Problem 56 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(f(x)=\frac{x-3}{x+1}\). Show that \(f(f(f(x)))=x\), provided \(x \neq \pm 1\).

Step-by-Step Solution

Answer

Given substitutions and simplifications, \(f(f(f(x))) = x\) shows the function is its own inverse every three iterations.

Step 1: Calculate f(f(x))

First, we need to substitute the function into itself. Start by finding \(f(f(x))\):\[f(x) = \frac{x-3}{x+1}\]To find \(f(f(x))\), replace \(x\) in \(f(x)\) with \(f(x)\):\[f(f(x)) = f\left(\frac{x-3}{x+1}\right) = \frac{\left(\frac{x-3}{x+1}\right) - 3}{\left(\frac{x-3}{x+1}\right) + 1}\]Simplify the numerator and denominator:Numerator: \[\frac{x-3}{x+1} - 3 = \frac{x-3 - 3(x+1)}{x+1} = \frac{x-3 - 3x - 3}{x+1} = \frac{-2x - 6}{x+1}\]Denominator:\[\frac{x-3 + x + 1}{x+1} = \frac{2x - 2}{x+1}\]So,\[f(f(x)) = \frac{\frac{-2x - 6}{x+1}}{\frac{2x - 2}{x+1}} = \frac{-2x - 6}{2x - 2}\]

Step 2: Simplify f(f(x))

Now, simplify the function \(f(f(x)) = \frac{-2x - 6}{2x - 2}\):Factor the numerator and denominator:Numerator: \(-2(x + 3)\) Denominator: \(2(x-1)\)This gives:\[f(f(x)) = \frac{-2(x+3)}{2(x-1)} = \frac{-(x+3)}{x-1}\]The simplified version of \(f(f(x))\) is \(\frac{-(x+3)}{x-1}\).

Step 3: Calculate f(f(f(x)))

Now substitute \(f(f(x))\) into \(f(x)\) again to calculate \(f(f(f(x)))\):Given \(f(x) = \frac{x-3}{x+1}\) and \(f(f(x)) = \frac{-(x+3)}{x-1}\), substitute \(f(f(x))\) into \(f(x)\):\[f(f(f(x))) = f\left(\frac{-(x+3)}{x-1}\right) = \frac{\frac{-(x+3)}{x-1} - 3}{\frac{-(x+3)}{x-1} + 1}\]Simplify the numerator and denominator:Numerator: \[\frac{-(x+3) - 3(x-1)}{x-1} = \frac{-(x+3) - 3x + 3}{x-1} = \frac{-4x}{x-1}\]Denominator: \[\frac{-(x+3) + (x-1)}{x-1} = \frac{-x-3 + x - 1}{x-1} = \frac{-4}{x-1}\]Thus:\[f(f(f(x))) = \frac{\frac{-4x}{x-1}}{\frac{-4}{x-1}} = x\]

Step 4: Confirm that f(f(f(x))) = x

After simplifying, we find that \(f(f(f(x))) = x\). This verifies that substituting \(x\) through the function three times brings us back to the original \(x\) as expected, provided \(x eq \pm 1\), as the function is undefined for those values due to division by zero in the denominators at various substitution stages.

Key Concepts

Composite Function

A composite function involves applying one function to the result of another function. To fully understand, envision it as a series of functions processed in a sequence. If you have two functions, let's call them \( f(x) \) and \( g(x) \), a composite function \( (f \circ g)(x) \) means you first apply \( g(x) \), then apply \( f(x) \) to the result. In other words, calculate \( g(x) \) and then substitute that into \( f(x) \).

In our exercise, we're dealing with \( f(f(x)) \) and \( f(f(f(x))) \). Start by substituting the output of \( f(x) \) back into itself. Doing this multiple times demonstrates how repeated applications of the same function can simplify back to the original input, a fascinating property of certain functions. This highlights the concept of function iteration, which is repeatedly applying a function.

Composite functions test your understanding of functions as objects and challenge your ability to manipulate and simplify these objects. As seen in the exercise, crafting three applications of \( f \) eventually simplifies to returning the original \( x \), showing an interesting characteristic of this specific function.

Rational Function

Rational functions are quotients of two polynomials. They appear often due to their versatile nature in modeling various phenomena. The general form is \( \frac{P(x)}{Q(x)} \) where both \( P(x) \) and \( Q(x) \) are polynomials. In our exercise, the function \( f(x) = \frac{x-3}{x+1} \) is a rational function. The polynomial in the numerator is \( x-3 \) and in the denominator, it's \( x+1 \).

Understanding the behavior of rational functions involves recognizing points where the function is undefined. This can occur at values making the denominator zero. For \( f(x) \), it includes \( x = -1 \). In addition, when evaluating rational relationships, one cannot ignore these undefined points, which are crucial when considering the domain of the function.

Hence, in the exercise, ensuring that \( x eq \pm 1 \) is necessary to avoid division by zero errors that make the rational function invalid at these points. Rational functions can illustrate complex behaviors but can still be carefully managed with algebraic strategies.

Algebraic Simplification

Algebraic simplification is a critical process when dealing with complex functions, especially rational ones. It involves reducing expressions to their simplest forms. The main aim is to clarify the expressions, making them more manageable and comparable with other functions.

The exercise provides ample practice in this. Starting with \( f(f(x)) = \frac{\frac{-2x - 6}{x+1}}{\frac{2x - 2}{x+1}} \), simplification involves canceling out common terms and factoring when possible. As shown, simplifying reduces it to \( \frac{-(x+3)}{x-1} \). This requires recognizing factorable patterns and proceeding step by step to ensure mistakes are avoided.

Each step in simplification is vital as errors can compound quickly when managing complex rational expressions. Simplifying not only makes calculations clearer but also aids in understanding the underlying structure of the function. Therefore, practice ensures proficiency in this necessary skill, honing one's ability to see through the layers of complexity in algebraic expressions.