Show that f(x) = 3x - 4 is bijective.

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Discrete Math

**Problem Statement:**

Show that \( f(x) = 3x - 4 \) is bijective.

**Explanation:**

To demonstrate that a function is bijective, we need to prove that it is both injective (one-to-one) and surjective (onto).

1. **Injective (One-to-One):** A function is injective if for every pair of different inputs, the function produces different outputs.

   - Consider two elements \( x_1 \) and \( x_2 \) such that \( f(x_1) = f(x_2) \).
   - This implies \( 3x_1 - 4 = 3x_2 - 4 \).
   - Simplifying gives \( 3x_1 = 3x_2 \), which implies \( x_1 = x_2 \).
   - Since \( x_1 = x_2 \), the function is injective.

2. **Surjective (Onto):** A function is surjective if for every element in the codomain, there is a preimage in the domain.

   - Let \( y \) be an element in the codomain.
   - We need to find an \( x \) such that \( f(x) = y \).
   - So, \( 3x - 4 = y \).
   - Solving for \( x \), we get \( x = \frac{y + 4}{3} \).
   - For every real number \( y \), there is a real number \( x = \frac{y + 4}{3} \) such that \( f(x) = y \).
   - Thus, the function is surjective.

Since \( f(x) = 3x - 4 \) is both injective and surjective, it is bijective.
Transcribed Image Text:**Problem Statement:** Show that \( f(x) = 3x - 4 \) is bijective. **Explanation:** To demonstrate that a function is bijective, we need to prove that it is both injective (one-to-one) and surjective (onto). 1. **Injective (One-to-One):** A function is injective if for every pair of different inputs, the function produces different outputs. - Consider two elements \( x_1 \) and \( x_2 \) such that \( f(x_1) = f(x_2) \). - This implies \( 3x_1 - 4 = 3x_2 - 4 \). - Simplifying gives \( 3x_1 = 3x_2 \), which implies \( x_1 = x_2 \). - Since \( x_1 = x_2 \), the function is injective. 2. **Surjective (Onto):** A function is surjective if for every element in the codomain, there is a preimage in the domain. - Let \( y \) be an element in the codomain. - We need to find an \( x \) such that \( f(x) = y \). - So, \( 3x - 4 = y \). - Solving for \( x \), we get \( x = \frac{y + 4}{3} \). - For every real number \( y \), there is a real number \( x = \frac{y + 4}{3} \) such that \( f(x) = y \). - Thus, the function is surjective. Since \( f(x) = 3x - 4 \) is both injective and surjective, it is bijective.
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