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. 
 A: Let $S_a$ be the set of solutions of the inequality $|f(x)|\leq 1/4$, or equivalently, $-1/4\leq f(x)\leq 1/4$. 
Clearly, the requirement of the problem is equivalent to $S$ containing two points with distance $2$. 
The leading coefficient of $f$ is positive, so the function tends to $\infty$ as $x\rightarrow \pm \infty$. 
So the set of solutions $S_a^+$ of the inequality $f(x)\leq 1/4$ is a bounded interval (possibly empty or degenerate), whose endpoints are the solutions of the equation $f(x)=1/4$. 
Clearly, the set of solutions $S_a^-$ of the inequality $f(x)\leq -1/4$ is also a bounded interval contained in $S_a^+$, and then $S= S_a^+\setminus S_a^-$. 
Thus if an $x$ with the desired properties exists, then the length of $S_a^+$ is at least $2$, i.e., there are exactly two real solutions of the equation $f(x)=1/4$, and the difference of the two solutions is at least $2$. 
First we show that this is equivalent to $a^2\geq 7$:
There exist two solutions of $x^2-ax+3/4=0$, namely $x_1< x_2$, iff $D= a^2-3 >0$, and then $x_2-x_1=\sqrt{D}$. 
So the exact condition for the above interval to have length at least $2$ is that $D\geq 4$, i.e., $a^2-3\geq 4$, or equivalently, $a^2\geq 7$.
So $a^2\geq 7$ is necessary. However, if the equation $f(x)=-1/4$ has real solution(s), then $a^2\geq 7$ might not be enough. 
Indeed, if the solutions of this equation are $y_1<y_2$, then clearly $x_1<y_1<y_2<x_2$, and if $y_2-y_1>2$, then we might end up with an $S= S_a^+\setminus S_a^-$ that does not contain two points of distance $2$. In this case, $S=[x_1, y_1]\cup [y_2, x_2]$, so it contains two points of distance $2$ iff the $a^2\geq 7$ condition is met, and furhtermore 


*

*either $y_2-y_1\leq 2$, or 

*$y_2-y_1> 2$ and $y_1-x_1\geq 2$. (The two intervals $[x_1, y_1]$ and $[y_2, x_2]$ have equal lengths, hence the condition $y_1-x_1\geq 2$.) 


Similarly to the above calculation, we have that $f(x)=-1/4$ has a real solution iff $a^2\geq 5$, so this actually holds when the necessary condition $a^2\geq 7$ holds. 
Furthermore, the difference of the two soluions of $f(x)=-1/4$ is at least  $2$ iff $a^2>9$. 
To sum up: 


*

*$a^2\geq 7$ is necessary, 

*If $7\leq a^2\leq 9$, then there are two numbers in $x$ exactly $2$ units apart, so this is fine. 

*If $a^2>9$, then the exact condition to have a solution is $y_1-x_1\geq 2$. That is $\frac{a-\sqrt{a^2-5}}{2}-\frac{a-\sqrt{a^2-3}}{2}\geq 2$, or equivalently, $\sqrt{a^2-3}-\sqrt{a^2-5}\geq 2$. 
This inequality has no solution (easy). 


So the final answer: $7\leq a^2\leq 9$.
A: 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 $a<0$. 
A: 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 $[\sqrt{7},3]\cup[-3,-\sqrt{7}]$. 
A: Considering the problem geometrically, then
$$
f(x) = x^{\,2}  - ax + 1
$$
is a vertical parabola, convex (upward concave), with axis $x=a/2$ and vertex in $V=(a/2,1-a^2/4)$.
Thus the parameter $a$ is then just a translation parameter.

So, from the above and with the help of the sketch it is clear that, to satisfy the requirements we shall have
$$
\eqalign{
  & \exists \,x\;:\;\left\{ \matrix{
  \left| {f(x)} \right| \le 1/4 \hfill \cr 
  \left| {f(x + 2)} \right| \le 1/4 \hfill \cr}  \right.\quad  \Leftrightarrow   \cr 
  &  \Leftrightarrow \quad \left\{ \matrix{
  2 \le \overline {A_{\,1} A_{\,2} }  \hfill \cr 
   - 1/4 \le V_{\,y}  \le 1/4 \hfill \cr}  \right.\quad  \cup \quad \left\{ \matrix{
  2 \le \overline {A_{\,1} A_{\,2} }  \hfill \cr 
  V_{\,y}  \le  - 1/4 \hfill \cr 
  \overline {B_{\,1} B_{\,2} }  \le 2 \hfill \cr}  \right. \cr} 
$$
which means
$$
\left\{ \matrix{
  2 \le \overline {A_{\,1} A_{\,2} }  \hfill \cr 
  \overline {B_{\,1} B_{\,2} }  = \emptyset \quad  \vee \quad \overline {B_{\,1} B_{\,2} }  \le 2 \hfill \cr}  \right.
$$
i.e.
$$
\left\{ \matrix{
  2 \le \sqrt {a^{\,2}  - 3}  \hfill \cr 
  a^{\,2}  - 5 \le 0\quad  \vee \quad \sqrt {a^{\,2}  - 5}  \le 2 \hfill \cr}  \right.
$$
and finally
$$ \bbox[lightyellow] {  
\left\{ \matrix{
  7 \le a^{\,2}  \hfill \cr 
  a^{\,2}  \le 9 \hfill \cr}  \right.\quad  \Rightarrow \quad \sqrt 7  \le \left| a \right| \le 3
}$$
A: Introduce a new horizontal coordinate $t:=x-{a\over2}$. Then we have to analyze how the parabolas
$$\gamma_c:\quad y=\hat f(t):=t^2+c,\qquad c:=1-{a^2\over4}\ ,\tag{1}$$
intersect the strip $|y|\leq{1\over4}$.
To begin with: If $-{5\over4}\leq c\leq-{3\over4}$ then $\hat f(\pm1)=1+c\in\bigl[-{1\over4},{1\over4}\bigr]$, hence such a $c$ is admissible. If $c>-{3\over4}$ then 
$\gamma_c$ is completely above the line $y={1\over4}$, or intersects it in a chord of length $<2$. Such a $c$ is not admissible. If $c<-{5\over4}$ then $\gamma_c$ intersects both horizontals $y=\pm{1\over4}$ in chords of length $>2$, and there are no two points $(t_i,y_i)\in\gamma_c$ on the intermediate arcs that satisfy $|t_2-t_1|=2$ (not even on the same side of the $y$-axis).
The conclusion is that the admissible $c$-interval is $\bigl[-{5\over4},-{3\over4}\bigr]$, and this translates via $(1)$ into the condition $$\sqrt{7}\leq|a|\leq 3$$ for the parameter $a$.
A: 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.
A: 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.
A: 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}
A: My Work: I will approach in a very short way.we have been given
 two characteristic equations of a function and their range .
 $$f(x)=x^2-ax+1$$
 $$\text{and, we know}$$
 $$\lvert x^2-ax+1 \rvert \le \dfrac{1}{4}$$
 $$\lvert (x+2)^2-a(x+2)+1 \rvert \le \dfrac{1}{4}$$
from the absolute value,we can write-
 $$-\dfrac{1}{4}\le x^2-ax+1 \le \dfrac{1}{4}$$
 $$-\dfrac{1}{4}\le (x+2)^2-a(x+2)+1 \le \dfrac{1}{4}$$
Now, Let's see how do these two inequalities come to our help? We have to find the range
 of $a$ for which these two inequalities held.It is certain that we will find a point of a if we work on
upper limit of these two inequalities and will find another point if we work on lower limit of these two inequalities.
to make my life easy ,I am counting them as equations.[Look,I am just using the equations to determine the peak values of  "$a$"]
So,if we take upper limit,we get-
$$x^2-ax+1 = \dfrac{1}{4}.......(1)$$
$$(x+2)^2-a(x+2)+1  = \dfrac{1}{4}......(2)$$
now if we solve these two equation we will get ,$a^2=7$
Again, if we take lower limit,we get-
$$ x^2-ax+1=-\dfrac{1}{4}.........(3)$$
$$ (x+2)^2-a(x+2)+1=-\dfrac{1}{4}..........(4)$$
If we solve these two equations we will get $a^2=9$
Hence, $ 7\le a^2 \le 9$
A: 
Let $$f(x)=x^2-ax+1.$$
  Find the range of all possible $a$ so that there exist $x$ with 
  $$|f(x)|\le\dfrac{1}{4},\quad |f(x+2)|\le\dfrac{1}{4}.$$

There is a solution if and only if one of the two equations $x^2-ax+1=\pm\dfrac14$ has two roots $r,s$ and $|r-s|\ge 2$ and the solution of $f(x+2)=f(x)$ satisfy $\left|f\left(\frac a2-1\right)\right|\le \frac 14.$
Part 1
$$\left|f\left(\frac a2-1\right)\right|\le \frac 14\iff 7\le a^2 \le 9.$$
Part 2
Let's work with $x^2-ax+1=\dfrac14.$ We have that $$x^2-ax+1=\dfrac 14\iff x=\dfrac{-a\pm\sqrt{a^2-3}}{2}.$$ So, there are two roots if and only if $a^2>3.$ Now
$$\left|\dfrac{a+\sqrt{a^2-3}}{2}-\dfrac{a-\sqrt{a^2-3}}{2}\right|\ge 2\iff \sqrt{a^2-3}\ge 2.$$ That is, $a^2\ge 7.$ So $a$ must satisfy three conditions: $a^2> 3,$ $a^2\ge 7$ and $7\le a^2\le 9.$ 
Part 3
If we work with $x^2-ax+1=-\dfrac14$ we get that $$x^2-ax+1=-\dfrac 14\iff x=\dfrac{-a\pm\sqrt{a^2-5}}{2}.$$ So, there are two roots if and only if $a^2>5.$ Now
$$\left|\dfrac{a+\sqrt{a^2-5}}{2}-\dfrac{a-\sqrt{a^2-5}}{2}\right|\ge 2\iff \sqrt{a^2-5}\ge 2.$$ That is, $a^2\ge 9.$ So $a$ must satisfy three conditions: $a^2> 5,$ $a^2\ge 9$ and $7\le a^2\le 9.$ 
Part 4
Thus the solution is
$$a\in [-3,-\sqrt 7]\cup [\sqrt 7,3].$$
