Sets, Relations & Functions
Invertible Functions
Grade 11
Question:
<p>Let \(f : \mathbb{N} \to Y\) be a function defined as \(f(x) = 4x + 3\), where \(Y = \{y \in \mathbb{N} : y = 4x + 3 \text{ for some } x \in \mathbb{N}\}\). Show that <i>f</i> is invertible and its inverse is</p>
<p>\(g(y) = \dfrac{3y+4}{3}\)</p>
<p>\(g(y) = 4 + \dfrac{y+3}{4}\)</p>
<p>\(g(y) = \dfrac{y+3}{4}\)</p>
<p>\(g(y) = \dfrac{y-3}{4}\)</p>
Step-by-Step Solution
Key Concept: For f to be invertible, it must be bijective (one-one and onto). Since Y is defined as the range of f itself, f is automatically onto Y. Verify f is one-one by checking injectivity, then find the inverse by solving y = 4x + 3 for x in terms of y.
<p><strong>Step 1: Verify f is One-One (Injective)</strong></p><p>Assume f(x₁) = f(x₂) where x₁, x₂ ∈ ℕ</p><p>4x₁ + 3 = 4x₂ + 3</p><p>4x₁ = 4x₂</p><p>x₁ = x₂</p><p>∴ f is one-one</p><p><strong>Step 2: Verify f is Onto (Surjective)</strong></p><p>By definition, Y = {y ∈ ℕ : y = 4x + 3 for some x ∈ ℕ}, which is precisely the range of f.</p><p>Every element in Y is of the form 4x + 3 for some x ∈ ℕ, so f: ℕ → Y is onto.</p><p>∴ f is onto Y</p><p><strong>Step 3: Find the Inverse Function</strong></p><p>Since f is bijective, f⁻¹ exists.</p><p>Let y = 4x + 3, where x ∈ ℕ and y ∈ Y</p><p>Solving for x: 4x = y - 3</p><p>x = (y - 3)/4</p><p><strong>Step 4: Express f⁻¹</strong></p><p>f⁻¹: Y → ℕ is defined by:</p><p><strong>f⁻¹(y) = (y - 3)/4</strong></p><p>Verification: f(f⁻¹(y)) = f((y-3)/4) = 4·(y-3)/4 + 3 = y ✓</p><p>∴ f is invertible with inverse f⁻¹(y) = (y - 3)/4</p>
Correct Answer: D