<p>For any <em>x</em> = <em>a</em> ≥ 5, <em>f</em>(<em>a</em>) = \(\sqrt{a-5}\) ≥ 0. What is the range of the function?</p>
Step-by-Step Solution
Key Concept: The range of a function is the set of all possible output values. Since f(a) = √(a-5) where a ≥ 5, as 'a' varies from 5 to ∞, the expression (a-5) varies from 0 to ∞, making √(a-5) vary from 0 to ∞.
<p><strong>Step 1:</strong> Identify the domain constraint: x = a ≥ 5</p><p><strong>Step 2:</strong> Find minimum output: When a = 5, f(5) = √(5-5) = √0 = 0</p><p><strong>Step 3:</strong> Find maximum output: As a → ∞, we have (a-5) → ∞, so f(a) = √(a-5) → ∞</p><p><strong>Step 4:</strong> Since √(a-5) is continuous and strictly increasing for a ≥ 5, it takes all values in [0, ∞)</p><p><strong>Step 5:</strong> The range includes 0 as the minimum value achieved at a = 5</p><p>∴ Range = [0, ∞) and the minimum value in the range is 0</p>
Correct Answer: 0