1. Determine the maximum value of the expression

$$
\left|8 x^{2}-2 x-3\right|
$$

on the interval $x \in[0,1]$.
1. Let's consider the function $f(x)=8 x^{2}-2 x-3$. Its zeros are at points $x_{1/2}=\frac{2 \pm \sqrt{4+96}}{16}=\frac{2 \pm 10}{16}$, i.e., $x_{1}=-\frac{1}{2}$ and $x_{2}=\frac{3}{4}$, which means that the function $f(x)$ is negative on the segment $[0,1]$ for $x \in\left[0, \frac{3}{4}\right)$ and positive for $x \in\left(\frac{3}{4}, 1\right]$. From the properties of quadratic functions, we know that its minimum is achieved at the point
$\frac{x_{1}+x_{2}}{2}=\frac{1}{4}=\frac{1}{8}$, and this minimum is $8 \cdot \frac{1}{64}-2 \cdot \frac{1}{8}-3=\frac{1}{8}-$ $\frac{1}{4}-3=-\frac{1}{8}-3$; thus, the maximum value of $|f(x)|$ for $x \in\left[0, \frac{3}{4}\right)$ is $3+\frac{1}{8}$. For $x \in\left(\frac{3}{4}, 1\right]$, the function $f(x)$ is positive and increasing, so it achieves its maximum at $x=1$, and this maximum is $8 \cdot 1^{2}-2 \cdot 1-3=3$; this is also the maximum of $|f(x)|$ for $x \in\left(\frac{3}{4}, 1\right]$. Therefore, the maximum value of the expression in the problem on the interval $x \in[0,1]$ is $3+\frac{1}{8}$ and is achieved at $x=\frac{1}{8}$.