Inverse of the function $f(x)=\sqrt{x-3}-\sqrt{4-x}+1$ I am trying to find the inverse of the function $$f(x)=\sqrt{x-3}-\sqrt{4-x}+1$$
First of all its domain is $[3,4]$
As far as my knowledge is concerned, since $f'(x)>0$, it is monotone increasing in $[3,4]$, so it is injective. Also the Range is $[0,2]$. So $$f:[3,4]\to [0,2],\:\:y=f(x)=\sqrt{x-3}-\sqrt{4-x}+1$$ is Invertible.
Now to find $f^{-1}(x)$, we need to express $x$ in terms of $y$.
We have:
$$\begin{aligned}
&(y-1)^2=1-2 \sqrt{(x-3)(4-x)} \\
&\Rightarrow \quad 2\sqrt{(x-3)(4-x)}=1-(y-1)^2 \\
&\Rightarrow \quad 4(x-3)(4-x)=4 y^2+y^4-4 y^3 \\
&\Rightarrow \quad 4x^2-28 x+y^4-4 y^3+4 y^2+48=0
\end{aligned}$$
Which is a quadratic in $x$. But how to decide which root of $x$?

We want to solve
$$y=\sqrt{x-3}-\sqrt{4-x}+1$$
for $x$. Start by making the substitution $x=u^2+3$ (with $u\geq0$) so that
$$y=u-\sqrt{1-u^2}+1.$$
Rearranging and squaring we get
$$u^2-2u(y-1)+(y-1)^2=1-u^2.$$
Which can be rewritten as
$$u^2-u(y-1)+\frac{(y-1)^2-1}{2}=0.$$
Applying the appropriate quadratic formula we easily get that
$$u=\frac{y-1\pm\sqrt{2-(y-1)^2}}{2}.$$
We recall that $u=\sqrt{x-3}$, giving us that
$$\sqrt{x-3}=\frac{y-1\pm\sqrt{2-(y-1)^2}}{2}.$$
Since the left-hand side is non-negative, we must have the same for the right-hand side, and so we can replace the $\pm$ with $+$ on your given range for $y$ (check this for yourself). This it follows that
$$x=\left(\frac{y-1+\sqrt{2-(y-1)^2}}{2}\right)^2+3.$$
Thus, if
$$f(x)=\sqrt{x-3}-\sqrt{4-x}+1,$$
then
$$f^{-1}(x)=\left(\frac{x-1+\sqrt{2-(x-1)^2}}{2}\right)^2+3$$
(you can obviously try to simplify this if you want to).
