# How do you graph and solve abs(x+1)<=11?

Jun 7, 2016

Solution:
$- 12 \le x \le 10$

#### Explanation:

Start from a definition of the absolute value.
For non-negative real $x$ the value of $\left\mid x \right\mid$ is $x$.
For negative real $x$ the value of $\left\mid x \right\mid$ is $- x$.

In short:
if $x \ge 0$, $| x | = x$
if $x < 0$, $| x | = - x$

Using this definition, let's divide the domain of the $| x + 1 |$ in two intervals:
1. $x + 1 \ge 0$, where $| x + 1 | = x + 1$
2. $x + 1 < 0$, where $| x + 1 | = - \left(x + 1\right)$

Case 1. Looking for solutions in interval $x + 1 \ge 0$ (or, equivalently, $x \ge - 1$)
Since in this interval $| x + 1 | = x + 1$, our inequality looks like
$x + 1 \le 11$ and the solution is $x \le 10$.
Together with a boundary of the interval, we have come to a solution:
$- 1 \le x \le 10$

Case 2. Looking for solutions in interval $x + 1 < 0$ (or, equivalently, $x < - 1$)
Since in this interval $| x + 1 | = - \left(x + 1\right)$, our inequality looks like
$- \left(x + 1\right) \le 11$ and the solution is $x \ge - 12$.
Together with a boundary of the interval, we have come to a solution:
$- 12 \le x < - 1$

So, two intervals are the solution to an original inequality:
$- 1 \le x \le 10$ and
$- 12 \le x < - 1$

These can be combined into
$- 12 \le x \le 10$

Here is a graph of $y = | x + 1 |$

graph{|x+1| [-20, 20, -15, 15]}
It illustrates the solution