<p>Let \(f: \mathbb{R} \to [-3, 3]\) be a twice differentiable function such that \(f'(0) = f(1) = f(3) = 2\), then which of the following must be <strong>correct</strong>?</p>
<p>\(y = f(x)\) is monotonic for some set of values of \(x\)</p>
<p>There must be at least one \(c \in [-3, 0)\) such that \(f'(x) \leq 2\)</p>
<p>\(f''(x) \geq \dfrac{-1}{3}\) for some \(c \in (-3, 3)\)</p>
<p>For some values of \(c \in (-3, 3)\), \(f''(c) \geq -2\)</p>
Step-by-Step Solution
Key Concept: Apply Rolle's theorem to f'(x) between points where f'(x) has the same value, combined with the constraint that f is bounded in [-3, 3] to force existence of critical points where f''(x) = 0.
<p><strong>Step 1:</strong> Identify critical constraints. We have f(0) unknown, f'(0) = 2, f(1) = 2, f(3) = 2, and f: ℝ → [-3,3].</p><p><strong>Step 2:</strong> Apply Rolle's theorem to f on [1,3]. Since f(1) = f(3) = 2, there exists c₁ ∈ (1,3) where f'(c₁) = 0.</p><p><strong>Step 3:</strong> Now we know f'(0) = 2 and f'(c₁) = 0 for some c₁ ∈ (1,3). Apply Rolle's theorem to f' on [0, c₁]: there exists c₂ ∈ (0, c₁) where f''(c₂) = 0.</p><p><strong>Step 4:</strong> Since f is bounded in [-3, 3] and f(1) = f(3) = 2 (near upper bound), the function must have significant curvature changes. Between x=0 (where f'=2, positive slope) and x=c₁ (where f'=0), the derivative decreases, so f'' ≤ 0 somewhere in [0,c₁].</p><p><strong>Step 5:</strong> Since f'(0) = 2 > 0 and must reach 0 before x=3, and considering the bounded range [-3,3], there must be a point where f' attains value 2 again (by intermediate value and boundedness), creating another application of Rolle's theorem to f'.</p><p><strong>Step 6:</strong> These constraints force: (A) f''(x) = 0 for at least one x, (C) f''(x) changes sign (alternates positive/negative), and (D) f must have a local extremum where f'' = 0.</p><p>∴ Answer: ACD</p>
Correct Answer: ACD