99. Find the increments of the argument and the function $y=2 x^{2}+1$, if the argument $x$ changes from 1 to 1.02.
Solution. $1^{0}$. Find the increment of the argument: $\Delta x=1.02-1=$ $=0.02$.

$2^{0}$. Find the value of the function at the old value of the argument, i.e., at $x=1: y=2 \cdot 1^{2}+1=3$.

$3^{\circ}$. Find the value of the function at the new value of the argument, i.e., at $x=1+0.02=1.02$:

$$
y_{\mathrm{N}}=y+\Delta y=2 \cdot 1.02^{2}+1=2 \cdot 1.0404+1=3.0808
$$

$4^{0}$. By subtracting the original value of the function from the new one, we find the increment of the function: $\Delta y=3.0808-3=0.0808$.

We can also arrive at the same result in a different way. First, we find the increment of the given function in general form:

$$
\Delta y=2(x+\Delta x)^{2}+1-\left(2 x^{2}+1\right)=4 x \Delta x+2(\Delta x)^{2}
$$

Now, substituting the values of $x$ and $\Delta x$ here, we get

$$
\Delta y=4 \cdot 1 \cdot 0.02+2 \cdot 0.004=0.0808
$$