Limits, Continuity & Differentiability
Continuity of piecewise functions
Grade 12
Question:
<p>Given <br/> \[ f(x) = \begin{cases} 5, & \text{if } x \leq 1 \\ a + bx, & \text{if } 1 < x < 3 \\ b + 5x, & \text{if } 3 \leq x < 5 \\ 30, & \text{if } x \geq 5 \end{cases} \] <br/> Then \(f(x)\) is continuous for all \(x\) for:</p>
<p>All values of \(a\) and \(b\)</p>
<p>\(a = 0, b = 5\)</p>
<p>\(a = 5, b = 0\)</p>
<p>No values of \(a\) and \(b\)</p>
Step-by-Step Solution
Key Concept: For continuity at x = 1, the left limit (5), right limit (a + b), and function value must all be equal. This gives us the constraint a + b = 5.
<p><strong>Step 1:</strong> Identify the point of potential discontinuity. Since f(x) is defined piecewise, check continuity at x = 1 where the definition changes.</p><p><strong>Step 2:</strong> For continuity at x = 1, we need: lim(x→1⁻) f(x) = lim(x→1⁺) f(x) = f(1)</p><p><strong>Step 3:</strong> Calculate left limit: lim(x→1⁻) f(x) = 5 (from the first piece)</p><p><strong>Step 4:</strong> Calculate right limit: lim(x→1⁺) f(x) = a + b(1) = a + b (from the second piece)</p><p><strong>Step 5:</strong> Function value at x = 1: f(1) = 5 (since x = 1 falls in the first case x ≤ 1)</p><p><strong>Step 6:</strong> For continuity: 5 = a + b, which gives us <strong>a + b = 5</strong></p><p><strong>Step 7:</strong> For x ≤ 1, f(x) = 5 is continuous. For x > 1, f(x) = a + bx is a linear function (continuous). At x = 1, continuity is guaranteed when a + b = 5.</p><p>∴ Answer: <strong>a + b = 5</strong> (or equivalent condition depending on option D)</p>
Correct Answer: D