Limits, Continuity & Differentiability
Differentiability
Grade 12
Question:
<p>A function \( f(x) \) satisfies the relation \( f(x+y) = f(x) + f(y) + xy(x+y) \) \( \forall \, x, y \in R \). If \( f'(0) = -1 \), then</p>
<p>(a) \( f(x) \) is a polynomial function</p>
<p>(b) \( f(x) \) is an exponential function</p>
<p>(c) \( f(x) \) is twice differentiable for all \( x \in R \)</p>
<p>(d) \( f'(3) = 8 \)</p>
Step-by-Step Solution
Key Concept: Use the functional equation to find f'(x) by computing the derivative from first principles at x=0, then use the relation f(x+h) - f(x) = f(h) + xh(2x+h) to establish that f'(x) is constant plus a linear term. The constraint f'(0) = -1 determines the specific form of f(x).
<p><strong>Step 1: Find f(0)</strong></p><p>Setting x = y = 0: f(0) = f(0) + f(0) + 0 ⟹ f(0) = 0</p><p><strong>Step 2: Differentiate the functional equation</strong></p><p>Given: f(x+y) = f(x) + f(y) + xy(x+y)</p><p>Differentiate with respect to y: f'(x+y) = f'(y) + x(x+y) + xy</p><p>Set y = 0: f'(x) = f'(0) + x² = -1 + x²</p><p><strong>Step 3: Find f(x) by integration</strong></p><p>f'(x) = x² - 1</p><p>f(x) = ∫(x² - 1)dx = x³/3 - x + C</p><p>Since f(0) = 0: C = 0</p><p>Therefore: f(x) = x³/3 - x</p><p><strong>Step 4: Verify the solution</strong></p><p>f(x+y) = (x+y)³/3 - (x+y) = (x³ + 3x²y + 3xy² + y³)/3 - x - y</p><p>f(x) + f(y) + xy(x+y) = x³/3 - x + y³/3 - y + x²y + xy²</p><p>= (x³ + y³)/3 + x²y + xy² - x - y = (x³ + 3x²y + 3xy² + y³)/3 - x - y ✓</p><p><strong>Step 5: Check properties</strong></p><p>• f'(x) = x² - 1, so f'(0) = -1 ✓</p><p>• f is differentiable everywhere (polynomial)</p><p>• f is continuous everywhere</p><p>∴ Answer: ACD (likely involving continuity at 0, differentiability, and f'(0)=-1)</p>
Correct Answer: ACD