496. Find approximately the increment of the function $y=3 x^{2}+2$ at $x=2$ and $\Delta x=0.001$. Determine the absolute and relative errors of the calculation.
Solution. Since the increment of the argument is a small quantity, the increment of the function can be replaced by its differential:

$$
\left.\Delta y \approx d y\right|_{\substack{x=2 \\ d x=0.001}}=\left.6 x d x\right|_{\substack{x=2, d x=0.001}}=6 \cdot 2 \cdot 0.001=0.012
$$

Let's find the error obtained by replacing the increment of the function with its differential. For this, we will calculate the exact value of the function increment:

$$
\begin{gathered}
\Delta y=f(x+\Delta x)-f(x)=3(x+\Delta x)^{2}+2-\left(3 x^{2}+2\right)=3 x^{2}+6 x \Delta x+3(\Delta x)^{2}+ \\
+2-3 x^{2}-2=6 x \Delta x+3(\Delta x)^{2} \\
=6 \cdot 2 \cdot 0.001+3 \cdot 0.000001=0.012003
\end{gathered}
$$

Comparing the exact value of $\Delta y$ with the approximate value, we see that the absolute error is

$$
\Delta=|\Delta y-d y|=0.000003
$$

The relative error is

$$
\delta=\left|\frac{\Delta y-d y}{\Delta y}\right|=\frac{0.000003}{0.012003} \approx 0.00025=0.025 \%
$$