inverse function of a squared

Question

i'm stuck on trying to find the inverse of the following function:

$f($$x) = {\sqrt{x^2-4x}}$

What i did so far is:

$y = {\sqrt{x^2-4x}}$

Swap

$x = {\sqrt{y^2-4y}}$

Remove the root by squaring both sides

$x^2 = {y^2-4y}$

$y^2 - 4y - x^2 = 0$

and here is where i'm stuck, how should i proceed?

Answer 1

To find its inverse, let's find the domain of $f(x)$ for start. Since it involves a square root, the root's argument must be non-negative, thus :

$$D_f = \{x \in \mathbb R : x^2-4x \geq 0 \} = x \in(-\infty,0]\cup[4,+\infty) $$

Now, let's solve for $x$ :

$$f(x) = \sqrt{x^2-4x} \Leftrightarrow y = \sqrt{x^2-4x} \Leftrightarrow y^2 =x^2-4x $$

$$\Leftrightarrow$$

$$x^2-4x-y^2 = 0$$

You can solve this equation with respect to $x$ now, by letting the determinant be $\Delta = 4^2 +4y^2$ which is always positive, thus it has two unique and real solutions :

$$x_{1,2} = \frac{4\pm\sqrt{4^2+4y^2}}{2}$$

Now, it is important to know the domain of your given function when you want to inverse. The domain will lead you to rejecting one of the two signs of the proposed solution. The domain of the inverse function will be the range of the initial function. Since a root is always non-negative, then the domain of the inverse must be $[0,\infty)$ and that can lead you to determining $f^{-1}(x)$ by keeping this in mind and by interchaning $x$ and $y$ in the final expression yielded.

Answer 2

Assuming domain is $\mathbb{R} \setminus(0,4)$ and co-domian is $[0,\infty) $ $y^2+4=x^2-4x+4\implies y^2+4=(x-2)^2\implies x=2\pm\sqrt{y^2+4}$