Problem 50 - Calculus with Concepts in Calculus

Question

Let \(f(x)=1 / x\). Show that \(f(f(x))=x\) for all \(x \neq 0\).

Step-by-Step Solution

Answer

Demonstrating \(f(f(x))=x\) involves showing that the reciprocal of the reciprocal of \(x\) is \(x\).

Step 1: Understand the Function Composition

To solve the problem, we need to understand what the function composition \(f(f(x))\) means. We are given \(f(x) = \frac{1}{x}\), which means that \(f\) takes an input \(x\) and outputs the reciprocal of \(x\).

Step 2: Apply the Inner Function

Start by applying \(f\) to \(x\):\[ f(x) = \frac{1}{x} \]This gives us an expression for the inner function application.

Step 3: Apply the Outer Function

Now take the result of \(f(x) = \frac{1}{x}\) and substitute it back into another application of \(f\):\[ f(f(x)) = f\left( \frac{1}{x} \right) \]Apply the definition of \(f\) again, which is \(f(x) = \frac{1}{x}\).

Step 4: Simplify the Expression

Apply the rule of \(f\) to the input \(\frac{1}{x}\):\[ f\left( \frac{1}{x} \right) = \frac{1}{\left( \frac{1}{x} \right)} \]Simplify the expression:\[ \frac{1}{\left( \frac{1}{x} \right)} = x \]Thus, we find that \(f(f(x)) = x\).

Step 5: Conclusion

The function composition \(f(f(x)) = x\) is true for all \(x eq 0\), as we have verified by simplification. This demonstrates that applying the function \(f\) twice returns the original value \(x\).

Key Concepts

Reciprocal Function

A reciprocal function is essentially a function that takes the reciprocal or multiplicative inverse of a number. For example, if you have a function defined as \( f(x) = \frac{1}{x} \), applying this function to a number \( x \) will give you \( \frac{1}{x} \) as a result.

By understanding the reciprocal function, we can see how this simple operation of inverting a number plays a crucial role in many mathematical problems. This concept of reciprocal is not only limited to simple numbers but is widely used in calculus and algebra to solve equations, analyze graphs, and even in real-world applications like calculating rates.

One essential fact about reciprocal functions is that they exist everywhere except at \( x = 0 \), because division by zero is undefined. This is a key understanding when working with reciprocal functions and applies directly to domain restrictions.

Function Application

Function application means applying a function to an input or argument to produce an output. In mathematical terms, if you have a function \( f \) and an input \( x \), then the application is denoted as \( f(x) \).

Each application of a function results in a specific output. In our exercise, the function we are dealing with is \( f(x) = \frac{1}{x} \), so the application gives us the reciprocal of \( x \).

Understanding function application is important because you may often need to apply a function multiple times. For instance, in the problem above, we see a double application: applying \( f \) to \( x \) and then applying \( f \) to the result, which simplifies the expression to \( f(f(x)) = x \). This intricate process shows how function application and composition can work together to reveal deeper mathematical properties.

Domain Restrictions

Domain restrictions are limitations on the inputs for which a function is defined. For any function, the domain is the set of all possible input values. However, for some functions like the reciprocal function \( f(x) = \frac{1}{x} \), domain restrictions are vital because there are values that cannot be included.

In our case, \( f(x) \) is not defined at \( x = 0 \) because you cannot divide by zero. This means the domain of \( f \) is all real numbers except zero (\( x eq 0 \)).

Recognizing domain restrictions is crucial when working with functions, especially in complex expressions involving multiple functional applications. These restrictions must always be respected, as ignoring them can lead to undefined results and errors in calculations. In any mathematical analysis, always ensure that the expressions and numbers used are within the appropriate domain of the functions involved.