# Question #1cf68

Oct 31, 2016

$x \in \left[- 1 , 1\right]$

#### Explanation:

The absolute value function $| x |$ is defined by

$| x | = \left\{\begin{matrix}x \mathmr{if} x \ge 0 \\ - x \mathmr{if} x < 0\end{matrix}\right.$

As such, we will consider three cases:

Case 1: $x < - 1$

$\implies x + 1 < 0 \mathmr{and} x - 1 < 0$

$\implies | x + 1 | = - \left(x + 1\right) \mathmr{and} | x - 1 | = - \left(x - 1\right)$

$\implies - \left(x + 1\right) - \left(x - 1\right) \le 2$

$\implies - x - 1 - x + 1 \le 2$

$\implies - 2 x \le 2$

$\implies 2 x \implies 2$

$\implies x \ge 1$

As this contradicts the premise that $x < - 1$, there are no solutions on $\left(- \infty , - 1\right)$

Case 2: $- 1 \le x \le 1$

$\implies x + 1 \ge 0 \mathmr{and} x - 1 \le 0$

$\implies | x + 1 | = x + 1 \mathmr{and} | x - 1 | = - \left(x - 1\right)$

$\implies x + 1 - \left(x - 1\right) \le 2$

$\implies x + 1 - x + 1 \le 2$

$\implies 2 \le 2$

As this is true in all cases, every value in $\left[- 1 , 1\right]$ is a solution.

Case 3: $x > 1$

$\implies x + 1 > 0 \mathmr{and} x - 1 > 0$

$\implies | x + 1 | = x + 1 \mathmr{and} | x - 1 | = x - 1$

$\implies x + 1 + x - 1 \le 2$

$\implies 2 x \le 2$

$\implies x \le 1$

This contradicts the premise that $x > 1$, meaning there are no solutions on $\left(1 , \infty\right)$.

Taken together, we have the solution set as $\left[- 1 , 1\right]$. The inequality will hold for any $x$ in that interval, and will not for any $x$ outside of it.