<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] To start off a lot of time can be saved by only handling $a\geq0$ since both $|x^2-ax+1|\leq \frac{1}{4}$ and $|(x+2)^2-a(x+2)+1|\leq\frac{1}{4}$ are odd inequalities all of solutions that we find for $a\geq0$ will be the same except with opposite signs.
As a first step we should try to isolate $a$ from both of the inequalities. Starting with $|x^2-ax+1|\leq \frac{1}{4}$ we remove the absolute value and rewrite the single inequality as a double inequality, $-\frac{1}{4}\leq x^2-ax+1\leq \frac{1}{4}$ from there working step by step we can solve for $a$.
$$-\frac{1}{4}\leq x^2-ax+1\leq \frac{1}{4}$$
$$-\frac{5}{4}\leq x^2-ax\leq-\frac{3}{4}$$
$$-\frac{5}{4x}\leq x-a\leq-\frac{3}{4x}$$
$$-\frac{5}{4x}-x\leq -a\leq-\frac{3}{4x}-x$$
$$\frac{5}{4x}+x\geq a\geq\frac{3}{4x}+x$$
For the second inequality a similar path can be taken to obtain $$\frac{5}{4(x+2)}+x+2\geq a \geq \frac{3}{4(x+2)}+x+2$$
From here we can find the overlap of these two double inequalities and the range of $a$ for which they overlap which will be our solution. The upper bound of the over lap can be found by finding the intersection of the corresponding upper bonds of each double inequality.
$$\frac{5}{4x}+x=\frac{5}{4(x+2)}+x+2$$
$$\frac{5}{4x}=\frac{5}{4(x+2)}+2$$
Assuming that $x\neq0$,$x\neq-2$
$$\frac{5(x+2)}{4x}=\frac{5}{4}+2(x+2)$$
$$\frac{5(x+2)}{4}=\frac{5x}{4}+2x(x+2)$$
$$\frac{5x+10}{4}=\frac{5x}{4}+2x^2+4x$$
$$\frac{5x+10}{4}=\frac{5x}{4}+2x^2+4x$$
$$5x+10=5x+8x^2+16x$$
$$0=8x^2+16x-10$$
$$0=4x^2+8x-5$$
Using the quadratic formula
$$x=\frac{-8\pm\sqrt{64-4(4)(-5)}}{8}$$
$$x=\frac{-8\pm\sqrt{144}}{8}$$
$$x=\frac{-8\pm12}{8}$$
$$x=-1\pm1.5$$
$$x=\{.5,-2.5\}$$
We can discard the negative solution for now since for now we are only considering $a\geq0$ which also means that $x\geq0$.
Using the upper bound from the first inequality to calculate the corresponding value of $a$ gives us $a=3$ as our maximum for $a$'s range.
Working through the same problem for our lower bound gives the values $x=\{\frac{1}{2}\sqrt{7}-1,-\frac{1}{2}\sqrt{7}-1\}$, discarding the negative one for the same reason as before and plugging in $\frac{1}{2}\sqrt{7}-1$ into the lower bound equations of either the first or second double inequality gives us $a=\sqrt{7}$ as our lower bound for the range of $a$. That gives us the range for $a\geq0$ as $[\sqrt{7},3]$. Since both inequalities are odd we can extrapolate that for $a\leq0$ the range is $[-3,-\sqrt{7}]$, combining the two gives use that the range for $a$ is $[\sqrt{7},3]\cup[-3,-\sqrt{7}]$. 
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