Showing that a function is one-to-one How can I show if the following function is one-to-one? $$y = x^2 - x \ (x \ge \frac{1}{2})$$
A function is one-to-one if $f(x_1) = f(x_2)$ then $x_1 = x_2$. So I put in $x_1$ and $x_2$ in the function and tried to see if $x_1(x_1 - 1) = x_2(x_2 - 1)$ made sense. However, I am stuck right here: how do I know that $x_1 = x_2$ in this situation? I think I might be missing something quite simple...please help!
 A: From $x_1^2-x_1=x_2^2-x_2$ we get $x_1^2-x_2^2=x_1-x_2$, hence $(x_1-x_2)(x_1+x_2)=x_1-x_2$.
Now assume that $x_1 \ne x_2$. Then it follows that $x_1+x_2=1$. Since $x_1,x_2 \ge 1/2$, we derive $x_1=x_2 =1/2$. Conclusion ?
A: Hint:
Note that the derivative is $2x-1$ hence is positive for $x>0.5$.
This implies that the function is strictly increasing there.
A: We can complete the square to write
\begin{align*}
y & = x^2 - x\\
  & = x^2 - x + \frac{1}{4} - \frac{1}{4}\\
  & = \left(x - \frac{1}{2}\right)^2 - \frac{1}{4}
\end{align*}
Can you take it from here?  
A: If
$$
x_1^2-x_1=x_2^2-x_2,
$$
then you can write
$$
x_1^2-x_1-(x_2^2-x_2)=0
$$
What happens if you solve this second degree equation with respect to $x_1$? Since $x_1,x_2\geq \frac{1}{2}$, we may discard the root with negative discriminant and note that the discriminant will be positive. You will have
$$
x_1=\dfrac{1+ \sqrt{1-4x_2+4x_2^2}}{2} = \dfrac{1+\sqrt{(-1+2x_2)^2}}{2} = \dfrac{1-1+2x_2}{2}=x_2
$$
Therefore the function is injective.
A: One strategy is to note that y can be rewritten:
$y =$
$x(x+1)=$
$((x-\frac{1}{2})+\frac{1}{2}))(x-\frac{1}{2})-\frac{1}{2})=$
$(x-\frac{1}{2})^2-\frac{1}{2}$
Since $x>\frac{1}{2}$, $(x-\frac{1}{2})$ is positive, so you're squaring a positive number, which is one-to-one, and subtracting a constant is also one-to-one.
