Problem 57 - Intermediate Algebra

Question

Show that each pair of functions are inverses. $$ f(x)=\frac{2}{x-3}, f^{-1}(x)=\frac{2}{x}+3 $$

Step-by-Step Solution

Answer

The functions are inverses as both conditions \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \) are satisfied.

Step 1: Understanding Inverse Functions

Two functions are inverses of each other if the composition of one function with the other results in an identity function, meaning \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \). We will verify both conditions.

Step 2: Verify \( f(f^{-1}(x)) = x \)

Start with \( f(f^{-1}(x)) \). Substitute \( f^{-1}(x) = \frac{2}{x} + 3 \) into \( f(x) \):\[f\left(\frac{2}{x} + 3\right) = \frac{2}{\left(\frac{2}{x} + 3\right) - 3}\]Simplify the expression inside the parentheses:\[\frac{2}{x} + 3 - 3 = \frac{2}{x}\]Thus, the function becomes:\[f\left(\frac{2}{x} + 3\right) = \frac{2}{\frac{2}{x}} = x\]This verifies \( f(f^{-1}(x)) = x \).

Step 3: Verify \( f^{-1}(f(x)) = x \)

Now, substitute \( f(x) = \frac{2}{x-3} \) into \( f^{-1}(x) \):\[f^{-1}\left(\frac{2}{x-3}\right) = \frac{2}{\left(\frac{2}{x-3}\right)} + 3\]Simplify the expression:\[\frac{2}{\frac{2}{x-3}} = x - 3\]Thus, the function becomes:\[f^{-1}\left(\frac{2}{x-3}\right) = (x - 3) + 3 = x\]This verifies \( f^{-1}(f(x)) = x \).

Step 4: Conclusion

Since both conditions \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \) have been satisfied, we conclude that the functions \( f(x) = \frac{2}{x-3} \) and \( f^{-1}(x) = \frac{2}{x} + 3 \) are indeed inverses of each other.

Key Concepts

Function Composition

Function composition is a fundamental concept in mathematics. It involves applying one function to the results of another to produce a new function. This is typically denoted as \((f \circ g)(x) = f(g(x))\). In the context of inverse functions, composition is used to verify if two functions are truly inverses of each other. When you compose a function with its inverse, you should end up with the original input value. For example, if \(f(x)\) is a function and \(f^{-1}(x)\) is its inverse, then both \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\) must hold true. This composition process effectively "undoes" the function, demonstrating the reversibility inherent in inverse functions. It's like having a key and a lock; only the correct combination will "unlock" or lead you back to the starting point.

Identity Function

An identity function is a special type of function that returns its input without change. It is represented as \(I(x) = x\). The identity function is central to the concept of inverse functions. Such a function acts as a neutral element in function composition. For instance, when verifying if two functions are inverses, the successful composition both ways—\(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\)—ensures that each function returns the identity function. In simple terms, putting an output of an inverse function back into its original function, or vice-versa, should leave you where you started, preserving the identity of the input. Just as 0 is the identity element for addition, and 1 is for multiplication, the identity function acts as a neutral "middle ground" in composition.

Algebraic Functions

Algebraic functions encompass polynomial and rational functions, making them a common feature in mathematics education. These functions involve operations like addition, subtraction, multiplication, division, and the extraction of roots applied to variables. The gateway to understanding these is through solving for one variable in terms of another. In solving for inverses, one often manipulates algebraic expressions to swap the roles of the input (\(x\)) and the output (\(f(x)\)). For example, given \(f(x) = \frac{2}{x-3}\), we found its inverse by manipulating algebraic structures to express \(x\) in terms of \(y\) (where \(y = f(x)\)), finally yielding \(f^{-1}(x) = \frac{2}{x} + 3\). Arithmetic operations must be carefully managed to ensure functions and inverses maintain the correct relationships, a task in which understanding algebra is crucial.