Problem 25 - College Algebra

Question

The functions are all one-to-one. For each function, a. Find an equation for \(f^{-1}(x),\) the inverse function. b. Verify that your equation is correct by showing that $$f\left(f^{-1}(x)\right)=x \text { and } f^{-1}(f(x))=x$$ $$ f(x)=\frac{7}{x}-3 $$

Step-by-Step Solution

Answer

The inverse of the function \( f(x) = \frac{7}{x} - 3 \) is \( f^{-1}(x) = \frac{7}{x + 3} \). This was confirmed by verifying that \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \).

Step 1: Finding the inverse of the function

The first step is to replace \( f(x) \) with \( y \), to get \( y = \frac{7}{x} - 3 \). The inverse is found by switching \( y \) and \( x \), yielding \( x = \frac{7}{y} - 3 \). Solving this for \( y \) using algebraic manipulation gives: \n1. First add 3 to both sides to obtain \( x + 3 = \frac{7}{y} \).\n2. Next, taking the reciprocal of both sides provides: \( y = \frac{7}{x+3} \) . So \( f^{-1}(x) = \frac{7}{x+3} \).

Step 2: Verifying the inverse of the function

The second step involves showing that \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \). For the first equation, start by replacing \( f^{-1}(x) \) in the function: \( f(f^{-1}(x)) = f\left(\frac{7}{x+3}\right) \). Substituting \( \frac{7}{x+3} \) for \( x \) in \( f(x) \), we get: \( f\left(\frac{7}{x+3}\right) = \frac{7}{(\frac{7}{x+3})} - 3 = x+3-3 \), which simplifies to x. For \( f^{-1}(f(x)) \), start off with x replaced by \( f(x) \) in \( f^{-1}(x) \): \( f^{-1}(f(x)) = f^{-1}\left(\frac{7}{x} - 3\right) \). Substitute \( \frac{7}{x} - 3 \) for \( x \) in \( f^{-1}(x), \) which gives: \( f^{-1}\left(\frac{7}{x} - 3\right) = \frac{7}{(\frac{7}{x} - 3)+3} = \frac{7}{(\frac{7}{x})} = x \). As expected, \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \), hence \( f^{-1}(x) = \frac{7}{x+3} \) is indeed the inverse function.

Key Concepts

Understanding One-to-One Functions

One-to-one functions are fundamental in the realm of mathematics. Their core property is that each element of the range is paired with exactly one unique element of the domain. This uniqueness is crucial when it comes to finding inverse functions. An easy way to think of a one-to-one function is to envision it as a perfect matching between two sets, where no item is left out or paired with more than one partner.
To check if a function is one-to-one, we can use what is known as the Horizontal Line Test. If every horizontal line intersects the graph of the function at most once, then the function is one-to-one. This must be true for an inverse function to exist, because if multiple inputs gave the same output, the inverse would not be able to distinguish which input to return.

• If a function is not one-to-one, an inverse that is itself a function does not exist.
• The inverse of a one-to-one function reverses the roles of the domain and range.
• Finding an inverse essentially involves swapping the input (x) with the output (y) and solving for the new output.

The exercise provided reflects such a scenario, where the initial function is given, and we are tasked with deriving its inverse, signified by the notation \( f^{-1}(x) \). By ensuring the function is one-to-one, we can confidently proceed to find its inverse.

Algebraic Manipulation to Find Inverses

Algebraic manipulation plays a pivotal role when it comes to finding the inverse of a function. It's through careful rearrangement and operations that we unravel the original function, effectively 'undoing' its effect to reveal its inverse.
Here are the steps often involved in this process:

• Replace the function notation \( f(x) \) with \( y \) to make the equations easier to work with.
• Swap the roles of \( x \) and \( y \), essentially reversing the function.
• Isolate \( y \) on one side of the equation through algebraic processes such as adding, subtracting, multiplying, dividing, and factoring.

The demonstration of this process is in our exercise, where the given function \( f(x) = \frac{7}{x}-3 \) is manipulated step by step to isolate \( y \) and thus find the inverse \( f^{-1}(x) \). These steps involve adding 3 to both sides and then taking the reciprocal to solve for \( y \). The resulting inverse function, \( f^{-1}(x) = \frac{7}{x+3} \), unlocks the original input values when applied to the outputs of the main function.

Verifying Inverses through Function Composition

Function composition is a method where the output of one function becomes the input of another. To verify the correctness of an inverse function, we compose it with the original function. If the composition returns the original input, the inverse is correct.
By composition, we mean using the output of one function as the input of another, which is denoted as \( f(g(x)) \) or \( g(f(x)) \). In the context of inverse functions, if we compose a function \( f \) with its inverse \( f^{-1} \), and both compositions \( f(f^{-1}(x)) \) and \( f^{-1}(f(x)) \) yield an identity function (which results in \( x \)), then the inverse function is correct.

Example of Verification

In the given exercise, after deriving the inverse, two compositions are checked: \( f(f^{-1}(x)) \) and \( f^{-1}(f(x)) \). The algebra shows that both compositions simplify to the identity function, meaning the input is echoed back unchanged, thereby confirming the accuracy of the inverse. Such verifications are paramount because they ensure the two functions are indeed inverses of each other, allowing them to 'undo' each other's operations, thereby testing our initial algebraic manipulations.