Applications of Derivatives
Differentiation and finding constants
Grade None
Question:
<p>We have \(f(x) = \alpha\log(x) + \beta x^2 + x\). If \(f'(-1) = 0\) and \(f'(2) = 0\), find \(\alpha\) and \(\beta\).</p><p>The equations are \(\alpha + 2\beta - 1 = 0\) and \(\alpha + 8\beta + 2 = 0\). Which values satisfy these?</p>
<p>\(\alpha = \dfrac{1}{2},\ \beta = -\dfrac{1}{4}\)</p>
<p>\(\alpha = 2,\ \beta = -\dfrac{1}{2}\)</p>
<p>\(\alpha = -\dfrac{1}{2},\ \beta = \dfrac{1}{4}\)</p>
<p>\(\alpha = -2,\ \beta = \dfrac{1}{2}\)</p>
Step-by-Step Solution
Key Concept: Differentiate f(x) using standard derivatives, apply the given conditions f'(-1) = 0 and f'(2) = 0 to create a system of linear equations, then solve simultaneously to find α and β.
<p><strong>Step 1: Find f'(x)</strong></p><p>f(x) = α log(x) + βx² + x</p><p>f'(x) = α/x + 2βx + 1</p><p><strong>Step 2: Apply condition f'(-1) = 0</strong></p><p>f'(-1) = α/(-1) + 2β(-1) + 1 = 0</p><p>-α - 2β + 1 = 0</p><p>α + 2β = 1 ... (equation 1)</p><p><strong>Step 3: Apply condition f'(2) = 0</strong></p><p>f'(2) = α/2 + 2β(2) + 1 = 0</p><p>α/2 + 4β + 1 = 0</p><p>Multiply by 2: α + 8β + 2 = 0 ... (equation 2)</p><p><strong>Step 4: Solve the system</strong></p><p>From equation 1: α = 1 - 2β</p><p>Substitute into equation 2: (1 - 2β) + 8β + 2 = 0</p><p>1 - 2β + 8β + 2 = 0</p><p>6β + 3 = 0</p><p>β = -1/2</p><p>Then: α = 1 - 2(-1/2) = 1 + 1 = 2</p><p><strong>Verification:</strong> Equation 1: 2 + 2(-1/2) = 2 - 1 = 1 ✓; Equation 2: 2 + 8(-1/2) + 2 = 2 - 4 + 2 = 0 ✓</p><p>∴ <strong>α = 2, β = -1/2</strong></p>
Correct Answer: A