91Ó°ÊÓ

Inverse functions are pairs of functions that undo each other. For instance, if you have a function \( f(x) \), its inverse \( f^{-1}(x) \) will return the original input when applied in succession. In mathematical terms, this can be expressed as:

• \( (f \, \circ \, f^{-1})(x) = x \)
• \( (f^{-1} \, \circ \, f)(x) = x \)

To understand this fully, take the given exercise as an example. Here, \( f(x) = \frac{1}{x} \) and \( g(x) = \frac{1}{x} \). To show \( f \) and \( g \) are inverses, we can prove their compositions result in the identity function which maps every element to itself. Both compositions \( (f \circ g)(x) \) and \( (g \circ f)(x) \) yield \( x \), confirming that \( f \) and \( g \) are indeed inverses of each other.

An identity function is a critical concept in mathematics. It is a function that always returns the same value that was used as its input. Formally, it is represented as: \( I(x) = x \). This function plays a vital role in verifying inverse functions.

In the given exercise, we see that the composition of \( f \) and \( g \), where both \( f \) and \( g \) are defined as \( f(x) = \frac{1}{x} \), results in the identity function:

• \((f \circ g)(x) = x\)
• \((g \circ f)(x) = x\)

This means that when you apply \( f \) after \( g \), or \( g \) after \( f \), you get the original input back. Thus, the identity function confirms that the given functions are inverses.

Proof in algebra involves using well-established rules and formulas to verify mathematical statements. Here is how it works in the given exercise:

First, we check the composition \((f \circ g)(x)\). Starting with \( g(x) = \frac{1}{x} \), we substitute \( g(x) \) into \( f \), obtaining:
\ f(g(x)) = f\left(\frac{1}{x}\right) = \frac{1}{\frac{1}{x}} = x\

Next, we do the same for \((g \circ f)(x)\). Here, we substitute \( f(x) \) into \( g \), giving:
\ g(f(x)) = g\left(\frac{1}{x}\right) = \frac{1}{\frac{1}{x}} = x\

By showing both results are \( x \), we prove using algebra that \(f\) and \(g\) are indeed inverse functions and that their compositions result in the identity function. This step-by-step proof strengthens our understanding of inverse and identity functions in algebra.