I am trying to figure out whether a function is one-to-one or not. $$y = x^2-x\,(x\ge\frac{1}{2})$$
I supposed $f(x_1) = f(x_2)$. Then $x_1^2-x_1=x_2^2-x_2$ and $x_1(x_1-1)=x_2(x_2-1)$. To prove that $x_1=x_2$ I need to get rid of the $x$'s, but I can't find a way to do that. Please help!
$x_1^2-x_1=x_2^2-x_2$
$\Rightarrow x_1^2-x_2^2=x_1-x_2$
$\Rightarrow (x_1-x_2)(x_1+x_2) = x_1-x_2$
$\Rightarrow (x_1-x_2)(x_1+x_2-1) = 0$
$\Rightarrow x_1=x_2 \quad$ or $\quad x_1 + x_2=1$
but if $x_1+x_2=1$, can $x_1$ and $x_2$ both be greater than $\frac{1}{2}$ ?
An alternative geometric method is to realise that the curve $y=x^2-x$ is a parabola that is symmetric about the line $x=\frac{1}{2}$.