Applications of Derivatives
Tangent to inverse function
Grade 12
Question:
<p>Let \(f(x) = x^3 + x + 1\) and \(g(x)\) be its inverse, then the equation of tangent to \(y = g(x)\) at \(x = 3\) is:</p>
<p>(a) \(x - 4y + 1 = 0\)</p>
<p>(b) \(x + 4y - 1 = 0\)</p>
<p>(c) \(4x - y + 1 = 0\)</p>
<p>(d) \(4x + y - 1 = 0\)</p>
Step-by-Step Solution
Key Concept: Use the derivative relationship for inverse functions: g'(x) = 1/f'(g(x)). First find g(3) by solving f(g(3)) = 3, then compute the tangent line using this derivative.
<p><strong>Step 1:</strong> Find g(3). Since g is the inverse of f, we need f(g(3)) = 3.</p><p>Let y = g(3). Then f(y) = 3, so y³ + y + 1 = 3, giving y³ + y - 2 = 0.</p><p>Testing y = 1: 1 + 1 - 2 = 0 ✓. Thus g(3) = 1.</p><p><strong>Step 2:</strong> Find g'(3) using the inverse derivative formula.</p><p>f'(x) = 3x² + 1, so f'(g(3)) = f'(1) = 3(1)² + 1 = 4.</p><p>Therefore, g'(3) = 1/f'(g(3)) = 1/4.</p><p><strong>Step 3:</strong> Write the tangent line equation at point (3, 1) with slope 1/4.</p><p>y - 1 = (1/4)(x - 3)</p><p>y = (1/4)x + 1/4</p><p>∴ Answer: A</p>
Correct Answer: A