To determine which functions are the same as their inverse functions, we need to examine each function and verify whether the function is self-inverse.

A function $f(x)$ is the same as its inverse if:

$f(f(x)) = x$

Let's go through each function and check this condition.

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

Find the inverse by solving $y = \frac{x - 1}{x + 5}$ for $x$:

$y(x + 5) = x - 1$ $yx + 5y = x - 1$ $yx - x = -1 - 5y$ $x(y - 1) = -1 - 5y$ $x = \frac{-1 - 5y}{y - 1}$ Thus, the inverse of $f(x)$ is $f^{-1}(x) = \frac{-1 - 5x}{x - 1}$, which is not equal to $f(x)$. Therefore, $f(x)$ is not self-inverse.

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

Find the inverse by solving $y = \frac{x - 2}{x - 1}$ for $x$:

$y(x - 1) = x - 2$ $yx - y = x - 2$ $yx - x = -y - 2$ $x(y - 1) = -y - 2$ $x = \frac{-y - 2}{y - 1}$ Thus, the inverse of $g(x)$ is $g^{-1}(x) = \frac{-x - 2}{x - 1}$, which is not equal to $g(x)$. Therefore, $g(x)$ is not self-inverse.

3. $h(x) = \frac{x + 3}{x - 2}$

Find the inverse by solving $y = \frac{x + 3}{x - 2}$ for $x$:

$y(x - 2) = x + 3$ $yx - 2y = x + 3$ $yx - x = 2y + 3$ $x(y - 1) = 2y + 3$ $x = \frac{2y + 3}{y - 1}$ Thus, the inverse of $h(x)$ is $h^{-1}(x) = \frac{2x + 3}{x - 1}$, which is not equal to $h(x)$. Therefore, $h(x)$ is not self-inverse.

4. $k(x) = \frac{x + 1}{x - 1}$

Find the inverse by solving $y = \frac{x + 1}{x - 1}$ for $x$:

$y(x - 1) = x + 1$ $yx - y = x + 1$ $yx - x = y + 1$ $x(y - 1) = y + 1$ $x = \frac{y + 1}{y - 1}$ Thus, the inverse of $k(x)$ is $k^{-1}(x) = \frac{x + 1}{x - 1}$, which is exactly $k(x)$. Therefore, $k(x)$ is self-inverse.

Conclusion:

The function $k(x) = \frac{x + 1}{x - 1}$ is the only function that is the same as its inverse.

Would you like any further details or explanation on the process?

Here are 5 related questions to explore further:

1. What properties must a function have to be self-inverse?
2. How do you determine the inverse of a rational function?
3. Can a function have more than one inverse?
4. Are all linear functions self-inverse?
5. How does symmetry relate to self-inverse functions?

Tip: If a function has symmetry over the line $y = x$, it is more likely to be self-inverse.