Limits, Continuity & Differentiability
Differentiability of piecewise functions
Grade 12
Question:
<p><strong>771.</strong> Let \(f: R \to R\) be a function defined by \(f(x) = \begin{cases} x^3 + 2x^2 + x + c, & \text{if } x \leq b \\ e^x, & \text{if } x > b \end{cases}\), where \(b\) and \(c\) are integers. If \(f(x)\) is differentiable \(\forall\, x \in R\), then find the value of \((b + c)\).</p>
Step-by-Step Solution
Key Concept: For f(x) to be differentiable everywhere, it must be continuous at x = b AND have equal left and right derivatives at x = b. This requires both the function values and derivatives to match at the junction point.
<p><strong>Step 1: Continuity at x = b</strong></p><p>For f to be continuous at x = b:</p><p>Left limit: lim(x→b⁻) f(x) = b³ + 2b² + b + c</p><p>Right limit: lim(x→b⁺) f(x) = eᵇ</p><p>Therefore: b³ + 2b² + b + c = eᵇ ... (1)</p><p><strong>Step 2: Differentiability at x = b</strong></p><p>For f to be differentiable at x = b, left derivative must equal right derivative:</p><p>Left derivative: f'(b⁻) = 3b² + 4b + 1</p><p>Right derivative: f'(b⁺) = eᵇ</p><p>Therefore: 3b² + 4b + 1 = eᵇ ... (2)</p><p><strong>Step 3: Solve for b</strong></p><p>From (2): 3b² + 4b + 1 = eᵇ</p><p>Testing integer values:</p><p>• b = 0: 3(0) + 4(0) + 1 = 1, and e⁰ = 1 ✓</p><p>• b = 1: 3(1) + 4(1) + 1 = 8, and e¹ ≈ 2.718 ✗</p><p>• b = -1: 3(1) + 4(-1) + 1 = 0, and e⁻¹ ≈ 0.368 ✗</p><p>So b = 0</p><p><strong>Step 4: Solve for c</strong></p><p>Substituting b = 0 into equation (1):</p><p>0³ + 2(0)² + 0 + c = e⁰</p><p>c = 1</p><p><strong>Step 5: Find (b + c)</strong></p><p>b + c = 0 + 1 = 1</p><p>∴ Answer: <strong>1</strong></p>
Correct Answer: 1