We need to analyze the given functions and sketch their graphs. I will address part (a) of the problem, which is $y = x - 2 + \frac{x(x^2 - 1)}{|x^2 - 1|}$.

1. Problem Description

We need to analyze the given functions and sketch their graphs. I will address part (a) of the problem, which is $y = x - 2 + \frac{x(x^2 - 1)}{|x^2 - 1|}$.

2. Solution Steps

First, let's simplify the expression. Note that $x^2-1 = (x-1)(x+1)$.

Also, recall that

$|x^2 - 1| = \begin{cases} x^2 - 1, & \text{if } x^2 - 1 \geq 0 \\ -(x^2 - 1), & \text{if } x^2 - 1 < 0 \end{cases}$

which means

$|x^2 - 1| = \begin{cases} x^2 - 1, & \text{if } x \in (-\infty, -1] \cup [1, \infty) \\ -(x^2 - 1), & \text{if } x \in (-1, 1) \end{cases}$

Now we can analyze the two cases:

Case 1: $x \in (-\infty, -1) \cup (1, \infty)$, $x \neq -1$, $x \neq 1$

Then $|x^2 - 1| = x^2 - 1$, and the expression becomes

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

Case 2: $x \in (-1, 1)$, $x \neq -1$, $x \neq 1$

Then $|x^2 - 1| = -(x^2 - 1)$, and the expression becomes

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

In summary, we have:

$y = \begin{cases} 2x - 2, & \text{if } x \in (-\infty, -1) \cup (1, \infty) \\ -2, & \text{if } x \in (-1, 1) \end{cases}$

The graph is a piecewise function. For $x < -1$ or $x > 1$, the graph is a line $y = 2x - 2$. For $-1 < x < 1$, the graph is a horizontal line $y = -2$. At $x=-1$ and $x=1$, the function is undefined.

3. Final Answer

The function is defined as:

$y = \begin{cases} 2x - 2, & \text{if } x \in (-\infty, -1) \cup (1, \infty) \\ -2, & \text{if } x \in (-1, 1) \end{cases}$

The graph consists of a line $y = 2x - 2$ for $x < -1$ and $x > 1$, and a horizontal line $y = -2$ for $-1 < x < 1$. The function is undefined at $x=-1$ and $x=1$.