Restrict a parabola and test an inverse composite

Sketch the downward parabola with vertex $(2,5)$, vertical intercept $(0,1)$ and included endpoint $(4,1)$; retain only inputs $x\leq4$. It extends downwards without bound to the left, so $R_f=(-\infty,5]$.

The horizontal line $y=3$ intersects the retained graph twice, at $x=2\pm\sqrt2$. Both are at most 4, so $f$ is not one-to-one and has no inverse function on its stated domain.

The largest possible restriction boundary is $k=2$. The branch on $(-\infty,2]$ is one-to-one; any larger boundary retains points on both sides of the vertex.

For $y=5-(x-2)^2$ with $x\leq2$, choose $x=2-\sqrt{5-y}$. Thus $g^{-1}:x\mapsto2-\sqrt{5-x}$, $x\in\mathbb R$, $x\leq5$.

Write $h(x)=\alpha+\dfrac{2\alpha+1}{x-2}$. Since $\alpha>0$, the numerator $2\alpha+1$ is nonzero, and $R_h=\mathbb R\setminus\{\alpha\}$. The graph has asymptotes $x=2$ and $y=\alpha$. The composite $h^{-1}g$ exists when $R_g=(-\infty,5]\subseteq D_{h^{-1}}=R_h$, which requires $\alpha>5$.

At $\alpha=6$, $h^{-1}(g(x))=-11$ implies $g(x)=h(-11)=\dfrac{-66+1}{-13}=5$. Hence $5-(x-2)^2=5$, giving $x=2$, which lies in the domain $x\leq2$.