Problem 33 - Precalculus with Limits

Question

In Exercises 23-34, show that \(f\) and \(g\) are inverse functions (a) algebraically and (b) graphically. \(f(x) = \frac{x-1}{x+5}\), \(g(x) = -\frac{5x+1}{x-1}\)

Step-by-Step Solution

Answer

The algebraic proofs \(f(g(x)) = x\) and \(g(f(x)) = x\) both hold true, and the graphs of \(f\) and \(g\) are reflections of each other across the line \(y = x\), confirming that \(f\) and \(g\) are indeed inverses of each other.

Step 1: Algebraic proof - \(f(g(x)) = x\)

First, compute the composition \(f(g(x))\): \n Substitute \(g(x)\) into \(f(x)\), so you get \(f(g(x)) = f(-\frac{5x+1}{x-1})\). This simplifies to \(f(-\frac{5x+1}{x-1}) = \frac{-\frac{5x+1}{x-1} - 1}{-\frac{5x+1}{x-1} + 5} = x\). This shows that for every \(x\) in the domain of \(g\), we have \(f(g(x)) = x\).

Step 2: Algebraic proof - \(g(f(x)) = x\)

Next, compute the composition \(g(f(x))\): \n Substitute \(f(x)\) into \(g(x)\), so you get \(g(f(x)) = g(\frac{x-1}{x+5})\). This simplifies to \(g(\frac{x-1}{x+5}) = -\frac{5(\frac{x-1}{x+5}) + 1}{\frac{x-1}{x+5} - 1} = x\). This shows that for every \(x\) in the domain of \(f\), we have \(g(f(x)) = x\).

Step 3: Graphical proof

Graph both functions \(f(x) = \frac{x-1}{x+5}\) and \(g(x) = -\frac{5x+1}{x-1}\). If \(f\) and \(g\) are inverses, the graph of \(f\) should be the reflection of the graph of \(g\) across the line \(y = x\). By observing the graphs of the functions, you can see that this is indeed the case.

Key Concepts

Algebraic Proof

Understanding how to prove that two functions are inverses of each other algebraically is crucial in mathematics. The essence of this proof lies in function composition, which we can think of as plugging one function into another. For two functions, f and g, to be inverses, applying the composition in either order should yield the original input, i.e., f(g(x)) should equal x and likewise, g(f(x)) should equal x.

Let's delve into our example where for function f(x), we have f(x) = \(\frac{x-1}{x+5}\), and for function g(x), we have g(x) = -\(\frac{5x+1}{x-1}\). The algebraic proof proceeds by substitution. First, we check f(g(x)) by substituting g(x) into f, then we simplify the expression. If the result is x, then we have proven that the functions are inverses, at least in one direction.

Likewise, we test g(f(x)) by inserting f(x) into g. Again, after simplification, if we get x, this confirms that g is the inverse of f. In our exercise, both tests yielded x after simplification, confirming that the functions are indeed inverses of each other algebraically. This method is a powerful tool for establishing the inverse relationship between functions, demonstrating a deep consistency within the functions’ behavior.

Graphical Proof

The graphical approach to proving that two functions are inverses involves visualization, which is immensely helpful for gaining an intuitive understanding. When graphing the functions, if one function is the reflection of the other across the line y = x, then they are inverses. The line y = x acts as a mirror because for each point (a, b) on the graph of f, the reflection point (b, a) will be on the graph of g, if they are indeed inverses.

In our exercise, if we graph both functions, f(x) = \(\frac{x-1}{x+5}\) and g(x) = -\(\frac{5x+1}{x-1}\), we can visually inspect the symmetry relative to the line y = x. The visual confirmation comes from observing that every point on f has its reflection on g. The graphical proof is an excellent companion to the algebraic proof because it provides a concrete visual representation of what the algebraic operations imply, reinforcing the concept of inverse functions.

Function Composition

The idea of function composition is central to the concept of inverse functions and is a fundamental operation in algebra. Composition involves taking two functions and combining them such that the output of one becomes the input of the other. Mathematically, if we have two functions, f and g, the composition of f and g is denoted by f(g(x)), which means 'apply g to x, then apply f to the result.'

To prove that functions are inverses, we rely on the composition of functions in both possible orders: f(g(x)) and g(f(x)). If both of these compositions return the original value x for all x in the domain of the respective functions, then f and g are inverse functions.

It's also essential to understand that function composition is not necessarily commutative. That means f(g(x)) does not always equal g(f(x)). However, for inverse functions, it must hold true for each element in their domain. This reflects a kind of 'undoing' process: one function performs an operation, and the inverse function reverses it, bringing you back to your starting point.