Given the inverse of $$h(x) = 1 + x - \sqrt{1+2x},$$ i.e., $$h^{-1}(y)=y+\sqrt{2y},$$ I am trying to compute the inverse of $$\psi(t) = \frac{\nu}{c^2}h(\frac{ct}{\nu}).$$ My first step was to define $g(u) = \frac{c}{\nu}u$ and invert the composition $$(h\circ g)^{-1}(u) = (g^{-1} \circ h^{-1})(u) = \frac{\nu}{c}u + \frac{\nu}{c}\sqrt{2u}.$$ Then, using the same rule, I have tried
$$\psi^{-1}(u) = \frac{c^2}{\nu}h^{-1}(u) = \frac{c^2}{\nu} \frac{\nu}{c}u + \frac{c^2}{\nu} \frac{\nu}{c}\sqrt{2u} = c u +c \sqrt{2u}.$$ I know that the result is
$$\psi^{-1}(u) = \sqrt{2\nu v}+cu.$$ Could you please someone cast some light on what I am doing wrong?