402. Calculate with an accuracy of 0.000001 the real root of the equation $2-x-\lg x=0$.
Solution. To isolate the root,

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-148.jpg?height=278&width=395&top_left_y=326&top_left_x=100)

Fig. 80. Transform the equation to the form $\lg x=2-x$ and plot the curves $y=\lg x$ and $y=2-x$ (Fig. 80). From the graph, we determine that the desired root lies within the interval $[1.6; 1.8]$.

To check the conditions necessary for using the chord and tangent methods, we calculate the values of the function $f(x)=2 - x - \lg x$ at the endpoints of the found interval and find the derivatives $f'(x)$ and $f''(x)$:

$$
\begin{aligned}
f(1.6) & =2-1.6-0.2041=0.1959>0 \\
f(1.8) & =2-1.8-0.2553=-0.05530 \text { over the entire interval }[1.6 ; 1.8] .
\end{aligned}
$$

Having verified that the function $f(x)$ has different signs at the endpoints of the interval and that the derivatives $f'(x)$ and $f''(x)$ maintain their signs over the entire interval, we denote the endpoints of the interval: $a=1.6=\boldsymbol{\beta} ; b=1.8$ and apply the refining formulas ("):

$$
\begin{gathered}
a_{1}=1.6-\frac{(1.8-1.6) f(1.6)}{f(1.8)-f(1.6)}=1.6+0.1559=1.7559 \\
b_{1}=1.6-\frac{f(1.6)}{f_{i}^{\prime}(1.6)}=1.6+0.1540=1.7540
\end{gathered}
$$

We repeatedly apply the formulas ( ${ }^{*}$ ) until we obtain an interval $\left[b_{n}, a_{n}\right]$, the length of which will satisfy one of the conditions $(* *)$ :

$$
\begin{gathered}
a_{2}=1.7559-\frac{(1.7540-1.7559) f(1.7559)}{\ell(1.7540)-f(1.7559)}=1.75558 ; \\
b_{2}=1.7540-\frac{f(1.7540)}{f^{\prime}(1.7540)}=1.75557 \\
a_{3}=1.7555816 ; \quad b_{3}=1.7555807 .
\end{gathered}
$$

Here, the length of the interval $\left[b_{3}, a_{3}\right]$ is less than 0.000001; $a_{3}-b_{3}=$ $=0.0000009$. Therefore, the approximate value of the root of the given equation with an accuracy of 0.000001 is

$$
x_{0} \approx b_{3}=1.755581
$$

In problems 403-406, determine the number of real roots of the equation and compute the largest of them with an accuracy of 0.01.