Applications of Derivatives
Monotonicity
Grade 12
Question:
<p>Given that \(g(x) = 2f\!\left(\dfrac{x^2}{2}\right) + f(6 - x^2)\), \(\forall\, x \in R\) and \(f''(x) > 0\; \forall\, x \in R\), then:</p>
<p>(a) \(g(x)\) increases for \(x \in (-\infty, -2) \cup (0, 2)\)</p>
<p>(b) \(g(x)\) increases for \(x \in (-2, 0) \cup (2, \infty)\)</p>
<p>(c) \(g(x)\) decreases for \(x \in (-\infty, -2) \cup (0, 2)\)</p>
<p>(d) \(g(x)\) decreases for \(x \in (-2, 0) \cup (2, \infty)\)</p>
Step-by-Step Solution
Key Concept: Use the second derivative test on g(x) by computing g''(x) using the chain rule. Since f''(x) > 0 (f is strictly convex), the sign of g''(x) depends on the coefficients of f'' terms after differentiation.
<p><strong>Step 1:</strong> Find g'(x) using chain rule:</p><p>g'(x) = 2f'(x²/2) · x + f'(6 - x²) · (-2x) = 2x[f'(x²/2) - f'(6 - x²)]</p><p><strong>Step 2:</strong> Find g''(x) by differentiating g'(x):</p><p>g''(x) = 2[f'(x²/2) - f'(6 - x²)] + 2x[f''(x²/2) · x - f''(6 - x²) · (-2x)]</p><p>g''(x) = 2[f'(x²/2) - f'(6 - x²)] + 2x²f''(x²/2) + 4x²f''(6 - x²)</p><p><strong>Step 3:</strong> At x = 0: g''(0) = 2[f'(0) - f'(6)]</p><p><strong>Step 4:</strong> Since f''(x) > 0, f is strictly convex. For x² < 6 - x² (i.e., x² < 3), we have x²/2 < 6 - x², so f'(x²/2) < f'(6 - x²), making g''(0) < 0, indicating x = 0 is a local maximum.</p><p>At x² = 3: g''(x) = 2x²[f''(3/2) + 2f''(3/2)] > 0, indicating x = ±√3 are local minima.</p><p>∴ Answer: B</p>
Correct Answer: B