Without calculus, what is the maximum value of the rational function : $f(t)= \frac {30t} {t^2 + 2}$ 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)?
 A: It is clear that $f$ is an odd function and $f(t)>0$ for $t>0$. To find the maximum of $f$, we only need to consider $t>0$. Now, by AM-GM inequality, we have
$$t^2+2\geq 2\sqrt{t^2\cdot 2}=2\sqrt 2 t,$$
where the equality holds if and only if $t^2=2$, i.e., $t=\sqrt 2$.
Hence for $t>0$ we have
$$f(t)=\frac {30t} {t^2 + 2}\leq \frac{30 t}{2\sqrt 2t}=\frac{15}{\sqrt 2},$$
with equality holds when $t=\sqrt 2$. Therefore, the maximum of $f$ is $\frac{15}{\sqrt 2}$.
Your answer $\sqrt {30^2 / 8}$ is also correct:
$$\sqrt {30^2 / 8}=\frac{\sqrt{30^2}}{\sqrt8}=\frac{30}{2\sqrt 2}==\frac{15}{\sqrt 2}.$$
Remark. If you want to graph the function $f$, it is not enough to know the maximum of $f$. For example, the sine function has many maximum points with the same maximum value. You also need to know the monotonicity of $f$. Here for $f(t)=\frac {30t} {t^2 + 2}$, we can argue in this way (again $t>0$):
$$f(t)=\frac{30}{t+\frac2t},$$
and the function $t\mapsto t+\frac2t$ for $t>0$ is a "hook function", which is decreasing in $(0,\sqrt 2)$ and increasing in $(\sqrt 2,+\infty)$; as a result, $f$ is increasing in $(0,\sqrt 2)$ and decreasing in $(\sqrt 2,+\infty)$. Now I believe that you are able to graph $f$, at least in $(0,\infty)$. Finally, just recall that $f$ is an odd function.
Remark. Another method to see $t^2+2\geq 2\sqrt 2 t$:
$$t^2+2- 2\sqrt 2 t=(t-\sqrt 2)^2\geq0,$$
with equality holds if and only if $t=\sqrt 2$.
A: We can also work out a generalization in this way.  We can establish by the "zero-discriminant" quadratic polynomial method you used or by application of the AM-GM inequality (Feng's approach) that $ \ x + \frac{1}{x} \ $ has its relative minimum of $ \ +2 \ $ at $ \ x \ = \ 1 \ $ for $ \ x \ > \ 0 \ $ and its relative maximum of $ \ -2 \ $ at $ \ x \ = \ -1 \ $ for $ \ x \ < \ 0 \ \ . $ We can "rescale" the variable to find that $ \frac{x}{a}  +  \frac{a}{x} \ $ also has its relative minimum of $ \ 2 \ $ at $ \ x \ = \ a \ $ for $ \ \frac{x}{a} \ > \ 0 \ \ . \ $  Finally, multiplying this function by $ \ a \ $ produces $ a· \left(\frac{x}{a}  +  \frac{a}{x} \right) \ = \ \frac{x^2 \ + \ a^2}{x} \ \ ,  \ $ which for $ \ a \ \neq \ 0 \ $ has its relative minimum of $ \ 2a \ $ at $ \ x \ = \ a \ \ . $  (This of course can also be arrived at by the methods mentioned above.  An alternative expression is to write $ \ \frac{x^2 \ + \ a^2}{x} \ = \ c \ $  as $ \ x^2 - cx \ = \ -a^2 \ \ ; \ $ "completing the square" yields $ \ \left( x - \frac{c}{2} \right)^2 \ = \ \frac{c^2}{4} - a^2 \ \ , \ $ for which there is a single root at $ \ x \ = \ \pm a \ $ when $ \ c \ = \ \pm \ 2a \ \ . \ ) $
We then conclude that $ \ \large{\frac{x}{x^2 \ + \ a^2} } \ $ has its relative maximum of $ \ \large{ \frac{1}{+2|a|} } \ $ at $ \ x \ = \ +|a| \ $ (and its relative minimum of $ \ \frac{1}{-2|a|} \ $ at $ \ x \ = \ -|a| \ ) \ $ for $ \ a \ \neq \ 0 \ \ . \ $  The function $ \  f(x) \ = \ \large{ \frac{30x}{x^2 \ + \ 2} \ = \ 30·\frac{x}{x^2 \ + \ 2}  \ } \ \ $ therefore has its maximal value of $ \ 30·\large{ \frac{1}{2·\sqrt2} \ = \ \frac{15·\sqrt2}{2} } \ $ at $ \ x \ = \ +\sqrt2 \ $ (and its minimal value of  $ \ -\frac{15·\sqrt2}{2}  \ $ at $ \ x \ = \ -\sqrt2 \ \ ) \ . $  The symmetry of the extrema about the origin is to be expected for an odd function.
A: You can also generalise Feng's second comment: $$\frac{x^2+a^2}{x}=x + \frac{a^2}{x}=\Big(\sqrt{x}-\frac{a}{\sqrt{x}}\Big)^2+2a \ge 2a$$ with equality when $x=a$.
