67. Let's find the approximate roots of the equation $x^{3}-4 x+2=0$ with an accuracy of 0.01.
Solution. We will construct the graph of the function $f(x)=x^{3}-4 x+2$. We have:

1) $D(f)=\mathbb{R}$
2) $f^{\prime}(x)=3 x^{2}-4$
3) critical points:

$$
-\frac{2 \sqrt{3}}{3} \approx-1.1 ; \quad \frac{2 \sqrt{3}}{3} \approx 1.1
$$

We will create a table:

| $x$ | $-\infty<x<-1.1$ | $-1.1$ | $-1.1<x<1.1$ | 1.1 | $1.1<x<+\infty$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $f^{\prime}(x)$ | + | 0 | - | 0 | + |
| $f(x)$ | $\nearrow$ | $\approx 5.1$ | $\searrow$ | $-1.1$ | $\nearrow$ |
|  |  | $\max$ |  | $\min$ |  |

We construct the graph, finding several additional points (Fig. 37).

From the graph, we see that our equation has three real roots, belonging to the intervals $[-3 ;-2],[0 ; 1]$ and $[1 ; 2]$.

Let's find, for example, approximate values of the root belonging to the interval $[-3 ;-2]$. To the nearest integers, we have $x_{1} \approx-3$ (by deficiency), $x_{2} \approx-2$ (by excess). We will perform further trials using four-digit mathematical tables or a calculator:

| $x$ | $x^{3}$ | $-4 x$ | $x^{3}-4 x+2$ |
| :---: | :---: | :---: | :---: |
|  |  |  |  |
| $-2.6$ | $-17.58$ | 10.4 | $-5.18$ |
| $-2.3$ | $-12.17$ | 9.2 | $-0.97$ |
| $-2.2$ | $-10.65$ | 8.8 | 0.15 |
| $-2.29$ | $-12.01$ | 9.16 | $-0.85$ |
| $-2.28$ | $-11.85$ | 9.12 | $-0.73$ |
| $-2.22$ | $-10.94$ | 8.88 | $-0.06$ |
| $-2.21$ | $-10.79$ | 8.84 | 0.05 |

Therefore, $x \approx-2.22$ (by deficiency) and $x \approx-2.21$ (by excess).

Similar calculations can be used to obtain further significant digits.

In the same way, we can find the other two roots of the equation, belonging to the intervals $[0 ; 1]$ and $[1 ; 2]$.

If equation (1) is a fourth-degree equation or higher, then when finding the critical points of the function $y=f(x)$, we will obtain a cubic (or higher degree) equation. In such cases, to construct the graph, we should represent the expression $f(x)$ in the form $f(x)=g(x)-q(x)$, where $g(x)$ and $q(x)$ are polynomials.

Then solving equation (1) is equivalent to solving the equation

$$
g(x)=q(x)
$$

the solution of which is reduced to finding the abscissas of the points of intersection of the graphs of the functions $y=g(x)$ and $y=q(x)$.