497. Find the absolute and relative errors when replacing the increment of the function $y=x^{3}+2 x$ with its differential at the point $x=2$ for $\Delta x=0.1$.
Solution. $\Delta y=\left((x+\Delta x)^{3}+2(x+\Delta x)\right)-\left(x^{3}+2 x\right)=3 x^{2} \Delta x+$ $+3 x(\Delta x)^{2}+(\Delta x)^{3}+2 \Delta x$

$$
\begin{aligned}
& \left.\Delta y\right|_{\substack{x=2,0,1 \\
\Delta x=0,1}}=3 \cdot 2^{2} \cdot 0.1+3 \cdot 2 \cdot 0.1^{2}+0.1^{3}+2 \cdot 0.1=1.461 ; \\
& d y=y^{\prime} \Delta x=\left(3 x^{2}+2\right) \Delta x ;\left.\quad d y\right|_{\substack{x=2,0 \\
\Delta x=0,1}}=\left(3 \cdot 2^{2}+2\right) \cdot 0.1=1.4
\end{aligned}
$$

$$
|\Delta y-d y|=1.461-1.4=0.061 ; \quad \delta=\left|\frac{\Delta y-d y}{\Delta y}\right|=\frac{0.061}{1.461} \approx 0.042=4 \%
$$