Example 1. Calculate approximately the value of the function $f(x)=$ $=\sqrt{x^{2}+x+3}$ at $x=1.97$
Solution. Let $x_{0}=2$. Then $y_{0}=f(2)=\sqrt{2^{2}+2+3}=3$. Further,

$$
y^{\prime}=\left.\frac{2 x+1}{2 \sqrt{x^{2}+x+3}}\right|_{x=2}=\frac{5}{6}
$$

Since $\Delta x=x-x_{0}$, then $\Delta x=-0.03$ and, according to the formula in point $3^{\circ}$, we have $\sqrt{1.97^{2}+1.97+3} \approx 3+\frac{5}{6}(-0.03)=2.975$.