Applications of Derivatives
Tangents and Normals
Grade 12
Question:
<p>A function \(y = f(x)\) has a second order derivative \(f''(x) = 6(x-1)\). If its graph passes through the point (2, 1) and at that point the tangent to the graph is \(y = 3x - 5\), then the function is</p>
<p>\((x-1)^2\)</p>
<p>\((x-1)^3\)</p>
<p>\((x+1)^3\)</p>
<p>\((x+1)^2\)</p>
Step-by-Step Solution
Key Concept: Integrate the second derivative twice using the initial conditions from the point and tangent line to uniquely determine the function. The tangent line gives f'(2) = 3, and the point gives f(2) = 1.
<p><strong>Step 1:</strong> Start with f''(x) = 6(x - 1). Integrate once to find f'(x):</p><p>f'(x) = ∫6(x - 1)dx = 6 · (x²/2 - x) + C₁ = 3x² - 6x + C₁</p><p><strong>Step 2:</strong> Use the tangent line condition. At x = 2, the tangent is y = 3x - 5, so f'(2) = 3:</p><p>f'(2) = 3(4) - 6(2) + C₁ = 12 - 12 + C₁ = 3</p><p>Therefore C₁ = 3, giving f'(x) = 3x² - 6x + 3</p><p><strong>Step 3:</strong> Integrate again to find f(x):</p><p>f(x) = ∫(3x² - 6x + 3)dx = x³ - 3x² + 3x + C₂</p><p><strong>Step 4:</strong> Use the point condition f(2) = 1:</p><p>f(2) = 8 - 3(4) + 3(2) + C₂ = 8 - 12 + 6 + C₂ = 2 + C₂ = 1</p><p>Therefore C₂ = -1</p><p><strong>∴ Answer:</strong> f(x) = x³ - 3x² + 3x - 1 (which equals (x - 1)³)</p>
Correct Answer: B