To confirm that two functions $f(x)$ and $g(x)$ are inverses of each other, we need to show that both of the following are true:

1. $f(g(x)) = x$
2. $g(f(x)) = x$

We will go step by step to verify both conditions.

Given Functions:

• $f(x) = \frac{x - 4}{x}$
• $g(x) = -\frac{4}{x - 1}$

Part 1: Is $f(g(x)) = x$?

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

$f(g(x)) = f\left(-\frac{4}{x - 1}\right) = \frac{\left(-\frac{4}{x - 1}\right) - 4}{\left(-\frac{4}{x - 1}\right)}$

Simplify the numerator:

$\left(-\frac{4}{x - 1}\right) - 4 = \frac{-4 - 4(x - 1)}{x - 1} = \frac{-4 - 4x + 4}{x - 1} = \frac{-4x}{x - 1}$

Now, substitute this into the expression for $f(g(x))$:

$f(g(x)) = \frac{\frac{-4x}{x - 1}}{-\frac{4}{x - 1}} = x$

So, $f(g(x)) = x$. The answer for Part 1 is Yes.

Part 2: Is $g(f(x)) = x$?

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

$g(f(x)) = g\left(\frac{x - 4}{x}\right) = -\frac{4}{\frac{x - 4}{x} - 1}$

Simplify the denominator:

$\frac{x - 4}{x} - 1 = \frac{x - 4 - x}{x} = \frac{-4}{x}$

Now substitute this back into $g(f(x))$:

$g(f(x)) = -\frac{4}{\frac{-4}{x}} = x$

So, $g(f(x)) = x$. The answer for Part 2 is Yes.

Final Answer:

• Part 1: Yes
• Part 2: Yes

Both functions are indeed inverses of each other.

Would you like more details or have any questions?

Here are 5 related questions to explore further:

1. What are the general properties of inverse functions?
2. How can we algebraically find the inverse of a given function?
3. What is the geometric relationship between a function and its inverse on a graph?
4. How can we verify that a function is one-to-one before finding its inverse?
5. How does the domain and range of a function relate to the domain and range of its inverse?

Tip: Always check that the composition of a function with its inverse gives back the identity function, $x$, to verify they are truly inverses.