Limits, Continuity & Differentiability
Continuity and differentiability of piecewise functions
Grade 12
Question:
<p><strong>512.</strong> Let \(f(x) = x^3 - x^2 + x + 1\) and \(g(x) = \begin{cases} \max.\, f(t);\, 0 \leq t \leq x, & \text{for } 0 \leq x \leq 1 \\ 3 - x, & \text{for } 1 < x \leq 2 \end{cases}\), then \(g(x)\) is:</p>
<p>continuous for \(x \in [0, 2] - \{1\}\)</p>
<p>continuous for \(x \in [0, 2]\)</p>
<p>derivable for all \(x \in [0, 2]\)</p>
<p>derivable for all \(x \in [0, 2] - \{1\}\)</p>
Step-by-Step Solution
Key Concept: For g(x) = max{f(t): 0 ≤ t ≤ x}, you must find where f'(t) = 0 on [0,x] to identify critical points; the maximum is either at endpoints or critical points. At x = 1, both pieces must give the same value (g(1) = 2) for continuity, which they do.
<p><strong>Step 1: Analyze f(t) = t³ - t² + t + 1 on [0,1]</strong></p><p>f'(t) = 3t² - 2t + 1. Discriminant = 4 - 12 = -8 < 0, so f'(t) > 0 for all t. Thus f is strictly increasing on [0,1].</p><p><strong>Step 2: Determine g(x) on [0,1]</strong></p><p>Since f is strictly increasing on [0,1], max{f(t): 0 ≤ t ≤ x} = f(x) for x ∈ [0,1]. So g(x) = x³ - x² + x + 1 on [0,1].</p><p><strong>Step 3: Verify continuity at x = 1</strong></p><p>From left: g(1⁻) = 1 - 1 + 1 + 1 = 2<br/>From right: g(1⁺) = 3 - 1 = 2<br/>g(1) = 2 ✓ Continuous at x = 1.</p><p><strong>Step 4: Check differentiability at x = 1</strong></p><p>Left derivative: g'(1⁻) = f'(1) = 3(1)² - 2(1) + 1 = 3 - 2 + 1 = 2<br/>Right derivative: g'(1⁺) = -1<br/>Since 2 ≠ -1, g is not differentiable at x = 1.</p><p>∴ Answer: B</p>
Correct Answer: B