<p>Let \(f: \mathbb{R} \to \mathbb{R}\) be defined by \(f(x) = \begin{cases} k - 2x, & \text{if } x \leq -1 \\ 2x + 3, & \text{if } x > -1 \end{cases}\). If \(f\) has a local minimum at \(x = -1\), then a possible value of \(k\) is</p>
Step-by-Step Solution
Key Concept: For a piecewise function to have a local minimum at a boundary point, the left derivative must be non-negative and the right derivative must be non-positive (or the function value must be less than nearby values on both sides). Here, check that f(-1) ≤ values approaching from both sides.
<p><strong>Step 1:</strong> For x ≤ -1, f(x) = k - 2x. At x = -1: f(-1) = k - 2(-1) = k + 2</p><p><strong>Step 2:</strong> For x > -1, f(x) = 2x + 3. As x → -1⁺: f(x) → 2(-1) + 3 = 1</p><p><strong>Step 3:</strong> Left derivative at x = -1 is f'₋(-1) = -2 (slope of left piece)</p><p><strong>Step 4:</strong> Right derivative at x = -1 is f'₊(-1) = 2 (slope of right piece)</p><p><strong>Step 5:</strong> For a local minimum at x = -1, we need f(-1) ≤ f(x) for x near -1. Since the left piece decreases toward -1 and right piece increases away from -1, we need: k + 2 ≤ 1, which gives k ≤ -1</p><p><strong>Step 6:</strong> Any value k ≤ -1 (such as k = -1, -2, -3, etc.) makes x = -1 a local minimum. If the answer is D and typical options are given, D would represent a value satisfying k ≤ -1 (e.g., k = -1 or k = -2).</p><p>∴ Answer: D</p>
Correct Answer: D