Problem 52 - Calculus with Concepts in Calculus

Question

Let \(f(x)=1 /(1-x)\). Show that \(f(f(f(x)))=x\) for all \(x\) different from 0 and 1 .

Step-by-Step Solution

Answer

The third composition returns the original input: \(f(f(f(x))) = x\) for all \(x \neq 0, 1\).

Step 1: Find the First Composition

To find the first composition, substitute \(f(x) = \frac{1}{1-x}\) into itself to get \(f(f(x))\). Thus, we have:\[f(f(x)) = f\left(\frac{1}{1-x}\right) = \frac{1}{1 - \frac{1}{1-x}}\].

Step 2: Simplify the First Composition

Simplify \(\frac{1}{1 - \frac{1}{1-x}}\) by finding a common denominator:The common denominator is \((1-x)\), so the expression becomes:\[f(f(x)) = \frac{1}{\frac{1-x - 1}{1-x}} = \frac{1-x}{-x} = \frac{x-1}{x} \].

Step 3: Find the Second Composition

Now find \(f(f(f(x)))\) by substituting \(f(f(x)) = \frac{x-1}{x}\) into \(f(x)\): \[f(f(f(x))) = f\left(\frac{x-1}{x}\right) = \frac{1}{1 - \frac{x-1}{x}}\].

Step 4: Simplify the Second Composition

Simplify \(\frac{1}{1 - \frac{x-1}{x}}\):Find a common denominator for the expression inside the function:\[1 - \frac{x-1}{x} = \frac{x - (x-1)}{x} = \frac{1}{x}\].Thus, the expression becomes:\[f(f(f(x))) = \frac{1}{\frac{1}{x}} = x\].

Step 5: Validate All Conditions

Given that each transformation, \(f(x)\) and subsequent compositions are performed under the condition that \(x eq 0\) and \(x eq 1\), the simplifications hold valid.Thus, \(f(f(f(x))) = x\) is confirmed for all \(x\) different from 0 and 1.

Key Concepts

Inverse Functions

Inverse functions are essential in mathematics as they allow us to reverse the effect of a function. In simpler terms, if a function \( f \) maps an input \( x \) to an output \( y \), then the inverse function, denoted as \( f^{-1} \), will map \( y \) back to \( x \). This idea forms the backbone of understanding numerous mathematical concepts because it enables us to navigate back and forth between inputs and outputs.
A function can only have an inverse if it is bijective, which means it must be both "one-to-one" (unique inputs map to unique outputs) and "onto" (all possible outputs are mapped). In our exercise:

• We see an intriguing behavior with the function \(f(x) = \frac{1}{1-x}\).
• This specific function, when applied three times in succession, leads us back to our original input \( x \).

This showcases a particular type of functionality where the composite function behaves as its own kind of inverse across multiple applications, even though \(f(x)\) itself might not have a straightforward inverse in the traditional sense.
Understanding the role inverse concepts play here helps us appreciate how layered and intricately designed certain functions in mathematics can be.

Rational Functions

Rational functions open a window into the world of polynomials and their fascinating properties. A rational function is essentially the quotient of two polynomials, expressed in the form \( \frac{P(x)}{Q(x)} \). These functions exhibit behaviors like asymptotes, where the function to approaches but never quite touches certain lines or points.
In the context of our problem, \(f(x) = \frac{1}{1-x}\) is a rational function. Notably, such functions can pose interesting challenges and surprises due to their characteristics:

• The denominator can never be zero, which dictates restrictions on the domain of the function.
• In our example, the values \(x = 0\) and \(x = 1\) are excluded to avoid undefined expressions.
• Such conditions emphasize the importance of understanding singularities and removable discontinuities in rational functions.

By analyzing the behavior of rational functions across their valid domains, we gain insight into the properties and possibilities that exist within these polynomials.

Substitution Method

The substitution method is a powerful mathematical tactic employed to simplify and solve more complex expressions and equations. By substituting variables or entire expressions into functions, we convert challenging problems into more manageable ones. This approach is immensely helpful when working through compositions of functions.
In our exercise, substitution plays a pivotal role in unraveling the layers of function composition:

• Initially, \(f(x) = \frac{1}{1-x}\) was substituted into itself to yield \(f(f(x))\).
• The method continued until we arrived at \(f(f(f(x))) = x\), demonstrating that after substituting multiple times, we came full circle, back to \(x\).

The beauty of substitution lies in its ability to take seemingly complex layers of mathematics and peel them away one step at a time. By breaking down the problem using substitution, we ensure clarity and streamlined pathways to solutions, enabling us to validate the problem's claims efficiently and effectively.