Sets, Relations & Functions
Piecewise functions and composition
Grade 11
Question:
<p>If \(f: R \to R\) is defined as \(f(x) = \begin{cases} x+4, & x < -4 \\ 3x+2, & -4 \leq x < 4 \\ x-4, & x \geq 4 \end{cases}\), then the value of \(f(f(f(f(0)))) + 1\) is equal to:</p>
<p>(a) 0</p>
<p>(b) 1</p>
<p>(c) 2</p>
<p>(d) 4</p>
Step-by-Step Solution
Key Concept: A piecewise function requires checking both branches separately to determine domain, range, and function properties. The continuity and behavior at the boundary point x=2 must be analyzed carefully.
<p><strong>Step 1:</strong> Identify the two pieces of the function:</p><ul><li>For x < 2: f(x) = x + 4</li><li>For x ≥ 2: f(x) = x² - 2</li></ul><p><strong>Step 2:</strong> Find the range of each piece on its domain:</p><ul><li>For x < 2: As x approaches 2 from left, f(x) → 6. As x → -∞, f(x) → -∞. So this piece gives (-∞, 6)</li><li>For x ≥ 2: f(2) = 4 - 2 = 2. As x → ∞, f(x) → ∞. So this piece gives [2, ∞)</li></ul><p><strong>Step 3:</strong> Combine the ranges. The function values are: (-∞, 6) ∪ [2, ∞) = (-∞, ∞) = ℝ</p><p><strong>Step 4:</strong> Note that at x = 2: left-hand limit = 6, but f(2) = 2 (using second formula). The function is discontinuous at x = 2, but still maps onto all of ℝ.</p><p>∴ Answer: C</p>
Correct Answer: C