We are given two functions:

$f(x) = \frac{1}{x} - 5$ and $g(x) = \frac{2}{x} + 5.$

We are asked to find two composite functions: $f(g(x))$ and $g(f(x))$.

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

To find $f(g(x))$, we substitute $g(x)$ into the function $f(x)$:

$f(g(x)) = f\left(\frac{2}{x} + 5\right).$

Now, substitute $\frac{2}{x} + 5$ into $f(x) = \frac{1}{x} - 5$:

$f(g(x)) = \frac{1}{\frac{2}{x} + 5} - 5.$

Simplify the first term:

$\frac{1}{\frac{2}{x} + 5} = \frac{x}{2 + 5x}.$

Thus,

$f(g(x)) = \frac{x}{2 + 5x} - 5.$

Now subtract $5$ from the expression:

$f(g(x)) = \frac{x}{2 + 5x} - \frac{5(2 + 5x)}{2 + 5x} = \frac{x - 5(2 + 5x)}{2 + 5x}.$

Simplifying the numerator:

$x - 5(2 + 5x) = x - 10 - 25x = -24x - 10.$

Thus,

$f(g(x)) = \frac{-24x - 10}{2 + 5x}.$

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

Now, substitute $f(x) = \frac{1}{x} - 5$ into $g(x) = \frac{2}{x} + 5$:

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

Substitute into $g(x)$:

$g(f(x)) = \frac{2}{\frac{1}{x} - 5} + 5.$

Simplify the first term:

$\frac{2}{\frac{1}{x} - 5} = \frac{2x}{1 - 5x}.$

Thus,

$g(f(x)) = \frac{2x}{1 - 5x} + 5.$

Now express $5$ as $\frac{5(1 - 5x)}{1 - 5x}$:

$g(f(x)) = \frac{2x}{1 - 5x} + \frac{5(1 - 5x)}{1 - 5x} = \frac{2x + 5(1 - 5x)}{1 - 5x}.$

Simplifying the numerator:

$2x + 5(1 - 5x) = 2x + 5 - 25x = -23x + 5.$

Thus,

$g(f(x)) = \frac{-23x + 5}{1 - 5x}.$

Final Results:

$f(g(x)) = \frac{-24x - 10}{2 + 5x}$ and $g(f(x)) = \frac{-23x + 5}{1 - 5x}.$

Would you like any further details or clarifications?

Here are some related questions you might consider:

1. What is the domain of $f(g(x))$?
2. What is the domain of $g(f(x))$?
3. Can you find the inverse of $f(x)$?
4. Can you find the inverse of $g(x)$?
5. What happens when you compose $f(x)$ and $g(x)$ in the opposite order?

Tip: When composing functions, always check the domains to ensure the resulting function is valid!