Let \(f(x) = y\), then we have

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

Let \(f^{1}(x) = x; x = 2y-1\), replacing y with x,

\(f^{1}(x) = 2x - 1 \implies f^{1}(-2) = 2(-2) -1= -5\)