For a function $f:(-∞; 1]$ $\rightarrow$ ${R}$ $$f(x)=x^2+mx+1$$ Determine $m$ for which $f$ is injective(~stricly monotonous), without use of calculus.
Based on my attempts, I've already restrained the interval of $m$ to $[-2;+∞)$ or even $$[\frac{1+ 17^{1/2}} {2};+∞)$$ though I' m unsure it is correct, and am confident it can be more rigorously restrained.
Note that $$ x^2 + mx + 1 = \left(x + \frac m2\right)^2 - \frac{m^2}4 + 1, $$ hence, $f$ attains its minimal value in $x = -\frac m2$. Therefore, $f$ is injective iff $-\frac m2\ge 1$, or $m \le -2$.