The problem is to simplify the expression $y = \frac{|x+1| - |x-1|}{2}$.

1. Problem Description

The problem is to simplify the expression $y = \frac{|x+1| - |x-1|}{2}$.

2. Solution Steps

We need to consider different cases for the values of $x$ based on where the expressions inside the absolute values change signs. The critical points are $x=-1$ and $x=1$. This divides the number line into three intervals: $x < -1$, $-1 \le x \le 1$, and $x > 1$.

Case 1: $x < -1$

In this case, $x+1 < 0$ and $x-1 < 0$, so $|x+1| = -(x+1)$ and $|x-1| = -(x-1)$. Then

$y = \frac{-(x+1) - (-(x-1))}{2} = \frac{-x-1 + x - 1}{2} = \frac{-2}{2} = -1$.

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

In this case, $x+1 \ge 0$ and $x-1 \le 0$, so $|x+1| = x+1$ and $|x-1| = -(x-1)$. Then

$y = \frac{(x+1) - (-(x-1))}{2} = \frac{x+1 + x - 1}{2} = \frac{2x}{2} = x$.

Case 3: $x > 1$

In this case, $x+1 > 0$ and $x-1 > 0$, so $|x+1| = x+1$ and $|x-1| = x-1$. Then

$y = \frac{(x+1) - (x-1)}{2} = \frac{x+1 - x + 1}{2} = \frac{2}{2} = 1$.

Therefore, the function $y$ can be expressed as a piecewise function:

$y = \begin{cases} -1, & x < -1 \\ x, & -1 \le x \le 1 \\ 1, & x > 1 \end{cases}$

3. Final Answer

$y = \begin{cases} -1, & x < -1 \\ x, & -1 \le x \le 1 \\ 1, & x > 1 \end{cases}$