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. 

As is clear, this question is invariant under horizontal displacement i.e. you can replace $f(x)=(x-\dfrac{a}{2})^2+1-\dfrac{a^2}{4}$ with $f(x)=x^2+1-\dfrac{a^2}{4}$. Here after we use the latter notation. Define $b=1-\dfrac{a^2}{4}$therefore $$f(x)=x^2+b$$assume such a $x_0$ exists. We cannot have both $x_0$ and $x_0+2$ are positive or both negative. To show that if they are both positive we have $$-0.25<x_0^2+b<0.25\qquad(1)\\-0.25<(x_0+2)^2+b<0.25\qquad(2)$$but this is a contradiction since $$(x_0+2)^2+b=x_0^2+b+4x_0+4>-0.25+4x_0+4>3.75>0.25$$to the same argument they can't both be negative so we have $$x_0<0\\x_0+2>0$$the function is strictly decreasing when $x<0$ and strictly increasing when $x>0$ therefore the inequalities $(1)$ and $(2)$ impose that $$-\sqrt{0.25-b}\le x_0\le -\sqrt{-0.25-b}\\\sqrt{-0.25-b}\le x_0+2\le\sqrt{0.25-b}$$or equivalently $$-\sqrt{0.25-b}\le x_0\le -\sqrt{-0.25-b}\\\sqrt{-0.25-b}-2\le x_0\le \sqrt{0.25-b}-2$$both intervals are of equal length therefore such a $x_0$ exists and satisfies both inequalities iff the intervals intersect i.e.$$\text{either }\quad -\sqrt{0.25-b}\le \sqrt{-0.25-b}-2\le -\sqrt{-0.25-b}\\\text{or }\quad-\sqrt{0.25-b}\le \sqrt{0.25-b}-2\le -\sqrt{-0.25-b}$$which leaves us with $$-1.25\le b\le-1-\dfrac{1}{64}\\\text{or}\\-1-\dfrac{1}{64}\le b\le -0.75$$$$-1.25\le b \le -0.75$$which means that $$7\le a^2\le 9$$which means that $$\Large\sqrt 7\le |a|\le3$$You can see a sketch of the answer for $b=-1.25$ and $b=-0.75$ below

where $x_0$ and $x_0+2$ are at the ends of the blue line.
