Applications of Derivatives
Mean Value Theorem
Grade 12
Question:
<p>\(f : [1, 4] \to [7, 44]\) is a surjective, twice differentiable function such that \(f''(x) > 0\), \(\forall\, x \in (1, 4)\), then the equation \((f'(x))^2 - \dfrac{49}{9}\) has</p>
<p>at least one root in (1, 4)</p>
<p>exactly one root in (1, 4)</p>
<p>at least two roots in (1, 4)</p>
<p>at most two roots in (1, 4)</p>
Step-by-Step Solution
Key Concept: Since f is strictly convex (f''(x) > 0) and surjective from [1,4] to [7,44], f must be strictly increasing. For a strictly convex increasing function, f'(x) is strictly increasing, so (f'(x))² is also strictly increasing, making the equation (f'(x))² = 49/9 have at most one solution.
<p><strong>Step 1:</strong> Since f is surjective from [1,4] onto [7,44] and f''(x) > 0 on (1,4), f is strictly convex. A strictly convex function on a closed interval must be strictly monotone.</p><p><strong>Step 2:</strong> Since f(1) and f(4) must equal 7 and 44 (in some order), and f is strictly convex, f must be strictly increasing. Thus f(1) = 7 and f(4) = 44.</p><p><strong>Step 3:</strong> By the Mean Value Theorem: f'(c) = (f(4) - f(1))/(4-1) = (44-7)/3 = 37/3 for some c ∈ (1,4).</p><p><strong>Step 4:</strong> Since f''(x) > 0, f'(x) is strictly increasing on (1,4). We need to check if f'(x) = 7/3 has solutions: Since f'(c) = 37/3 for some c, and f' is strictly increasing, f'(x) < 37/3 for all x ∈ (1,4). As f' is strictly increasing and continuous, and 7/3 < 37/3, there exists exactly one x₀ ∈ (1,c) where f'(x₀) = 7/3.</p><p><strong>Step 5:</strong> Therefore (f'(x))² = 49/9 ⟹ f'(x) = ±7/3. Since f is increasing, f'(x) > 0, so only f'(x) = 7/3 is valid.</p><p>∴ The equation has <strong>exactly one solution</strong>.</p>
Correct Answer: A