# 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)?

• Suggest using wolfram or another computational engine to self check simple queries like this: wolframalpha.com/input?i=maximum+of+30x%2F%28x%5E2%2B2%29 Dec 4, 2022 at 0:42
• Hint: $t^2+2\geq 2\sqrt 2 t$ by AM-GM.
– Feng
Dec 4, 2022 at 0:44
• @Feng. Feel free, if you've got time for this, to convert your hint into a full answer, I've only recently heard about this inequality and I'd be very interested in seeing it put at work. Dec 4, 2022 at 0:47

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 $$t=\sqrt 2$$.

• Nice answer, thanks! Dec 4, 2022 at 9:17
• @VinceVickler You're welcome. I'm glad to help! :)
– Feng
Dec 4, 2022 at 9:23

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.

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 $$x=a$$.