Suppose I know that the function $f(x)=x²$ has a minimum value on $\mathbb R$.
Could I determine this munimum value algebraically ( without using calculus).
If I set $x² \geq m$, I only get this :
$x² \geq m$
$\iff \sqrt{x²} \geq \sqrt{m}$
$\iff |x| \geq \sqrt{m}$
$\iff (x \geq \sqrt{m}) \lor (-x \geq \sqrt{m}) $
$\iff (x \geq \sqrt{m}) \lor (x \leq\ -\sqrt{m} ) $
Does this tell me that $m=0$ , which is the minimum value of $x^2$ on $\mathbb R$?
Is there a way to determine algebraically the minimum value of some functions, at least on a given interval?
For the function $x^2$ the answer is very simple: by the rule of signs $$ x\neq 0\Rightarrow x^2>0,\qquad x=0\Rightarrow x^2=0$$ hence:
i) $0$ it the minimum value attained by the function $x^2$;
ii) $x=0$ is the only minimizer.
Of course there are many other similar cases (e.g. $x^{2n}$)