91Ó°ÊÓ

To understand how functions work together, we use the concept of function composition. This involves inserting the output of one function into the input of another. Imagine it as a chain, where each link represents a function. When composing functions, the notation is usually in the form \((f \circ g)(x) = f(g(x))\). This means you first apply the function \(g\) to \(x\), then apply the function \(f\) to the result.
In the given exercise, we have two functions \(f(x) = \frac{1}{x-1}\) and \(g(x) = \frac{1}{x} + 1\). To determine if they are inverses, we compose them: \(f(g(x))\) and \(g(f(x))\). If the outcome is just \(x\) in both cases, we move closer to confirming they are inverses. Think of function composition as a way to check if two functions "undo" each other, which is a key part of finding inverses.

The inverse function property states that for two functions to be inverses, both \(f(g(x)) = x\) and \(g(f(x)) = x\) must hold true. It's like if you perform an action with one function, the inverse function can "undo" that action and return you to the start.
In our exercise, we tested this by checking both function compositions. First, for \(f(g(x)) = x\), we inserted \(g(x) = \frac{1}{x} + 1\) into \(f(x)\) showing that \(f(g(x)) = x\). This means applying \(g\) and then \(f\) leads us directly back to our starting \(x\).

• This confirms one part of the inverse function property.

Then we did the reverse: substitute \(f(x)\) into \(g(x)\) to compute \(g(f(x)) = x\). This confirmed that \(g\) and \(f\) also "undo" each other when starting with \(f\). Both proofs affirm the inverse relationship between \(f\) and \(g\), precisely thanks to this crucial property.

Finding an inverse function is like peering through a mirror and retrieving the original image. It involves determining another function that exactly reverses every operation of the original.
To calculate an inverse, the goal is to switch the dependent and independent variables and solve. Start with replacing \(f(x)\) with \(y\), then solve for \(x\) to find \(g(y)\), the inverse function.

• For instance, with \(f(x) = \frac{1}{x-1}\), we'd write \(y = \frac{1}{x-1}\)
• Then, swap \(x\) and \(y\) and solve for \(y\) to discover \(x = \frac{1}{y-1}\), making adjustments if necessary.

The rigor lies in checking your work by ensuring \(f(g(x)) = x\) and \(g(f(x)) = x\), reinforcing that you indeed identified true inverse functions. This methodical approach guarantees you aren't merely guessing but actually proving functions neatly unwind each other.