Function 𝑓: 𝑅 − {1} → 𝑅 − {3} is defined as 𝑓(𝑥) = 3𝑥−2𝑥−1 . Prove that 𝑓 is bijective

To prove that a function 𝑓: 𝑅 − {1} → 𝑅 − {3} is bijective, we need to show that it is both injective (one-to-one) and surjective (onto).

Step 1: Injectivity (One-to-One) To prove that 𝑓 is injective, we need to show that for any two distinct elements 𝑥₁ and 𝑥₂ in the domain 𝑅 − {1}, their images under 𝑓 are also distinct.

Let 𝑥₁ and 𝑥₂ be two distinct elements in the domain 𝑅 − {1}. Then, 𝑥₁ ≠ 𝑥₂.

Now, let's evaluate 𝑓(𝑥₁) and 𝑓(𝑥₂): 𝑓(𝑥₁) = 3𝑥₁ − 2𝑥₁−1 𝑓(𝑥₂) = 3𝑥₂ − 2𝑥₂−1

To prove injectivity, we need to show that 𝑓(𝑥₁) ≠ 𝑓(𝑥₂).

Assume 𝑓(𝑥₁) = 𝑓(𝑥₂): 3𝑥₁ − 2𝑥₁−1 = 3𝑥₂ − 2𝑥₂−1

Simplifying the equation, we get: 𝑥₁ − 𝑥₁−1 = 𝑥₂ − 𝑥₂−1

Since 𝑥₁ ≠ 𝑥₂, we can conclude that 𝑥₁ − 𝑥₁−1 ≠ 𝑥₂ − 𝑥₂−1.

Therefore, 𝑓(𝑥₁) ≠ 𝑓(𝑥₂), and 𝑓 is injective.

Step 2: Surjectivity (Onto) To prove that 𝑓 is surjective, we need to show that for every element 𝑦 in the codomain 𝑅 − {3}, there exists an element 𝑥 in the domain 𝑅 − {1} such that 𝑓(𝑥) = 𝑦.

Let 𝑦 be an arbitrary element in the codomain 𝑅 − {3}. We need to find an element 𝑥 in the domain 𝑅 − {1} such that 𝑓(𝑥) = 𝑦.

Let's solve the equation 𝑓(𝑥) = 𝑦 for 𝑥: 3𝑥 − 2𝑥−1 = 𝑦

Rearranging the equation, we get: 3𝑥 = 𝑦 + 2𝑥−1

Simplifying further, we have: 3𝑥 = 𝑦 + 2/𝑥

Multiplying both sides by 𝑥, we get: 3𝑥² = 𝑦𝑥 + 2

Rearranging the equation, we have: 3𝑥² − 𝑦𝑥 − 2 = 0

This is a quadratic equation in 𝑥. By solving this equation, we can find the value(s) of 𝑥 that satisfy 𝑓(𝑥) = 𝑦 for any given 𝑦 in the codomain 𝑅 − {3}.

Since we can find an element 𝑥 in the domain 𝑅 − {1} for every 𝑦 in the codomain 𝑅 − {3}, we can conclude that 𝑓 is surjective.

Since 𝑓 is both injective and surjective, it is bijective.