Let's go step by step to solve these compositions.

Given:
$f(x) = \frac{1}{x} - 5$
$g(x) = \frac{1}{x} - 5$

We need to determine:
(a) $f(g(x))$
(b) $g(f(x))$

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(a) $f(g(x))$

By definition of function composition, $f(g(x))$ means substituting $g(x)$ into $f(x)$.

So:
$f(g(x)) = f\left(\frac{1}{x} - 5\right)$

Now substitute $\frac{1}{x} - 5$ in place of $x$ in $f(x) = \frac{1}{x} - 5$:
$f\left(\frac{1}{x} - 5\right) = \frac{1}{\frac{1}{x} - 5} - 5$

To simplify, let $u = \frac{1}{x} - 5$. Then: $\frac{1}{u} = \frac{1}{\frac{1}{x} - 5} = \frac{x}{1 - 5x}$

Thus: $f(g(x)) = \frac{x}{1 - 5x} - 5$

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(b) $g(f(x))$

Similarly, $g(f(x))$ means substituting $f(x)$ into $g(x)$.

So:
$g(f(x)) = g\left(\frac{1}{x} - 5\right)$

Now substitute $\frac{1}{x} - 5$ in place of $x$ in $g(x) = \frac{1}{x} - 5$:
$g\left(\frac{1}{x} - 5\right) = \frac{1}{\frac{1}{x} - 5} - 5$

Following the same steps as above, we get: $g(f(x)) = \frac{x}{1 - 5x} - 5$

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Final Results:

(a) $f(g(x)) = \frac{x}{1 - 5x} - 5$
(b) $g(f(x)) = \frac{x}{1 - 5x} - 5$

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Do you need further clarifications or a more detailed breakdown of the steps?

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Here are 5 related questions to explore further:

1. What are the domains of $f(x)$ and $g(x)$?
2. How would the compositions change if $f(x) = \frac{1}{x + 5}$?
3. What is the inverse function of $f(x) = \frac{1}{x} - 5$?
4. Are $f$ and $g$ bijective functions?
5. How do compositions behave when both functions are identical?

Tip: When simplifying complex fractions, multiply the numerator and denominator by the least common denominator to avoid mistakes.