<p>If <span>\(f :( -\infty , 2 ] \to ( -\infty , 4 ]\)</span>, where <span>\(f ( x ) = x ( 4 - x )\)</span>, then <span>\(f ^{-1}( x )\)</span> is given by:</p>
<p>(a) <span>\(2 - \sqrt{4 - x}\)</span></p>
<p>(b) <span>\(2 + \sqrt{4 - x}\)</span></p>
<p>(c) <span>\(-2 + \sqrt{4 - x}\)</span></p>
<p>(d) <span>\(-2 - \sqrt{4 - x}\)</span></p>
Step-by-Step Solution
Key Concept: To find the inverse function, swap x and y in the equation y = f(x), then solve for y. Since the domain of f is (-∞, 2], we must choose the branch of the inverse that maps back to this domain.
<p><strong>Step 1: Set up the inverse function equation</strong></p><p>Let y = f(x) = x(4-x) = 4x - x²</p><p>To find f⁻¹, swap x and y: x = 4y - y²</p><p><strong>Step 2: Rearrange as a quadratic in y</strong></p><p>x = 4y - y²</p><p>y² - 4y + x = 0</p><p><strong>Step 3: Apply the quadratic formula</strong></p><p>y = (4 ± √(16 - 4x))/2 = (4 ± √(4(4-x)))/2 = (4 ± 2√(4-x))/2 = 2 ± √(4-x)</p><p><strong>Step 4: Determine the correct branch</strong></p><p>We have two possible inverses: y = 2 + √(4-x) and y = 2 - √(4-x)</p><p>Since the domain of f is (-∞, 2], the range of f⁻¹ must be (-∞, 2].</p><p>For y = 2 + √(4-x): When x = 4, y = 2 + 0 = 2. When x → -∞, y → +∞. This gives range [2, ∞), which is wrong.</p><p>For y = 2 - √(4-x): When x = 4, y = 2 - 0 = 2. When x → -∞, y → -∞. This gives range (-∞, 2], which is correct.</p><p><strong>Step 5: Verify the answer</strong></p><p>If f⁻¹(x) = 2 - √(4-x), then f(f⁻¹(x)) should equal x.</p><p>f(2 - √(4-x)) = (2 - √(4-x))[4 - (2 - √(4-x))] = (2 - √(4-x))(2 + √(4-x)) = 4 - (4-x) = x ✓</p><p><strong>∴ Answer:</strong> a</p>
Correct Answer: a