x_{n}-3$ | $f^{\prime}\left(x_{n}\right)=$  $=3 x_{n}^{2}+1$ | $\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$ | $x_{n+1}=x_{n}-$  $-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 0 | 1.25 | 1.953125 | 0.203125 | 5.6875 | 0.035714 | 1.214286 |
| 1 | 1.214286 | 1.790452 | 0.004738 | 5.42347 | 0.000874 | 1.213412 |
| 2 | 1.213412 | 1.786590 | 0.000002 | 5.417107 | 0.0000004 | 1.213412 |