We are given the functions:

$f(x) = \frac{8}{x - 7}$ and $g(x) = \frac{8}{x} + 7$

Step 1: Find $f(g(x))$

We substitute $g(x)$ into $f(x)$:

$f(g(x)) = f\left( \frac{8}{x} + 7 \right)$ Now, substitute $\frac{8}{x} + 7$ into the formula for $f(x)$:

$f\left( \frac{8}{x} + 7 \right) = \frac{8}{\left( \frac{8}{x} + 7 \right) - 7}$

Simplify the denominator:

$f(g(x)) = \frac{8}{\frac{8}{x} + 7 - 7} = \frac{8}{\frac{8}{x}} = x$

So, $f(g(x)) = x$.

Step 2: Find $g(f(x))$

We now substitute $f(x)$ into $g(x)$:

$g(f(x)) = g\left( \frac{8}{x - 7} \right)$ Substitute $\frac{8}{x - 7}$ into the formula for $g(x)$:

$g\left( \frac{8}{x - 7} \right) = \frac{8}{\frac{8}{x - 7}} + 7$

Simplify:

$g(f(x)) = x - 7 + 7 = x$

So, $g(f(x)) = x$.

Step 3: Conclusion

Since both $f(g(x)) = x$ and $g(f(x)) = x$, the functions $f$ and $g$ are inverses of each other.

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Would you like further details or clarifications on this solution?

Here are 5 related questions for practice:

1. Find the inverse of $f(x) = \frac{10}{x + 2}$.
2. Verify if $f(x) = 2x + 1$ and $g(x) = \frac{x - 1}{2}$ are inverses.
3. Find $f(g(x))$ for $f(x) = 3x - 4$ and $g(x) = \frac{x + 4}{3}$.
4. Prove that $f(x) = \frac{1}{x}$ is its own inverse.
5. Solve $f(g(x)) = 3$ for $f(x) = \frac{6}{x + 1}$ and $g(x) = \frac{6}{x - 1}$.

Tip: If $f(g(x)) = g(f(x)) = x$, then $f$ and $g$ are guaranteed to be inverses.