Limits, Continuity & Differentiability
Discontinuity of piecewise functions
Grade 12
<p>Given <br/> \[ f(x) = \begin{cases} |x| + [x], & -1 \leq x < 1 \\ x + |x|, & 1 \leq x < 2 \\ x + [x], & 2 \leq x < 3 \end{cases} \] <br/> Find the number of points where \(f(x)\) is discontinuous.</p>
Step-by-Step Solution
Key Concept: Discontinuity occurs where left and right limits differ or don't equal the function value; identify critical points where absolute value or greatest integer function changes behavior (at x = -1, 0, 1, 2) and check each carefully.
<p><strong>Step 1: Identify critical points</strong></p><p>The function changes behavior at x = 0, 1, 2 (where [x] jumps). Also check endpoint x = -1.</p><p><strong>Step 2: Analyze at x = -1 (left endpoint)</strong></p><p>Only right limit matters: f(−1⁺) = |−1| + [−1] = 1 + (−1) = 0, and f(−1) = 0. Continuous from the right.</p><p><strong>Step 3: Analyze at x = 0</strong></p><p>Left limit: lim(x→0⁻) |x| + [x] = 0 + (−1) = −1</p><p>Right limit: lim(x→0⁺) |x| + [x] = 0 + 0 = 0</p><p>Since −1 ≠ 0, <strong>f is discontinuous at x = 0</strong></p><p><strong>Step 4: Analyze at x = 1</strong></p><p>Left limit: lim(x→1⁻) |x| + [x] = 1 + 0 = 1</p><p>Right limit: lim(x→1⁺) |x| + [x] = 1 + 1 = 2</p><p>Since 1 ≠ 2, <strong>f is discontinuous at x = 1</strong></p><p><strong>Step 5: Analyze at x = 2</strong></p><p>Left limit: lim(x→2⁻) |x| + [x] = 2 + 1 = 3</p><p>Right limit: lim(x→2⁺) |x| + [x] = 2 + 2 = 4</p><p>Since 3 ≠ 4, <strong>f is discontinuous at x = 2</strong></p><p>∴ Answer: <strong>3</strong> points of discontinuity (x = 0, 1, 2)</p>
Correct Answer: 3