You want to find the composition of the functions $f(x) = \frac{x - 2}{x + 2}$ and $g(x) = \frac{1}{x}$, specifically $\left( f \circ g \right)(x)$. This notation represents $f(g(x))$.

To find $\left( f \circ g \right)(x)$, substitute $g(x) = \frac{1}{x}$ into the function $f(x)$:

$(f \circ g)(x) = f\left(g(x)\right) = f\left(\frac{1}{x}\right)$

Given that $f(x) = \frac{x - 2}{x + 2}$, substitute $\frac{1}{x}$ for $x$:

$f\left(\frac{1}{x}\right) = \frac{\frac{1}{x} - 2}{\frac{1}{x} + 2}$

Now, let's simplify this expression:

$f\left(\frac{1}{x}\right) = \frac{\frac{1 - 2x}{x}}{\frac{1 + 2x}{x}}$

Since the denominator $x$ cancels out:

$f\left(\frac{1}{x}\right) = \frac{1 - 2x}{1 + 2x}$

Therefore, the composition $\left( f \circ g \right)(x)$ is:

$\left( f \circ g \right)(x) = \frac{1 - 2x}{1 + 2x}$

Do you have any questions or would you like more details on this topic?

---

Here are some related questions you might find helpful:

1. What is the domain of the composition $\left( f \circ g \right)(x)$?
2. How do you find the inverse of $\left( f \circ g \right)(x)$?
3. Can you determine the range of $\left( f \circ g \right)(x)$?
4. What happens if you compute the composition in the reverse order, $\left( g \circ f \right)(x)$?
5. How would you solve for $x$ if $\left( f \circ g \right)(x) = 0$?

Tip: Always remember that when composing functions, you must consider the domains of both functions to ensure the composition is valid over the appropriate intervals.