The image presents three different functions, and based on the text at the end of the image, it seems like we are expected to find the domains. Let's find the domains of these functions.

1. Problem Description

The image presents three different functions, and based on the text at the end of the image, it seems like we are expected to find the domains. Let's find the domains of these functions.

2. Solution Steps

Function 1: $f(x) = \frac{3x - 1}{2x - 6}$

To find the domain, we need to find the values of $x$ for which the function is defined. Since it is a rational function, the denominator cannot be zero. So,

$2x - 6 \neq 0$

$2x \neq 6$

$x \neq 3$

The domain is all real numbers except $x = 3$.

Function 2: $f(x) = \frac{x^2 - 2x + 1}{x^2 - x - 2}$

Again, the denominator cannot be zero. So,

$x^2 - x - 2 \neq 0$

$(x - 2)(x + 1) \neq 0$

$x \neq 2$ and $x \neq -1$

The domain is all real numbers except $x = 2$ and $x = -1$.

Function 3: $f(x) = \begin{cases} \frac{1}{2}x + 1 & \text{if } x \le 2 \\ 3 - x & \text{if } x > 2 \end{cases}$

This is a piecewise function. The first piece is defined for $x \le 2$, and it is a linear function, so it is defined for all $x$ in this interval. The second piece is defined for $x > 2$, and it is also a linear function, so it is defined for all $x$ in this interval.

Therefore, the function is defined for all real numbers.

3. Final Answer

Function 1: The domain is all real numbers except $x=3$.

Function 2: The domain is all real numbers except $x=2$ and $x=-1$.

Function 3: The domain is all real numbers.