Let $a$ be a real number and $P(x)=x^{2}-2 a x-a^{2}-\frac{3}{4}$. Find all values of $a$ such that $|P(x)| \leqslant 1$ for all $x \in[0,1]$.
The quadratic polynomial reaches its minimum at $x=a$, so on the interval of interest when $a \in[0,1]$. Apart from this point, it can only be a maximum or minimum at the boundaries of the interval, i.e., at -1 or 1. Therefore, it is sufficient to check when each of the three values $P(0), P(1)$, and $P(a)$ if $0 \leqslant a \leqslant 1$ lies between -1 and 1. For $P(0)=-a^{2}-\frac{3}{4}$, this is true when $-\frac{1}{2} \leqslant a \leqslant \frac{1}{2}$. $P(1)=\frac{1}{4}-2 a-a^{2}$ is less than or equal to 1 when $a \leqslant-\frac{3}{2}$ or $a \geqslant-\frac{1}{2}$, and greater than or equal to -1 when $-\frac{5}{2} \leqslant a \leqslant \frac{1}{2}$. Finally, for $0 \leqslant a \leqslant 1, P(a)=-2 a^{2}-\frac{3}{4}$ is always $<1$ and is $\geqslant-1$ when $|a| \leqslant \frac{\sqrt{2}}{4}$. The intersection of all these intervals gives the solution interval: $a \in\left[\frac{-1}{2}, \frac{\sqrt{2}}{4}\right]$.