We are given the functions:

1. $f(x) = \frac{1 - x}{x}$, with $x \neq 0$.
2. $g(x) = x + 2$.

The problem asks for $(g \circ f)(x)$, which is the composition $g(f(x))$.

Step 1: Find $f(x)$

We have: $f(x) = \frac{1 - x}{x} = \frac{1}{x} - 1.$

Step 2: Substitute $f(x)$ into $g(x)$

Since $(g \circ f)(x) = g(f(x))$, we substitute $f(x) = \frac{1 - x}{x}$ into $g(x) = x + 2$:

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

Step 3: Simplify $g(f(x))$

To add $\frac{1 - x}{x}$ and $2$, we write $2$ as $\frac{2x}{x}$:

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

Answer

Thus, $(g \circ f)(x) = \frac{1 + x}{x}$.

Looking at the answer choices, the correct answer is: $\text{D. } \frac{1}{x} + 1 = \frac{1 + x}{x}.$

Would you like more details on any step, or have any questions?

Here are some related questions to explore further:

1. How do we find the domain of a composite function?
2. What are the properties of composite functions?
3. Can we break down composite functions involving rational expressions further?
4. How would the answer change if $g(x)$ were different?
5. How do we find the inverse of a composite function?

Tip: When working with compositions, always carefully substitute and simplify, especially if fractions are involved.