We are given the functions:

$f: \mathbb{R} \setminus \{0\} \to \mathbb{R}, \quad f(x) = \frac{1}{2} \left( \frac{1}{x} + 1 \right)$

$g: \mathbb{R} \setminus \left\{\frac{1}{2} \right\} \to \mathbb{R}, \quad g(x) = \frac{1}{2x - 1}$

We need to:

1. Compute $g \circ f$, meaning $g(f(x))$.
2. Determine the range of $f(x)$ and $g \circ f(x)$.

Step 1: Compute $g \circ f$

By definition,

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

Substituting into $g(x)$:

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

Simplify the denominator:

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

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

Thus,

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

This shows that $g \circ f(x) = x$, which means it is the identity function.

Step 2: Determine the Range of Each Function

Range of $f(x)$

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

As $x \to 0^+$, we have $\frac{1}{x} \to +\infty$, so $f(x) \to +\infty$.

As $x \to 0^-$, we have $\frac{1}{x} \to -\infty$, so $f(x) \to -\infty$.

For large positive and negative $x$, we see that:

$\frac{1}{x} \to 0 \Rightarrow f(x) \to \frac{1}{2}.$

Thus, the range of $f(x)$ is:

$\mathbb{R} \setminus \left\{\frac{1}{2}\right\}.$

Range of $g \circ f(x)$

Since we found that $g \circ f(x) = x$, it covers all of $\mathbb{R} \setminus \{0\}$, meaning its range is:

$\mathbb{R} \setminus \{0\}.$

---

Final Answer:

1. $g \circ f(x) = x$ (Identity function).
2. Range of $f(x)$: $\mathbb{R} \setminus \{ \frac{1}{2} \}$.
3. Range of $g \circ f(x)$: $\mathbb{R} \setminus \{0\}$.

Do you need a more detailed explanation or further clarifications?

Here are five related questions to explore further:

1. What is $f \circ g(x)$, and does it also give the identity function?
2. How would the domain and range change if $f(x)$ or $g(x)$ were modified slightly?
3. Can you determine an inverse function $f^{-1}(x)$ explicitly?
4. What happens to the composition if we change $f(x)$ to $\frac{1}{x+1}$?
5. How do these transformations relate to real-world functions in physics or engineering?

Tip: Always check the domain restrictions carefully when composing functions to avoid undefined values!