Functions
General
Grade 12
Question:
A function <span class="math-inline">f</span> has domain <span class="math-inline">[-1, 2]</span> and range <span class="math-inline">[0, 1]</span>. The domain and range respectively of the function <span class="math-inline">g</span> defined by <span class="math-inline">g(x) = 1 - f(x + 1)</span> is
(A) <span class="math-inline">[-1, 1]</span> ; <span class="math-inline">[-1, 0]</span>
(B) <span class="math-inline">[-2, 1]</span>; <span class="math-inline">[0, 1]</span>
(C) <span class="math-inline">[0, 2]</span>; <span class="math-inline">[-1, 0]</span>
(D) <span class="math-inline">[1, 3]</span>; <span class="math-inline">[-1, 0]</span>
Step-by-Step Solution
Key Concept: To find the domain of g(x) = 1 - f(x+1), we need x+1 to be in f's domain. To find the range of g(x), we apply the transformation 1 - y to f's entire range.
<p><strong>Step 1: Find the domain of g(x) = 1 - f(x+1)</strong></p><p>For g(x) to be defined, we need (x+1) to be in the domain of f.</p><p>Domain of f: [-1, 2]</p><p>Therefore: -1 ≤ x+1 ≤ 2</p><p>Solving for x: -1-1 ≤ x ≤ 2-1</p><p>Domain of g: <strong>[-2, 1]</strong></p><p><strong>Step 2: Find the range of g(x) = 1 - f(x+1)</strong></p><p>Since f(x+1) ranges over the same values as f(x) (the transformation x → x+1 doesn't change the range of f), we have:</p><p>f(x+1) ∈ [0, 1]</p><p><strong>Step 3: Apply the transformation y = 1 - t where t ∈ [0, 1]</strong></p><p>When t = 0: y = 1 - 0 = 1</p><p>When t = 1: y = 1 - 1 = 0</p><p>Since y = 1 - t is a decreasing linear function, as t varies from 0 to 1, y varies from 1 down to 0.</p><p>Range of g: <strong>[0, 1]</strong></p><p><strong>∴ Answer: Domain = [-2, 1]; Range = [0, 1]</strong></p><p><strong>The answer is B</strong></p>
Correct Answer: B