<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] Let $f(x) = x^2 -ax + 1$ and let $g(x) = f(x+2) = x^2 - (a-4)x + (5-2a)$

Find all $x$ such that $|f(x)|\le\frac{1}{4}$ and $|g(x)|\le\frac{1}{4}$
If $f(x)$ and $g(x)$ are to share that property at the same point, that point must be the one point that they have in common. 
\begin{align}
   f(x) &= g(x) \\
   x^2 -ax + 1 &= x^2 - (a-4)x + (5-2a) \\
   -4x &= 4-2a \\
    x &= -1 + \dfrac{a}{2} \\
    y &= 2-\dfrac{a^2}{4}
\end{align}
The point that they have in common is the point 
$\left(-1 + \dfrac a2,2-\dfrac{a^2}{4}  \right)$.
So we need to solve
\begin{array}{c}
   \left| 2-\dfrac{a^2}{4} \right| \le \dfrac 14 \\
   |8 - a^2| \le 1 \\
   -1 \le a^2 - 8 \le 1 \\
   7 \le a^2 \le 9 \\
   a \in (-3, -\sqrt 7) \cup (\sqrt 7, 3)
\end{array}
And finally, $a = 2x + 2 \implies$
\begin{array}{c}
   -3 \le  2x+2 \le -\sqrt 7 \\
   -\dfrac 52 \le x \le -\dfrac{\sqrt 7 + 2}{2} \\
\hline
   \sqrt 7 \le 2x+2 \le 3 \\
   \dfrac{\sqrt 7 - 2}{2} \le x \le \dfrac 12 \\
\hline
   x \in \left[-\dfrac 52, -\dfrac{\sqrt 7 + 2}{2} \right]
   \cup \left[\dfrac{\sqrt 7 - 2}{2} ,\dfrac 12  \right]
\end{array}
</s>