Find a formula for the inverse of the function. $f(x) = x^2 − x$, $x \le1$. I'm a bit confused by this inequality.
I'm trying to solve for the inverse function, I get $x = y^2 - y$ but I don't know what to do from there.
Help would be greatly appreciated.
 A: This is not even injective $(f(0)=f(1))$
A: Note that, if you specify the right interval $x\leq \frac{1}{2}$, then just solve the quadratic equation
$$ y=x^2-x \implies x^2-x-y=0 .$$
for $x$ and then replace $x$ by $y$.
Note:

For a function to have a composition inverse, it has to be one to one and onto.

A: This is my first post here so please forgive me if I am not using proper syntax.
You are given the equation EQ1: $y = x^2 - x$.
Try to put it into a form $y = (x + b)^2 + c$ for some b and c.  This makes solving for x easy.
$y = (x + b)^2 + c \implies y = x^2 + 2bx + b^2 + c$
So you get
$2b = 1$ and $b^2 + c = 0$
Solve for $b$ and $c$, you'll get $\,\,y = (x - 1/2)^2 - 1/4$.  Solve this equation for x.
EQ2: $(y + 1/4)^{1/2} + 1/2 = x$
Then you have to find out what range the inverse is valid for.  This is where the infinitesimal calculus kicks in.  For what range of $x$ and $y$ is it valid to say that EQ2 is the inverse of EQ1?  
Using derivatives you can find that the minimum value of y that EQ1 takes is $-\frac 14$, occurring at $x=\frac12$.  You are given $x \le 1$.  This corresponds to $\,y=0\,$ at $\,x=1\,$.  I'm not going into complete detail here, I recommend drawing a graph to see the complete detail.
All of this tells you that EQ1 is the inverse of EQ2 when $1/2 \le x \le 1$ and $-1/4 \le y \le 0$.
