<s>[INST] Find the range of $a$ such $|f(x)|\le\frac{1}{4},$ $|f(x+2)|\le\frac{1}{4}$ for some $x$. 
Let $$f(x)=x^2-ax+1.$$
  Find the range of all possible $a$ so that there exist $x$ with 
  $$|f(x)|\le\dfrac{1}{4},\quad |f(x+2)|\le\dfrac{1}{4}.$$

A sketch of my thoughts: I write 
$$f(x)=\left(x-\dfrac{a}{2}\right)^2+1-\dfrac{a^2}{4}\ge 1-\dfrac{a^2}{4}$$
so if $1-\dfrac{a^2}{4}>\dfrac{1}{4}$ or $-\sqrt{3}<a<\sqrt{3}$
this case impossible
But I don't know how to prove the other case, or if this there are better ideas. 
 [/INST] We first see that the case for $a$ is identical to the case for $-a$, so we consider only $a\geq 0$. Let's find the solutions to 
$$|f(x)|=\frac{1}{4}.$$
If $f(x)=\frac{1}{4}$ we have
$$x^2-ax+\frac{3}{4}=0$$
$$x = \frac{a\pm\sqrt{a^2-3}}{2},$$
which has real solutions iff $a^2\geq 3$. If $f(x)=-\frac{1}{4}$ we have
$$x^2-ax+\frac{5}{4}=0$$
$$x = \frac{a\pm\sqrt{a^2-5}}{2},$$
which has real solutions iff $a^2\geq 5$. We may now, as $a>0$, write the solution set to $|f(x)|\leq \frac{1}{4}$ as 
$$\left[\frac{a-\sqrt{a^2-3}}{2},\frac{a+\sqrt{a^2-3}}{2}\right]$$
if $3\leq a^2<5$, and
$$\left[\frac{a-\sqrt{a^2-3}}{2},\frac{a-\sqrt{a^2-5}}{2}\right]\bigcup\left[\frac{a+\sqrt{a^2-5}}{2},\frac{a+\sqrt{a^2-3}}{2}\right].$$
We have if $a^2<5$ a solution iff
$$\frac{a+\sqrt{a^2-3}}{2}-\frac{a-\sqrt{a^2-3}}{2}\geq 2$$
$$\sqrt{a^2-3}\geq 2$$
$$a^2\geq 7,$$
a contradiction. So, $a^2\geq 5$, and we need either the length of one interval to be $\geq 2$, or the distance between the "innermost" points to be $\leq 2$ and the distance between the "outermost" points to be $\geq 2$. In the first case, we have
$$\sqrt{a^2-3}-\sqrt{a^2-5}\geq 2$$
$$\frac{2}{\sqrt{a^2-3}+\sqrt{a^2-5}}\geq 2$$
$$\sqrt{a^2-3}+\sqrt{a^2-5}\leq 1.$$
However, $\sqrt{a^2-3}\geq \sqrt{5-3}=\sqrt{2}>1$, so this cannot occur. The other case is
$$\frac{a+\sqrt{a^2-5}}{2}-\frac{a-\sqrt{a^2-5}}{2}\leq 2$$
$$\sqrt{a^2-5}\leq 2$$
$$a^2\leq 9$$
$$a\leq 3,$$
as well as the first calculation we did (the distance between the two outermost points is algebraically identical to the distance between the two points in our $a^2<5$ case, which reduced to $a^2\geq 7$. So, our solution set is $\left[\sqrt{7},3\right]$ for $a>0$, and correspondingly $\left[-3,-\sqrt{7}\right]$ for $a<0$. 
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