Limits, Continuity & Differentiability
Continuity and Differentiability
Grade 12
Question:
<p>The function \(f(x) = |x-3|, x \geq 1\)<br>\(\frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4}, x < 1\)</p>
<p>(a) is continuous at \(x = 1\)</p>
<p>(b) is continuous at \(x = 3\)</p>
<p>(c) is differentiable at \(x = 1\)</p>
<p>(d) \(f'(3)\) exists</p>
Step-by-Step Solution
Key Concept: For piecewise functions, check continuity at the boundary point by verifying left-hand limit equals right-hand limit equals function value. The absolute value function |x-3| changes behavior at x=3, creating a potential discontinuity that must be verified against the quadratic piece.
<p><strong>Step 1:</strong> Identify the boundary point where the definition changes: x = 3</p><p><strong>Step 2:</strong> Calculate the right-hand limit (using |x-3| for x ≥ 1, approaching from right of 3):<br/>lim(x→3⁺) |x-3| = |3-3| = 0</p><p><strong>Step 3:</strong> Calculate the left-hand limit (using quadratic piece x < 3):<br/>lim(x→3⁻) [x²/4 - 3x/2 + 13/4] = 9/4 - 9/2 + 13/4 = 9/4 - 18/4 + 13/4 = 4/4 = 1</p><p><strong>Step 4:</strong> Since lim(x→3⁻) f(x) = 1 ≠ 0 = lim(x→3⁺) f(x), the function is <strong>discontinuous at x = 3</strong></p><p><strong>Step 5:</strong> Check differentiability: A function cannot be differentiable at a point where it is discontinuous</p><p>∴ Answer: C (The function is neither continuous nor differentiable at x = 3)</p>
Correct Answer: C