Applications of Derivatives
Inverse functions and derivatives
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), use the inverse function derivative formula: g'(3) = 1/f'(g(3)). First extract f(x) by differentiating the given functional equation, then find f'(x), identify g(3) by finding which x-value gives f(x)=3, and finally apply the inverse derivative formula.
<p><strong>Step 1:</strong> Differentiate the given equation with respect to x:</p><p>e^(-2x)·f'(x) + f(x)·(-2e^(-2x)) = 1 + 1/√(x⁶+1)</p><p>e^(-2x)[f'(x) - 2f(x)] = 1 + 1/√(x⁶+1)</p><p><strong>Step 2:</strong> At x=0, substitute into the original equation:</p><p>e⁰·f(0) = 0 + 3 + ∫₀⁰ dt/√(t⁶+1) = 3</p><p>So f(0) = 3</p><p><strong>Step 3:</strong> Since g is the reflection of f about y=x, g is the inverse function of f. Thus g(3) = 0 (since f(0)=3).</p><p><strong>Step 4:</strong> From the differentiated equation at x=0:</p><p>e⁰[f'(0) - 2f(0)] = 1 + 1/√(0+1)</p><p>f'(0) - 2(3) = 1 + 1 = 2</p><p>f'(0) = 8</p><p><strong>Step 5:</strong> Using the inverse function derivative formula:</p><p>g'(3) = 1/f'(g(3)) = 1/f'(0) = 1/8</p><p>∴ Answer: A</p>
Correct Answer: A