Given that \(f : x \to \frac{2x - 1}{x + 2}, x \neq -2\), find \(f^{-1}\), the inverse of f

A.

\(f^{-1} : x \to \frac{1+2x}{2-x}, x \neq 2\)

B.

\(f^{-1} : x \to \frac{1-2x}{x+2}, x \neq -2\)

C.

\(f^{-1} : x \to \frac{1-2x}{x-2}, x \neq 2\)

D.

\(f^{-1} : x \to \frac{1+2x}{x+2}, x \neq -2\)

Correct answer is A

\(f(x) = \frac{2x - 1}{x + 2}\)

\(y = \frac{2x - 1}{x + 2}\)

\(x = \frac{2y - 1}{y + 2} \implies x(y + 2) = 2y - 1\)

\(xy - 2y = -1 - 2x  \implies y = \frac{-1 - 2x}{x - 2}\)

\(f^{-1} : x \to \frac{1 + 2x}{2 - x} ; x \neq 2\)