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 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: For inverse functions, g'(x) = 1/f'(g(x)). First find g(3) by solving f(g(3)) = 3, then use the inverse derivative formula to find the slope of tangent to y = g(x) at x = 3.
<p><strong>Step 1:</strong> Find g(3) by solving f(g(3)) = 3.</p><p>Since g is the inverse of f, we need: f(a) = 3 where a = g(3).</p><p>f(a) = a³ + a + 1 = 3</p><p>a³ + a - 2 = 0</p><p>(a - 1)(a² + a + 2) = 0</p><p>Since a² + a + 2 > 0 for all real a, we get a = 1</p><p>Therefore g(3) = 1</p><p><strong>Step 2:</strong> Find the slope using g'(x) = 1/f'(g(x)).</p><p>f'(x) = 3x² + 1</p><p>f'(g(3)) = f'(1) = 3(1)² + 1 = 4</p><p>g'(3) = 1/f'(g(3)) = 1/4</p><p><strong>Step 3:</strong> Write the equation of tangent at point (3, 1) with slope 1/4.</p><p>y - 1 = (1/4)(x - 3)</p><p>y = (1/4)x - 3/4 + 1</p><p>y = (1/4)x + 1/4</p><p>Or: x - 4y + 1 = 0</p><p>∴ Answer: A</p>
Correct Answer: A