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 $x$?

Complete the square from where you left off:
$$\begin{align}
4x^2-28x&=-y^4+4y^3-4y^2-48\\
4x^2-28x+49&=49-y^4+4y^3-4y^2-48\\
(2x-7)^2&=49-y^4+4y^3-4y^2-48\\
2x-7&=\pm\sqrt{1-y^4+4y^3-4y^2}\\
\end{align}$$
Since $x$ comes from $[3,4]$, then $2x-7$ comes from $[-1,1]$, sometimes positive, sometimes negative. So it is not as simple as choosing the $+$ or the $-$ in that expression. This is my answer to your question "But how to decide which root of $x$?"
At $x=3.5$ is the boundary between positive values of $(2x-7)$ and negative. This corresponds to $y=1$. You can conclude:
$$f^{-1}(y)=\begin{cases}
3.5&y=1\\
\frac{7-\sqrt{1-y^4+4y^3-4y^2}}{2}&y\lt1\\
\frac{7+\sqrt{1-y^4+4y^3-4y^2}}{2}&y\gt1
\end{cases}$$
It turns out the radicand factors, which you could find using the rational root theorem: $$1-y^4+4y^3-4y^2=(1-y)^2(1+2y-y^2)$$
So the the cases simplify:
$$f^{-1}(y)=\begin{cases}
3.5&y=1\\
\frac{7-\lvert1-y\rvert\sqrt{1+2y-y^2}}{2}&y\lt1\\
\frac{7+\lvert1-y\rvert\sqrt{1+2y-y^2}}{2}&y\gt1
\end{cases}$$
But then all these cases are equivalent to @Angelo's comment, because $\pm \lvert1-y\rvert$ is the same as $-(1-y)$ when conditioning around $y=1$:
$$f^{-1}(y)=\frac{7-(1-y)\sqrt{1+2y-y^2}}{2}$$
or if you prefer:
$$f^{-1}(y)=\frac{7+(y-1)\sqrt{1+2y-y^2}}{2}$$
