Source : Stewart, Precalculus.
The original question is to graph the function : $f(t)= \frac {30t} {t^2 + 2}$.
Desmos construction : https://www.desmos.com/calculator/19kybcl3hc
It can be shown that $f(x)$ tends to $0$ as $t$ goes to infinity ( in both directions), so $y= 0$ is a horizonal asymptote.
Also, near zero, the function looks like $y = t$.
The part of the question I am interested in here is : what is the maximum value of $f(t)$, not using calculus?
Following a method shown by Sybermath ( ' Finding the maximum value of a rational function", YT ) , I attempted this approach:
(1) Set $f(t)= M$ , that is $f(t)= \frac {30t} {t^2 + 2}=M$. The question becomes: what is the maximum value of $M$?
(2) $\frac {30t} {t^2 + 2}=M$
$ \iff 30t = M(t^2 +2)$
$\iff 30t = Mt^2 +2M$
$\iff -Mt^2 +30 t -2M = 0$
(3) Since we only want real values of $t$ , we require
$\Delta = b^2 - 4 ac \geq 0 $
$\iff 30^2 - 4(-M)(-2M) \geq 0 $
$\iff 30^2 -8M^2 \geq 0$
$\iff M^2 \leq 30^2/8$
$\iff \sqrt{M^2} \leq \sqrt {30^2 / 8}$
$\iff |M|\leq \sqrt {30^2 / 8}$
$\iff -\sqrt {30^2 / 8} \leq M \leq \sqrt {30^2 / 8} \approx 10.6$
(3) So, the maximum value of $f(x)= M $ is $\sqrt {30^2 / 8} \approx 10.6.$
Is this answer correct? Are there other , desirably quicker methods to answer the question ( without calculus)?