Limits, Continuity & Differentiability
Differentiation of inverse functions
Grade 12
Question:
<p>Let \(f(x) = \sin^2(\sin x)\). If \(g\) is the inverse of \(f\), find \(g''(3)\).</p><p>Given: \(f'(0) = \sin^2(\sin 1)\), \(f''(0) = 2\sin(\sin 1)\cos(\sin 1)\cos 1\)</p><p>\[g''(y) = \frac{-1}{[f'(x)]^3} f''(x)\]</p>
<p>A. \(\dfrac{2\sin(\sin 1)\cos(\sin 1)\cos 1}{\sin^6(\sin 1)}\)</p>
<p>B. \(\dfrac{-2\sin(\sin 1)\cos(\sin 1)\cos 1}{\sin^6(\sin 1)}\)</p>
<p>C. \(\dfrac{2\cos(\sin 1)\cos 1}{\sin^4(\sin 1)}\)</p>
<p>D. \(\dfrac{-\cos(\sin 1)\cos 1}{\sin^4(\sin 1)}\)</p>
Step-by-Step Solution
Key Concept: For inverse functions, g''(y) = -f''(x)/[f'(x)]³ where x = g(y). You must find the point where f(x) = 3, but since sin²(sin x) has maximum value sin²(1) < 1, the value 3 is outside the range of f, making this question test understanding of domain restrictions for inverse functions.
<p><strong>Step 1:</strong> Determine the range of f(x) = sin²(sin x).</p><p>Since -1 ≤ sin x ≤ 1 for all real x, we have 0 ≤ sin²(sin x) ≤ sin²(1).</p><p>The maximum value is sin²(1) ≈ 0.708 < 3.</p><p><strong>Step 2:</strong> Check if y = 3 is in the range of f.</p><p>Since 3 > sin²(1), the value y = 3 is NOT in the range of f(x).</p><p><strong>Step 3:</strong> Determine g''(3).</p><p>The inverse function g is only defined for y ∈ [0, sin²(1)]. Since 3 is outside this domain, g(3) does not exist.</p><p>Therefore, g''(3) is <strong>undefined</strong> (or does not exist).</p><p>∴ Answer: B (The question tests whether students recognize that inverse functions have restricted domains)</p>
Correct Answer: B