Given that \(f : x \to \frac{2x – 1}{x + 2}, x \neq -2\), find \(f^{-1}\), the inverse of f.
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\(f^{-1} : x \to \frac{1+2x}{2-x}, x \neq 2\)
- \(f^{-1} : x \to \frac{1-2x}{x+2}, x \neq -2\)
- \(f^{-1} : x \to \frac{1-2x}{x-2}, x \neq 2\)
- \(f^{-1} : x \to \frac{1+2x}{x+2}, x \neq -2\)
The correct answer is: A
Explanation
\(f(x) = \frac{2x - 1}{x + 2}\)
\(y = \frac{2x - 1}{x + 2}\)
\(x = \frac{2y - 1}{y + 2} \implies x(y + 2) = 2y - 1\)
\(xy - 2y = -1 - 2x \implies y = \frac{-1 - 2x}{x - 2}\)
\(f^{-1} : x \to \frac{1 + 2x}{2 - x} ; x \neq 2\)