Solving inequalities involving absolute values Find all $a,b\in \mathbb{R}$
$$\max_{x\in[0,1]} |ax+b-x-1| < 1/2$$
I am confused about how to solve absolute value inequality with max function so I don't know how to start solving this problem.
 A: Just square to eliminate the absolute value. It is equivalent to solve for
$$\underset{0\le x\le 1}{\max}(ax+b-x-1)^2<\frac14.$$
Now $\;p(x)=(ax+b-x-1)^2\,$ is a quadratic polynomial and its maximum on $[0,1]$ is one of $p(0)$, $p(1)$. Hence the inequation in absolute value is equivalent to the system of inequations:
$$\begin{cases}
p(0)=(b-1)^2<\dfrac14,\\
p(1)=(a+b-2)^2<\dfrac14, 
\end{cases}\iff\begin{cases} |b-1|<\dfrac12,\\|a+b-2|<\dfrac12, \end{cases}
\iff\begin{cases} \dfrac12<b<\dfrac32,\\\dfrac32<a+b<\dfrac52. \end{cases}$$ 
These inequalities can be solved graphically in the $(a$,$b)$-plane. They're the points inside a parallelogram.
A: HINT.- The line $y=(a-1)x+(b-1)$ determines the absolute value $$\max_{x\in[0,1]} |ax+b-x-1|$$
Consider the square region determined by the points $$A(0,-\dfrac12),B=(1,-\dfrac12),C=(1,\dfrac12),D=(0,\dfrac12)$$ The solution is formed by the couples $(a,b)$ such that the lines $y=(a-1)x+(b-1)$   cut the two vertical sides $AD$ and $BC$ of the square excepting the four extremes $A,B,C$ and $D$. All other line gives
$$\max_{x\in[0,1]} |ax+b-x-1|\ge\dfrac12$$ as it is easy to prove.
