<p>If \(y = 2^{x(x-1)}\) and \(x \geq \dfrac{1}{2}\), find \(f^{-1}(x)\).</p>
<p>\(f^{-1}(x) = \dfrac{1 + \sqrt{1 + 4\log_2 x}}{2}\)</p>
<p>\(f^{-1}(x) = \dfrac{1 - \sqrt{1 + 4\log_2 x}}{2}\)</p>
<p>\(f^{-1}(x) = \dfrac{-1 + \sqrt{1 + 4\log_2 x}}{2}\)</p>
<p>\(f^{-1}(x) = \dfrac{1 + \sqrt{1 - 4\log_2 x}}{2}\)</p>
Step-by-Step Solution
Key Concept: To find the inverse, swap x and y, then solve for y by taking logarithms and recognizing that the resulting quadratic in the exponent must be solved carefully, keeping in mind the domain restriction on the original function.
<p><strong>Step 1:</strong> Start with y = 2^(x(x-1)) and swap variables to find the inverse: x = 2^(y(y-1))</p><p><strong>Step 2:</strong> Take log base 2 of both sides: log₂(x) = y(y-1) = y² - y</p><p><strong>Step 3:</strong> Rearrange as a quadratic: y² - y - log₂(x) = 0</p><p><strong>Step 4:</strong> Apply the quadratic formula: y = (1 ± √(1 + 4log₂(x)))/2</p><p><strong>Step 5:</strong> Since x ≥ 1/2 in the original function, we have y(y-1) ≥ -1/4 (minimum at y = 1/2), so the range is [1/2, ∞). For the inverse with domain x ∈ [1/2, ∞), we need y ≥ 1/2, so we take the positive sign:</p><p><strong>Step 6:</strong> f⁻¹(x) = (1 + √(1 + 4log₂(x)))/2, where x ≥ 1/2</p><p>∴ Answer: A</p>
Correct Answer: A