Let's create the tables of values for the given rational functions.

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

We will substitute $x = -2, -1, 0, 1, 2$ into the function:

$x$	$f(x) = \frac{2}{x + 1}$
-2	$f(-2) = \frac{2}{-2 + 1} = \frac{2}{-1} = -2$
-1	$f(-1) = \frac{2}{-1 + 1} = \frac{2}{0}$ (undefined)
0	$f(0) = \frac{2}{0 + 1} = \frac{2}{1} = 2$
1	$f(1) = \frac{2}{1 + 1} = \frac{2}{2} = 1$
2	$f(2) = \frac{2}{2 + 1} = \frac{2}{3} \approx 0.67$

Notice that at $x = -1$, the function is undefined because the denominator becomes zero.

2. $f(x) = \frac{3x}{x + 3}$

Now, we substitute the same $x$-values into the second function:

$x$	$f(x) = \frac{3x}{x + 3}$
-2	$f(-2) = \frac{3(-2)}{-2 + 3} = \frac{-6}{1} = -6$
-1	$f(-1) = \frac{3(-1)}{-1 + 3} = \frac{-3}{2} = -1.5$
0	$f(0) = \frac{3(0)}{0 + 3} = \frac{0}{3} = 0$
1	$f(1) = \frac{3(1)}{1 + 3} = \frac{3}{4} = 0.75$
2	$f(2) = \frac{3(2)}{2 + 3} = \frac{6}{5} = 1.2$

Summary:

1. For $f(x) = \frac{2}{x + 1}$, the function is undefined at $x = -1$, and the other values are straightforward.
2. For $f(x) = \frac{3x}{x + 3}$, the function is continuous, with various values for $x$.

Let me know if you want more details or have questions.

Here are some questions that might expand your understanding:

1. How does the behavior of each function change as $x \to -\infty$?
2. What happens to each function as $x \to \infty$?
3. Can you determine the asymptotes of each function?
4. How would you sketch the graph of each function?
5. What does the undefined point at $x = -1$ in the first function signify?

Tip: Rational functions often have vertical asymptotes where the denominator becomes zero. Identifying these can help you better understand their behavior.