<p><strong>50.</strong> Let \(f\) be an invertible function from \(R \to R\) satisfying the equation \[f^3(x) - (x^3 + 2)f^2(x) + (2x^3 + 1)f(x) - x^3 = 0.\] Then the value of \(f'(8) \times (f^{-1})'(8)\) is:</p>
Step-by-Step Solution
Key Concept: Recognize that the cubic equation in f(x) can be factored, and use the relationship (f⁻¹)'(y) = 1/f'(f⁻¹(y)) combined with implicit differentiation of the functional equation to find f'(8).
<p><strong>Step 1: Factor the functional equation</strong></p><p>Rewrite: f³(x) - (x³ + 2)f²(x) + (2x³ + 1)f(x) - x³ = 0</p><p>This factors as: [f(x) - x][f²(x) - 2f(x) + x³] = 0</p><p><strong>Step 2: Determine which solution gives an invertible function</strong></p><p>From f(x) - x = 0, we get f(x) = x (the identity function)</p><p>The quadratic f²(x) - 2f(x) + x³ = 0 gives f(x) = 1 ± √(1 - x³)</p><p>For f: ℝ → ℝ to be invertible, we need f(x) = x (since the quadratic roots don't define a function on all of ℝ)</p><p><strong>Step 3: Calculate the derivatives</strong></p><p>If f(x) = x, then f'(x) = 1 for all x</p><p>Therefore f'(8) = 1</p><p><strong>Step 4: Apply the inverse function derivative formula</strong></p><p>Since f(x) = x, we have f⁻¹(x) = x</p><p>Thus (f⁻¹)'(8) = 1</p><p><strong>Step 5: Find the product</strong></p><p>f'(8) × (f⁻¹)'(8) = 1 × 1 = 1</p><p>∴ Answer: D (which equals 1)</p>
Correct Answer: D