Limits, Continuity & Differentiability
Continuity of a function
Grade 12
Question:
<p>Let \(f(x) = 1 + 4x - x^2,\ \forall x \in R\)<br>
\[g(x) = \begin{cases} \max\{f(t);\ x \le t \le (x+1),\ 0 \le x < 3\} \\ \min\{(x+3);\ 3 \le x \le 5\} \end{cases}\]
Find the interval for continuity of \(g(x)\) for all \(x \in [0, 5]\).</p>
<p>\([0,3] \cup (3,5]\)</p>
<p>\([0,3] \cup (3,4]\)</p>
<p>\([0,2] \cup [3,5]\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: For g(x) = max{f(t) : x ≤ t ≤ x+1}, the maximum of a downward-opening parabola on interval [x, x+1] occurs either at the vertex (if it lies in the interval) or at one of the endpoints. Since f(t) = 1 + 4t - t² has vertex at t = 2, the location of this vertex relative to [x, x+1] determines which endpoint gives the maximum.
<p><strong>Step 1: Analyze f(t)</strong></p><p>f(t) = 1 + 4t - t² is a downward-opening parabola with vertex at t = 2 (from f'(t) = 4 - 2t = 0).</p><p>f(2) = 1 + 8 - 4 = 5</p><p><strong>Step 2: Find where maximum of f occurs on [x, x+1]</strong></p><p>For 0 ≤ x < 1: The interval [x, x+1] doesn't contain t = 2, and since f is increasing on (-∞, 2), the maximum is at t = x+1.</p><p>g(x) = f(x+1) = 1 + 4(x+1) - (x+1)² = 1 + 4x + 4 - x² - 2x - 1 = 3 + 2x - x²</p><p><strong>Step 3: For 1 ≤ x ≤ 2</strong></p><p>The vertex t = 2 lies in [x, x+1], so maximum is f(2) = 5.</p><p>g(x) = 5</p><p><strong>Step 4: Check continuity at x = 1</strong></p><p>lim(x→1⁻) g(x) = 3 + 2(1) - 1 = 4</p><p>g(1) = 5</p><p>Left and right limits differ, so g is discontinuous at x = 1.</p><p>∴ Answer: D</p>
Correct Answer: D