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. 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?