Limits, Continuity & Differentiability
Differentiation of inverse functions
Grade 12
Question:
<p>Let \(f: R \to R\) defined by \(f(x) = x^3 + 3x + 1\) and \(g\) be the inverse of \(f\), then the value of \(g''(5)\) equals:</p>
<p>(a) \(\dfrac{1}{6}\)</p>
<p>(b) \(\dfrac{-1}{6}\)</p>
<p>(c) \(\dfrac{1}{36}\)</p>
<p>(d) \(\dfrac{-1}{36}\)</p>
Step-by-Step Solution
Key Concept: Use implicit differentiation on the inverse function relationship f(g(x)) = x, then differentiate again to find g''(x) at a specific point. The key is finding g'(5) and g(5) first using the inverse relationship.
<p><strong>Step 1:</strong> Find g(5) by solving f(g(5)) = 5.</p><p>Let g(5) = a, then f(a) = 5.</p><p>So a³ + 3a + 1 = 5 → a³ + 3a - 4 = 0.</p><p>Testing: a = 1 gives 1 + 3 - 4 = 0. ✓ Thus g(5) = 1.</p><p><strong>Step 2:</strong> Find g'(5) using g'(x) = 1/f'(g(x)).</p><p>f'(x) = 3x² + 3, so f'(g(5)) = f'(1) = 3(1)² + 3 = 6.</p><p>Therefore, g'(5) = 1/6.</p><p><strong>Step 3:</strong> Differentiate g'(x) = 1/f'(g(x)) implicitly using the chain rule.</p><p>g''(x) = -f''(g(x))·g'(x) / [f'(g(x))]²</p><p>f''(x) = 6x, so f''(g(5)) = f''(1) = 6.</p><p>g''(5) = -(6)·(1/6) / 6² = -1/36.</p><p>∴ Answer: D</p>
Correct Answer: D