Find the inverse function of $y=x^2-2x$ where $x\gt 1$

Question

I already swapped the $x$ and $y$s to get $x=y^2-2y$

I then simplified this to get $x=y(y-2)$

But after that i don't know how to solve the equation for $y$

Answer 1

You have a quadratic equation there. It has two solutions: $x=1+\sqrt{y+1}$ and $x=1-\sqrt{y+1}$. But, since $x>1$, the second solution is not relevant here. So, the inverse is $x\mapsto1+\sqrt{x+1}$.

Answer 2

You could try this approach:
$$y+1=(x^2-2x+1)$$ $$\Rightarrow y+1=(x-1)^2$$ Swapping $x$ and $y$ we have: $$x+1=(y-1)^2$$ $$\Rightarrow1+ \sqrt{x+1}=y$$

Answer 3

$$\begin{array}{rcl}y& = &x^2-2x\\ y+1&=&x^2 -2x+1\\ y+1&=&(x-1)^2\end{array}$$ Since $x>1$ we can take the root. $$\pm\sqrt{y+1}=x-1$$ So $x=1\pm\sqrt{y+1}$. Again since $x>1$ only the plus sign is correct.

Answer 4

Complete the square on the right hand side: $$y=x^2-2x=x^2-2x+1-1=(x-1)^2-1$$ Swap $x$ and $y$: $$x=(y-1)^2-1$$ Solve for $y$: $$y=1\pm\sqrt{x+1}$$ To decide which sign you have to choose, you have to look what happens with the initial domain: $x>1$ implies $y>-1$. When you swap $x$ and $y$, you get $x>-1$ (which is obvious) and $y>1$. The result is, that you need the $+$ sign, in other words $$y=1+\sqrt{x+1}$$ is the inverse function.