Determine whether function is 1 to 1: $f(x) = x^2-2x$ I am following these steps: 
1) Write $y= f(x)$ 
2) Solve this equation for $x$ in terms of $y$ (if possible). 
3) To express $f^{-1}$ as a function of $x$, interchange $x$ and $y$.
The resulting equation is $y=f^{-1}(x)$
My work: 
$f(x) = x^2 - 2x$ 
$y = x^2 -2x $
$x= y^2-2y$ 
$x= y(y-2)$ This is where I am stuck?  
 A: It'll eventually depend on the domain of your function, but assuming it is $f\colon\mathbb{R}\to\mathbb{R}$ (the function is defined on the whole set of reals), then, observe that $f(x)=x(x-2)$. This means that $f(0)=0$ and $f(2)=0$. But both $0$ and $2$ belong to the domain of the function... Can you conclude?
A: Here is an alternative method to solve this problem. Recall the definition

Definition: $f$ is 1-1 on $A$ if whenever $f(x_1)=f(x_2)$ for $x_1,x_2\in A$ then $x_1=x_2$.

For our function, $f(x) = x^2-2x$, we find that (using $x^2-y^2 = (x-y)(x+y)$)
$$f(x_1)=f(x_2) \iff (x_1-x_2)(x_1+x_2-2) = 0$$
and since $x_2 = 2-x_1$ is a solution it follows that $f$ is not 1-1 on $\mathbb{R}$. 
If we restrict ourselves to an interval $A$ not containing $x=1$ (except possibly on the boundary) then the function is 1-1 on this interval because the two $x$ values mentioned above lies symmetric around $x=1$ so such intervals can only have one of them in it.
A: The idea is so solve the quadratic equation $y^2-2y-x=0$ by the quadratic formula,
with $a=1$, $b=-2$, and $c=x$.
A: $$y = x^2 -2x = (x-1)^2 -1$$
$$ \Rightarrow (x-1)^2 = y+1$$
$$\Rightarrow x = 1\pm \sqrt{y+1}$$
Hence, it is NOT one to one. 
For example, let $y=3$. Clearly, $x$ can take two values ($3$ or $-1$).
A: remember you are trying to solve for $y$ in the quadratic equation $$y^2 - 2y - x = 0.\tag 1$$  and you are assuming that $x$ is known. if you match the equation with your regular quadratic $$ay^2 + by + c = 0,$$  then the match is $$a = 1, b = -2, c = x. $$ the quadratic formula gives you $$y = \frac{2\pm\sqrt{4 + 4x}}2 = 1\pm \sqrt{1+x}$$ you will have to choose one of the signs for your inverse.
A: The derivative $f'(x)=(2x-2)$ is both positive and negative (in different regions of course) then f is also both increasing and decreasing. By continuity, there is an interval on which the function f is not 1-to-1 
