If \(g(x) = \frac{x + 1}{x - 2}, x \neq -2\), find \(g^{-1}(2)\).

Explanation

\(g(x) = \frac{x + 1}{x + 2}, x \neq 2\)

Let y = x, then \(g(y) = \frac{y + 1}{y + 2}\)

Let x = g(y), so that \(x = \frac{y + 1}{y + 2}\)

\(x(y + 2) = y + 1\)

\(xy + 2x = y + 1 \implies xy - y = 1 - 2x\)

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

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

\(g^{-1}(2) = \frac{1 - 2(2)}{2 - 1} = -3\)