To find the inverse function \(f^{-1}(x)\), we start with the equation:

\(y = \sqrt{\frac{x+3}{2}} + 1\)

Rearrange to make \(x\) the subject:

\(y - 1 = \sqrt{\frac{x+3}{2}}\)

Square both sides:

\((y - 1)^2 = \frac{x+3}{2}\)

Multiply both sides by 2:

\(2(y - 1)^2 = x + 3\)

Expand \((y - 1)^2\):

\(2(y^2 - 2y + 1) = x + 3\)

\(2y^2 - 4y + 2 = x + 3\)

Rearrange to solve for \(x\):

\(x = 2y^2 - 4y + 2 - 3\)

\(x = 2y^2 - 4y - 1\)

Thus, \(f^{-1}(x) = 2x^2 - 4x - 1\).

For the domain of \(f^{-1}\), since \(f(x) = \sqrt{\frac{x+3}{2}} + 1\) is defined for \(x \geq -3\), the range of \(f(x)\) is \(y \geq 1\). Therefore, the domain of \(f^{-1}\) is \(x \geq 1\).