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 $x$?
 A: 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).
A: 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:
$$f^{-1}(y)=\frac{7+(y-1)\sqrt{1+2y-y^2}}{2}$$
A: The equation of an inverse function graph $ f^{-1}(x) $ for given
$$y=f(x)=\sqrt{x-3}-\sqrt{4-x}+1$$
is
$  x=g(y)=\sqrt{y-3}-\sqrt{4-y}+1$
obtained by swapping/interchanging $(x,y)$ in its equations, explicit or implicit. The latter are shown for the green colored function and its inverse which is symmetrical to the line $ (x=y).$
As a simple example parabola equation $y=\sqrt x $ or any such function with square root has a positive sign only. When the negative sign is included full curve emerges. The blue parts are indicated for sake of completeness in this context as well.
The green line at  bottom right is for the given function and its inverse mirrored equation and graph in green can be seen at  top left. The blue lines may be ignored.
The implicit/contour plots exhibit their equation in Mathematica when the mouse is hovered on the four curves.

