I am having some problems getting started with this problem, as I never had to deal with an inequality that was between two values with absolute values. Any help is appreciated. The problem is find all values of $x$ in $\mathbb{R}$ that satisfy $4 < |x+2| + |x-1| < 5$. I keep trying to find cases with $x < -2$ or $x \geq 1$, but that is not getting me anywhere.
[Math] What are all values of $x$ in $\mathbb{R}$ that satisfy $4 < |x+2| + |x-1| < 5$
absolute valueinequalityreal-analysis
Best Answer
You can think of it as drawing the graph of such double-absolute value functions. So it is obvious that $-2$ and $1$ are two special points.
If $x \leq -2$, then $|x+2|=-x-2$ and $|x-1|=1-x$, so the $f(x) = -x-2+1-x = -1 - 2x $.
Solve the inequality of $4<-1-2x<5$ and combine with $x\leq-2$.
If $-2<x<1$, then $|x+2|=x+2$ and $|x-1|=1-x$ so the $f(x) = x+2+1-x=3$ which is constant.
If $x\geq 1$, then $|x+2|=x+2$ and $|x-1|=x-1$ so the $f(x) = x+ 2+x-1=2x+1$ which is grater than or equal to $3$. Similarly, solve the inequality $4<2x+1<5$ and combine it with $x\geq 1$.
Finally, combine all the possible cases and you will find the result. I think you just forget to consider the implied conditions $x\leq -2$ and $x\geq 1$.