Find a parameter $m$ of a second order polynomial for which the polynomial is injective on a given interval. I am given the function:
$f : \mathbb{R} \rightarrow \mathbb{R}$
$ f(x)=x^2-mx+2$
$ m \in \mathbb{R}$
And I am asked to find $m \in \mathbb{R}$ for which the function is injective on the interval $[-1, 1]$. Here is my though process:
Since the function is a second order polynomial, that means that for the function to be injective (that is, any parallel to the $Ox$ axis would cut the function in at most one point on the interval $[-1, 1]$) then the minimum point of the graph (minimum since the coefficient of the term $x^2$ is $1>0$) must have its $x$ coordinate either before or after the interval $[-1, 1]$. That would make the function on the interval $[-1, 1]$ either strictly increasing or decreasing and since the function is continuous, it would be injective on the interval. Also, if you would place the minimum somewhere inside the interval, you could cut the function in $2$ places with a parallel to the $Ox$ axis in the given interval. Since the minimum point has the coordinates:
$V( -\dfrac{b}{2a}, -\dfrac{\Delta}{4a} )$
That means that we must have:
$-\dfrac{b}{2a} \le -1$    or 
$-\dfrac{b}{2a} \ge 1$
That means:
$m \le -2$ or 
$m \ge 2$.
So I concluded that $m \in (- \infty, -2 ] \cup [2, \infty)$.
The problem is, this answer is wrong. The textbook says so but I was not given an explanation. I think I have to also consider the $y$ value of the minimum point, but I do not see how placing the $x$ value of $V$ in that interval could keep the function injective, regardless of how low/high the $y$ value would be.

Tthe position of the max/min of the quadratic $f(x)=Ax^2+Bx+C$ is $x_0=-\frac{B}{2A}$. If The function has to be one to one (injective) fo $x \in [a,b]$ it has to either monotonically increase or decrease in $[a,b]$, then $x_0=\le a ~or~ x_0 \ge b.$
So here $$x_0=-m/2 \le -1 ~or~ x_0=-m/2 \ge 1 \implies m \in (-\infty, 2] \cup [2, \infty).$$ Remarkably, here $\Delta=B^2-4AC$ does not play any role here.
