Question

Prove the function \(f: \mathbb{R}-\\{1\\} \rightarrow \mathbb{R}-\\{1\\}\) defined by \(f(x)=\left(\frac{x+1}{x-1}\right)^{3}\) is bijective.

Short answer

The function \(f(x)=\left(\frac{x+1}{x-1}\right)^{3}\) is a bijective function.

Step-by-step solution

Prove that the function is injective

To prove that a function is injective, it is necessary to show that for all \(x_{1}, x_{2}\) in the domain, if \(f(x_{1}) = f(x_{2})\), then \(x_{1} = x_{2}\). Start off by assuming that \(f(x_{1}) = f(x_{2})\) which implies \(\left(\frac{x_{1}+1}{x_{1}-1}\right)^{3} = \left(\frac{x_{2}+1}{x_{2}-1}\right)^{3}\). Then, eliminating the cube from both sides will give \(\frac{x_{1}+1}{x_{1}-1} = \frac{x_{2}+1}{x_{2}-1}\). Further simplification, using cross multiplication, leads to the conclusion that \(x_{1}=x_{2}\), hence proving that the function is indeed injective.

Prove that the function is surjective

To show that the function is surjective, it must be shown that for every \(y\) in the codomain, there exists an \(x\) in the domain so that \(f(x) = y\). To find such an \(x\), express \(x\) in terms of \(y\): After isolated \(x\), it can be written as \(x = \frac{y + 1}{y - 1}\). Now, substituting \(x\) in the function \(f\), will result in \(f(x) = y\), for all \(y\) in the range of \(f\). Thus, the function is surjective.

Concluding the proof

Having proven that the function is both injective and surjective, it can be concluded from the definitions that the function \(f(x)=\left(\frac{x+1}{x-1}\right)^{3}\) is indeed a bijective function.

Key concepts

Injective Function

An injective function, also known as a one-to-one function, ensures that different elements of the domain map to distinct elements of the codomain. To confirm that a given function is injective, an important process involves setting the outputs of two arbitrary inputs equal and deducing that the inputs themselves must be equal. This means for any pair of elements, say \(x_1 \eq x_2\) from the domain, the resulting function values \(f(x_1) \eq f(x_2)\) must be unique.
For instance, considering the function \(f(x)\) from the exercise, assuming \(f(x_1) = f(x_2)\) and following algebraic simplification leads to \(x_1 = x_2\), which satisfies the condition for injectivity. Thus, this implies that no two distinct elements in the domain have the same image, signifying that the function is indeed injective.

Surjective Function

A surjective function or an onto function requires every element in the codomain to be the function value of at least one element from the domain. Demonstrating surjectivity of a function involves showing that for every element \(y\) in the codomain, an element \(x\) can be found in the domain such that \(f(x) = y\).
In the given exercise, to prove the surjectivity of \(f\), we search for an \(x\) that satisfies \(f(x) = y\) for any \(y\) in the codomain. By algebraically re-arranging the function expression, we can isolate \(x\) in terms of \(y\) and verify that for each \(y\), there exists such an \(x\) in the domain. This ensures that each element in the codomain has a pre-image. Consequently, fulfilling the definition, the function is concluded to be surjective.

Bijectivity Proof

The proof of bijectivity of a function depends upon showing that the function is both injective and surjective. Essentially, a bijective function is a one-to-one correspondence between the domain and the codomain; for each element in the domain, there is a unique element in the codomain, and vice versa.
Based on the exercise, after separately establishing the injective and surjective nature of the function, these two properties integrate, demonstrating that every element of the domain pairs with exactly one unique element of the codomain, and each element of the codomain matches with exactly one element from the domain. Hence, the function \(f(x)=\left(\frac{x+1}{x-1}\right)^{3}\) is conclusively bijective, indicating a perfect one-to-one mapping throughout the set.

Function Domain and Range

Every function has a domain and a range, which are foundational concepts for understanding functions. The domain consists of all possible input values that can be fed into the function, while the range comprises all possible outputs that the function can produce.
Considering the function in our example, the domain excludes the number 1, which is expressed as \(\mathbb{R}-\{1\}\), as including it would cause division by zero. On the other end, the range is also \(\mathbb{R}-\{1\}\) due to the constraints set by the function definition. Identifying the domain and range is a crucial step before analyzing properties such as injectivity and surjectivity since these contexts guide where to look for elements that satisfy the conditions for both properties.