<p><i>f</i>(<i>x</i>) = max|2 sin <i>y</i> − <i>x</i>| where <i>y</i> ∈ ℝ. Determine the minimum value of <i>f</i>(<i>x</i>).</p>
Step-by-Step Solution
Key Concept: We need to find the value of x that minimizes the maximum distance from x to any value in the range of 2sin(y). The range of 2sin(y) is [-2, 2], so we're finding the point that minimizes the maximum deviation from the interval [-2, 2].
<p><strong>Step 1:</strong> Recognize that y ∈ ℝ means 2sin(y) ranges over all values in [-2, 2]. So we need f(x) = max{|2sin(y) - x| : y ∈ ℝ} = max distance from x to the interval [-2, 2].</p><p><strong>Step 2:</strong> For any fixed x, the maximum value of |2sin(y) - x| occurs at one of the endpoints of [-2, 2]. Thus:</p><p>f(x) = max{|−2 − x|, |2 − x|} = max{|x + 2|, |x − 2|}</p><p><strong>Step 3:</strong> Analyze by cases:</p><p>• If x < −2: f(x) = −(x + 2) and |x − 2| = 2 − x, so f(x) = max{−x − 2, 2 − x} = 2 − x (since −x − 2 < 2 − x when x < −2)</p><p>• If −2 ≤ x ≤ 2: f(x) = max{x + 2, 2 − x}. These are equal when x + 2 = 2 − x, giving x = 0. For x ∈ [−2, 0], f(x) = x + 2 (increasing). For x ∈ [0, 2], f(x) = 2 − x (decreasing).</p><p>• If x > 2: f(x) = x + 2 (both terms become x + 2 and x − 2 with x + 2 > x − 2)</p><p><strong>Step 4:</strong> The minimum occurs at x = 0, where f(0) = max{|0 + 2|, |0 − 2|} = max{2, 2} = 2.</p><p><strong>Step 5:</strong> Verify: At x = 0, for any y, we have |2sin(y) − 0| = |2sin(y)| ≤ 2, and the maximum value 2 is achieved when sin(y) = ±1.</p><p><strong>∴ Answer: The minimum value of f(x) is 2.</strong></p>
Correct Answer: 2