When is $f(x)=x^2-2x$ injective? Suppose $f(x)=x^2-2x$. Find the greatest interval $I_+$ with $2\in I_+$, such that $f$ is injective. Now, intuitively, I found that $I_+ = [1, \infty)$. But how to prove this?
This was my idea:
Let $A$ be an interval starting from $a$ to $b$, so $a<b$. When is $f|_A$ injective?
There are some possibilities. Since $a,b$ are real numbers, it holds that $a < 1 \vee a \ge 1$ and $b < 1 \vee b \ge 1$. Combining this, we have to check the following possibilities for $A$:
I) $a < 1, b < 1$
II) $a < 1, b \ge 1$
III) $a \ge 1, b \ge 1$
Now, we don't have to check I, because $2 \not \in A$. II is not injective. Again, here are two options: $1-a < b-1 \vee 1-a \ge b-1$. We will only look at $1-a < b-1$, the other case is analogue. Choose $x_1 = 1-\frac{1-a}{2}, x_2 = 1+\frac{1-a}{2}$. Then $f(x_1)=f(x_2) \Rightarrow x_1 \neq x_2$, so $f$ is not injective. So case II is not injective. III is injective; choose $x_3, x_4 \ge 1$. Then $f(x_3)=f(x_4) \Rightarrow x_3=x_4$.
So, only case III is injective and $2 \in A$ if $a < 2$. How can we make this interval as big as possible? Take $a = 1$ and any real number for $b$. So $A = [1,\infty) = I_+$. So $I_+$ must be the biggest interval for $f$ to be injective and $2 \in I_+$.
Am I missing some points? Can this be done in an easier way?
 A: Formally solving $y = x^2 - 2x$ gives $$x_{1,2} = 1 \pm \sqrt{1+y},$$ which is defined in $\mathbb{R}$ for $y \ge -1$. This can be used to separate the domain of $f(x)$ into two intervals. For $x \in [1,\infty)$ and $x \in (-\infty,1]$, the function $f(x)$ is injective (and invertible). The domain that contains $2$ is of course $ [1,\infty)$.
A: The intuition, I would guess, is that the graph of this function is a parabola and, thus, the function is 1-1 "to the right of" (or "to the left of") the vertex of its graph.  The $x$-coordinate of the vertex, here, is  $x=1$; so, we want the non-parenthetical case. 
Let's write $f$ in the form where we can take advantage of the above:
$$f(x)=(x-1)^2-1.$$
From this it  follows easily that $f$ is strictly monotone on $[1,\infty)$.
Indeed:
$$
1\le x<y \quad\Longrightarrow\quad  (x-1)^2<(y-1)^2 \quad\Longrightarrow\quad (x-1)^2-1<(y-1)^2-1,
$$
as claimed.
It follows that $f$ is  1-1  on $[1,\infty)$.
Also, it is easily seen that $f(1+h)=f(1-h)$ for all numbers $h$; so, $f$ is not 1-1 on any interval of the form $[a,\infty)$ for $a<1$.
A: Since $f$ is continuous, $f$ is injective on some interval if and only if $f$ is strictly increasing or decreasing on the interval. You can see that $f$ is strictly increasing when $x > 1$ and strictly decreasing when $x < 1$. Since you want to include $2$ on the interval, the greatest interval where $f$ is injective must be $[1, \infty[$.
