Sets, Relations & Functions
Piecewise functions and composition
Grade 11
<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>
Step-by-Step Solution
Key Concept: A piecewise function's range depends on analyzing each piece separately over its domain, then taking the union. Here, for x < -2 we get f(x) > 2, and for x ≥ -2 we get f(x) ≥ 0, so their union gives the range.
<p><strong>Step 1:</strong> Analyze the first piece where x < -2: f(x) = x + 4</p><p>As x ranges over (-∞, -2), the expression x + 4 ranges over (-∞, 2). Since x < -2 (strictly), we have f(x) < 2 (strictly), so this piece contributes (-∞, 2) to the range.</p><p><strong>Step 2:</strong> Analyze the second piece where x ≥ -2: f(x) = x²</p><p>For x ∈ [-2, ∞), the function x² achieves its minimum at x = -2 giving f(-2) = 4, and as x increases, x² → ∞. So this piece contributes [4, ∞) to the range.</p><p><strong>Step 3:</strong> Take the union of both pieces: (-∞, 2) ∪ [4, ∞)</p><p>Note: The interval (2, 4) is NOT in the range since no value of x produces outputs in this gap.</p><p>∴ Answer: C (Range = (-∞, 2) ∪ [4, ∞))</p>
Correct Answer: C