Finding the inverse of the function $f$ defined by $f(x) = x^2 - 4x$ for $x \geq 2$ Let $f$ be the function defined by $$f(x)=x^2-4x$$ for $x\ge 2$. Find the inverse function indicating its domain and the range. 
When I try to make $x$ the subject I get $+$and $-$ two answers for the inverse function. Is it the proper answer?
The answer I got is $$ x=2 
\pm \sqrt{y+4}.$$
 Is it correct and how do I determine the domain and range of the function? Thanks
 A: If You understand $f$ to be a function on $\mathbb{R}$ it is not invertible but the restrictions $f|_{\mathbb{R}^{\leq 2}}$ and $f|_{\mathbb{R}^{\geq 2}}$ are invertible with inverses $g_1:\mathbb{R}^{\geq-4}\rightarrow \mathbb{R}^{\leq 2},x\mapsto 2-\sqrt{x+4}$ and $g_2:\mathbb{R}^{\geq-4}\rightarrow \mathbb{R}^{\geq 2},x\mapsto 2+\sqrt{x+4} $  respectively.
A: yes we get $$x_{1,2}=2\pm \sqrt{4+y}$$ and it must be $$4+y\geq 0$$
A: The function $f$ defined by $f(x) = x^2 - 4x$ for $x \geq 2$ has domain $\text{Dom}_f = [2, \infty)$.  We can find its range by completing the square.
\begin{align*}
f(x) & = x^2 - 4x\\
     & = (x^2 - 4x + 4) - 4\\
     & = (x - 2)^2 - 4
\end{align*}
The graph of $y = (x - 2)^2 - 4$ is a parabola with vertex $(2, -4)$ that opens upwards.  The restriction that $x \geq 2$ means that the graph consists only of the right half of the parabola.  Since the graph is continuous and increases without bound as $x$ approaches $\infty$, the range of $f$ is $\text{Ran}_f = [-4, \infty)$. 
Solving for the inverse yields
\begin{align*}
y & = (x - 2)^2 - 4\\
y + 4 & = (x - 2)^2\\
\sqrt{y + 4} & = |x - 2|
\end{align*}
Since $x \geq 2$, $|x - 2| = x - 2$.  Thus, 
\begin{align*}
\sqrt{y + 4} & = x - 2\\
2 + \sqrt{y + 4} & = x
\end{align*}
Therefore, the inverse function is 
$$f^{-1}(x) = 2 + \sqrt{x + 4}$$
The domain of the inverse is $\text{Dom}_{f^{-1}} = [-4, \infty)$ and the range of the inverse is $\text{Ran}_{f^{-1}} = [2, \infty)$.
Notice that the domain of the function is the range of its inverse and the range of the function is the domain of its inverse.  Consequently, the graphs of $f$ and $f^{-1}$ are symmetric with respect to the line $y = x$, as shown in the figure below.

