Applications of Derivatives
Concavity and tangent lines
Grade 12
Question:
<p>Since \(f''(x) > 0\), the graph of \(f(x)\) is concave upward. Now \(f'(x) = \pm \frac{7}{3}\) having a solution would mean a line having slope \(\pm \frac{7}{3}\) touching the curve. The graph will strictly lie in the rectangle formed by \(x=1,\ x=4,\ y=7,\ y=14\). Which of the following are correct?</p>
<p>(a) \(f'(x) = \frac{7}{3}\) has a solution</p>
<p>(b) \(f'(x) = -\frac{7}{3}\) has no solution</p>
<p>(c) \(f(x)\) is strictly increasing</p>
<p>(d) \(f'(x) = \frac{7}{3}\) has a solution in \((1,4)\)</p>
Step-by-Step Solution
Key Concept: For a strictly convex function (f''(x) > 0) bounded in a rectangle, the slope of any tangent line is constrained by the slopes of the diagonals connecting boundary points. Since f'(x) must lie between the minimum slope (connecting bottom-left to top-right) and maximum slope (connecting top-left to bottom-right), we check if ±7/3 falls within this feasible range.
<p><strong>Step 1: Establish the constraint region.</strong> The graph lies strictly within the rectangle: 1 < x < 4 and 7 < y < 14.</p><p><strong>Step 2: Apply convexity.</strong> Since f''(x) > 0, f'(x) is strictly increasing on [1,4]. Thus f'(x) ∈ [f'(1), f'(4)].</p><p><strong>Step 3: Determine bounds on f'(x).</strong> The extreme slopes are determined by the rectangle corners. The minimum possible slope approaches (14-7)/(4-1) = 7/3 (top-right to bottom-left diagonal). The maximum slope could be unbounded near the boundaries, but the actual range of f'(x) depends on f'(1) and f'(4).</p><p><strong>Step 4: Check feasibility of f'(x) = 7/3.</strong> By the Mean Value Theorem, there exists c ∈ (1,4) where f'(c) = [f(4)-f(1)]/(4-1). Since the graph is bounded by the rectangle, [f(4)-f(1)]/(4-1) is constrained. The slope 7/3 represents the average slope across the interval, which must be achieved by continuity of f'(x).</p><p><strong>Step 5: Check feasibility of f'(x) = -7/3.</strong> A negative slope would mean f is decreasing. For a concave-up function on [1,4] contained in the bounded rectangle, a persistent negative slope is geometrically impossible if the function must stay between y = 7 and y = 14.</p><p>∴ Answer: <strong>AD</strong> (f'(x) = 7/3 has a solution; f'(x) = -7/3 does not)</p>
Correct Answer: AD