Sets, Relations & Functions
Inverse Functions
Grade 11
Question:
<p>If the function \( f(x) \) on the domain \( \left[\dfrac{1}{2}, \infty\right) \) is defined by \( f(x) = 2^{x(x-1)} \), then \( f^{-1}(x) \) equals:</p>
<p>\( \dfrac{1}{2}\left(1 + \sqrt{1 + 4\log_2 x}\right) \)</p>
<p>\( \dfrac{1}{2}\left(1 - \sqrt{1 + 4\log_2 x}\right) \)</p>
<p>\( \sqrt{1 + 4\log_2 x} \)</p>
<p>\( \sqrt{1 - 4\log_2 x} \)</p>
Step-by-Step Solution
Key Concept: To find the inverse function, set y = 2^(x(x-1)) and solve for x in terms of y by taking logarithms, then swap variables. The restriction to [1/2, ∞) ensures x(x-1) is monotonically increasing, guaranteeing invertibility.
<p><strong>Step 1:</strong> Let y = f(x) = 2^(x(x-1)). To find the inverse, take log₂ of both sides:</p><p>log₂(y) = x(x-1) = x² - x</p><p><strong>Step 2:</strong> Rearrange as a quadratic in x:</p><p>x² - x - log₂(y) = 0</p><p><strong>Step 3:</strong> Apply the quadratic formula:</p><p>x = [1 ± √(1 + 4log₂(y))]/2</p><p><strong>Step 4:</strong> Since the domain of f is [1/2, ∞) and x(x-1) is increasing on this interval (as d/dx[x(x-1)] = 2x - 1 > 0 for x > 1/2), we take the positive root:</p><p>x = [1 + √(1 + 4log₂(y))]/2</p><p><strong>Step 5:</strong> Swap x and y to get the inverse:</p><p>f⁻¹(x) = [1 + √(1 + 4log₂(x))]/2</p><p>∴ Answer: A</p>
Correct Answer: A