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. 

This solution may look different. :-) 
Note that $f(x)=(x-\frac{a}{2})^2+1-\frac{a^2}{4}$, 
$$2*\frac{1}{4} \ge |f(x) -f(x+2)| = |2*(2x+2-a)| \Rightarrow |(x+2-\frac{a}{2})+(x-\frac{a}{2})|\le \frac{1}{4},$$
so $x-\frac{a}{2}$ and $x+2-\frac{a}{2}$ must have different signs as their difference is 2.  It must be $x-\frac{a}{2} \le 0 \le x+ 2-\frac{a}{2}$, since $x-\frac{a}{2} < x+2-\frac{a}{2}.$
So $-x-2 + a -\frac{a}{2} \le 0$, and $f(-x-2+a) = f(x+2)$. Since $-1+ \frac{a}{2}$ is the average of $x$ and $-x-2+ a$, and both $x-\frac{a}{2}$ and $-x-2+a-\frac{a}{2}$ are negative, then we know $f(-1+\frac{a}{2})$ is between $f(x)$ and $f(-x-2+a)=f(x+2)$, so $|f(-1+\frac{a}{2})|\le 1/4$, which yields $$7 \le a^2 \le 9.$$ 
On the other hand, if $7\le a^2 \le 9$, let $x=-1 + a/2$, then $|f(x+2)|=|f(x)|\le 1/4.$  therefore, $7 \le a^2 \le 9$ is the solution.
