<p>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: Factor the functional equation as [f(x) - x][f²(x) - (x³ + 2)f(x) + x³] = 0 to determine that f(x) = x is the only invertible solution from ℝ → ℝ. Use the derivative relationship: (f⁻¹)'(y) = 1/f'(f⁻¹(y)).
<p><strong>Step 1: Factor the functional equation</strong></p><p>Rearrange: f³(x) - (x³ + 2)f²(x) + (2x³ + 1)f(x) - x³ = 0</p><p>Factor as: [f(x) - x][f²(x) - (x³ + 1)f(x) + x³] = 0</p><p><strong>Step 2: Determine which solution is invertible ℝ → ℝ</strong></p><p>The quadratic factor f²(x) - (x³ + 1)f(x) + x³ = 0 gives non-injective solutions (like piecewise functions). For f to be invertible from ℝ → ℝ, we must have:</p><p><strong>f(x) = x</strong></p><p><strong>Step 3: Apply the inverse function derivative formula</strong></p><p>For f(x) = x, we have f⁻¹(x) = x, so f'(x) = 1 for all x.</p><p>Therefore: f'(8) = 1</p><p>Using (f⁻¹)'(y) = 1/f'(f⁻¹(y)):</p><p>(f⁻¹)'(8) = 1/f'(f⁻¹(8)) = 1/f'(8) = 1/1 = 1</p><p><strong>Step 4: Calculate the product</strong></p><p>f'(8) × (f⁻¹)'(8) = 1 × 1 = <strong>1</strong></p><p>∴ Answer: A</p>
Correct Answer: A