Problem 9 - Calculus One Again

Question

Prove that \(f: x \rightarrow \frac{x-1}{x+1}\) is a bijection and find \(f^{-1} .\)

Step-by-Step Solution

Answer

The function is bijective, and the inverse is \(f^{-1}(y) = \frac{1+y}{1-y}\).

Step 1: Definition of Bijective Function

A function is bijective if it is both injective (one-to-one) and surjective (onto). We need to prove that the function \(f(x) = \frac{x-1}{x+1}\) is both injective and surjective.

Step 2: Prove Injectivity

To prove that \(f\) is injective, assume \(f(x_1) = f(x_2)\). This implies \(\frac{x_1-1}{x_1+1} = \frac{x_2-1}{x_2+1}\). Cross-multiply to get \((x_1-1)(x_2+1) = (x_2-1)(x_1+1)\). Simplify both sides to get \(x_1x_2 + x_1 - x_2 - 1 = x_2x_1 + x_2 - x_1 - 1\). Cancel terms to achieve \(x_1 = x_2\), proving injectivity.

Step 3: Prove Surjectivity

To prove \(f\) is surjective, we need to show that for every \(y\) in the codomain, there exists an \(x\) in the domain such that \(f(x) = y\). Solve \(y = \frac{x-1}{x+1}\) for \(x\). Multiply both sides by \(x+1\) to get \(y(x+1) = x-1\). Rearrange to \(yx + y = x - 1\), resulting in \(x(y-1) = -1 - y\). Thus, \(x = \frac{1+y}{1-y}\), ensuring every \(y\) has a corresponding \(x\), proving surjectivity.

Step 4: Determine the Inverse Function

The inverse function, \(f^{-1}(y)\), is derived from the equation \(x = \frac{1+y}{1-y}\). Therefore, \(f^{-1}(y) = \frac{1+y}{1-y}\). This expression is valid for all \(y\) except when \(y = 1\), since that would result in division by zero.

Key Concepts

Injective Function

An injective function, or one-to-one function, is a function where each element in the domain maps to a unique element in the codomain. This means that no two different elements in the domain are mapped to the same element in the codomain.

This can be thought of as a situation where if you were to look at elements in the codomain (output), each one would have only one partner in the domain (input).

• An example of proving a function is injective involves showing that if two inputs produce the same output, the inputs themselves must be the same.
• This was demonstrated in the solution step where we assumed that for the function \(f(x) = \frac{x-1}{x+1}\), if \(f(x_1) = f(x_2)\), then it implies \(x_1 = x_2\).

This cancellation of terms that leads back to the original inputs being equal confirms injectivity.
Understanding injective functions helps grasp the concept of bijective functions, as a bijective function must be both injective and surjective.

Surjective Function

A surjective function, also known as a onto function, ensures that every element in the function's codomain is mapped by at least one element in the domain. This means that the function covers the entire codomain.

For a function to be surjective, each possible output (in the codomain) must have a corresponding input (in the domain).
This is like making sure every seat (output) has been filled by a person (input).

• In the example solution, showing surjectivity involves proving that for any \(y\), there exists an \(x\) such that when \(f(x)\) is calculated, it equals \(y\).
• By rearranging the equation \(y = \frac{x-1}{x+1}\), it was proved that \(x = \frac{1+y}{1-y}\).

This ensures that for every \(y\), an \(x\) can be found, thus validating the function as surjective.
Knowing about surjective functions is crucial since a bijective function, which also encompasses this property, is essentially a perfect map from domain to codomain.

Inverse Function

An inverse function essentially reverses the mapping of the original function. For a function \(f\) to have an inverse, it must be bijective. When a function is both injective and surjective, it can have an orderly reverse path from codomain to domain.

The inverse function undoes the action of the function it inverts. If \(f(x) = y\), then \(f^{-1}(y) = x\).
This opposite mapping relationship allows us to find outputs in the domain for particular inputs in the codomain via the inverse function.

• In the exercise, the inverse was found by rearranging the function \(y = \frac{x-1}{x+1}\) to solve for \(x\), resulting in the expression \(f^{-1}(y) = \frac{1+y}{1-y}\).
• It's important to note constraints, such as \(y eq 1\), to avoid issues like division by zero.

Understanding inverse functions is important for various real-world applications, like solving equations and understanding the dynamics between variables.