What are all values of $x$ in $\mathbb{R}$ that satisfy $4 < |x+2| + |x-1| < 5$? 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.
 A: 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 $x\geq 1$.
A: I would think about it first in terms of positive numbers. Trying to minimize and maximum the inner expression gives you values of $x=\frac{3}{2}$ and $x=2$; however, these give you back the numbers $4$ and $5$, respectively. Thus, you may only use points very close to $x=\frac{3}{2}$ and $x=2$ (so the endpoints are open as you specify in your interval). So we have $\frac{3}{2}<x<2$ as valid $x$-values. Cool. Now you need to look at the negative numbers (because we are dealing with absolute values). When you do this, you will see that $x=-3$ and $x=-\frac{5}{2}$ yield $5$ and $4$ for the middle expression. Putting this all together, we see that
$$
4 < |x+2| + |x-1| < 5
$$
when $x\in\{\left(-3,-\frac{5}{2}\right)\cup \left(\frac{3}{2},2\right)\}$.
A: If $4 < |x+2| + |x-1| < 5$ then $\ \pm (x+2) \pm (x-1) < 5$. Note that for an opposite choice of sign the inequality is never satisfied since $(x+2) - (x-1)=3$. So the inequality implies 
$$ 4 < (x+2) + (x-1) < 5$$
or 
$$4 < -(x+2) -(x-1) < 5$$
The inequality $ 4 < (x+2) + (x-1) < 5$ is equivalent to $4 < 2x+1<5 \iff 1.5 < x < 2$ and on this interval both $x+2$, $x-1$ are $>0$ so we have part of the solution.
The inequality $ 4 < -(x+2) -(x-1) < 5$ is equivalent to $4 < -2x-1<5 \iff -3 < x < -2.5$ and on this interval both $x+2$, $x-1$ are $<0$ so we have another part of the solution. 
Therefore, the set of solutions is $(-3,-2.5) \cup (1.5,2)$
A: Think of it as a two variable inequality 4<|x+2|+|y-1|<5
Then pairs x,y may be represented by points, and the condition states that their taxi-cab distances to (-2,1) are bigger than 4, smaller than 5. Draw the two taxi-cab balls (same center (2,-1), different radius 4 and 5). The solution is the 'disk' left between the smaller and the bigger. Cut the disk with the line x=y. Then project on any axis.
