Applications of Derivatives
Inverse function derivative
Grade 12
Question:
<p>Let \(f\) be real-valued function such that \(e^{-2x}f(x) = x + 3 + \displaystyle\int_0^x \dfrac{dt}{\sqrt{t^6+1}}\) for all \(x \in (-1,1)\) and let \(y = g(x)\) be a function whose graph is reflection of the graph of \(y = f(x)\) w.r.t. line \(y = x\), then \(g'(3)\) is not equal to:</p>
<p>1</p>
<p>\(\dfrac{1}{2}\)</p>
<p>\(\dfrac{1}{4}\)</p>
<p>\(\dfrac{1}{8}\)</p>
Step-by-Step Solution
Key Concept: To find g'(3) where g is the inverse of f, use the inverse function derivative formula: g'(y) = 1/f'(x) where y = f(x). First, find f'(x) by differentiating the given functional equation, then determine which x-value gives f(x) = 3.
<p><strong>Step 1:</strong> Differentiate the given equation e^(-2x)f(x) = x + 3 + ∫₀ˣ dt/√(t⁶+1) with respect to x.</p><p><strong>Step 2:</strong> Using product rule on LHS: -2e^(-2x)f(x) + e^(-2x)f'(x) = 1 + 1/√(x⁶+1)</p><p><strong>Step 3:</strong> Simplify: e^(-2x)[f'(x) - 2f(x)] = 1 + 1/√(x⁶+1)</p><p><strong>Step 4:</strong> Find x such that f(x) = 3. From original equation: e^(-2x)·3 = x + 3 + ∫₀ˣ dt/√(t⁶+1). At x = 0: e⁰·f(0) = 0 + 3 + 0, so f(0) = 3.</p><p><strong>Step 5:</strong> Evaluate f'(0) using the differentiated equation: e⁰[f'(0) - 2·3] = 1 + 1/√(0+1), giving f'(0) - 6 = 1 + 1 = 2, so f'(0) = 8.</p><p><strong>Step 6:</strong> Since g is the inverse of f and f(0) = 3, we have: g'(3) = 1/f'(0) = 1/8</p><p>∴ Answer: The option NOT equal to 1/8 is correct (typically this would be given as choices)</p>
Correct Answer: A