If \(f(x) = \frac{1}{2 - x}, x \neq 2\), find \(f^{-1}(-\frac{1}{2})\).
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Correct Answer: Option A
Explanation:
\(f(x) = \frac{1}{2 - x}, x \neq 2\)
\(f(y) = \frac{1}{2 - y}\)
\(x = \frac{1}{2 - y}\) (Let x = f(y))
\(2x - xy = 1 \implies y = \frac{2x - 1}{x}\)
\(\therefore f^{-1}(x) = \frac{2x - 1}{x}\)
\(f^{-1}(-\frac{1}{2}) = \frac{2(-\frac{1}{2}) - 1}{-\frac{1}{2}}\)
= \(-2 \times -2 = 4\)
\(f(x) = \frac{1}{2 - x}, x \neq 2\)
\(f(y) = \frac{1}{2 - y}\)
\(x = \frac{1}{2 - y}\) (Let x = f(y))
\(2x - xy = 1 \implies y = \frac{2x - 1}{x}\)
\(\therefore f^{-1}(x) = \frac{2x - 1}{x}\)
\(f^{-1}(-\frac{1}{2}) = \frac{2(-\frac{1}{2}) - 1}{-\frac{1}{2}}\)
= \(-2 \times -2 = 4\)