Show that the range of the function $f(x)=(3x+2)/(x^2+5)$ is bounded 
Show that the set $\{\frac{3x+2}{x^2+5}|x\in \mathbb{R}\}$ is bounded.

A friend suggested that I should you Cauchy inequality to prove this. But I want to find a simpler way. Please help.
 A: Let $$\displaystyle y = \frac{3x+2}{x^2+5}\Rightarrow yx^2+5y=3x+2$$
So we get $$yx^2-3x+(5y-2) = 0$$
Now for real values of $y,$ values of $x$ must be real.
So its $$\bf{Discriminant\geq 0}$$
So we get $$\displaystyle 9-4y\cdot (5y-2)\geq 0$$
So we get $$\displaystyle 9-20y^2+8y\geq0$$
So we get $$\displaystyle 20y^2-8y-9\leq 0\Rightarrow \left(y+\frac{1}{2}\right)\cdot \left(y-\frac{9}{10}\right)\leq 0$$
So we get $$\displaystyle -\frac{1}{2}\leq y\leq \frac{9}{10}$$
A: $$f(x)=\frac{3x+2}{x^2+5}$$
is a continuous function on $\mathbb{R}$ for which
$$ \lim_{x\to \pm\infty}f(x) = 0,$$
hence $f$ is bounded. By computing $f'(x)$, we may check that the stationary points of $f(x)$ occur at $x=-3$ and $x=\frac{5}{3}$. If we compute the values of $f(x)$ at such points, we get:
$$-\frac{1}{2}\leq f(x) \leq \frac{9}{10}.$$
Anyway, the Cauchy-Schwarz inequality is a pretty fast way to go:
$$ \left|3x+2\right| = \left|3x+\frac{2}{\sqrt{5}}\cdot\sqrt{5}\right|\leq \sqrt{9+\frac{4}{5}}\sqrt{x^2+5} $$
hence:
$$ \left| f(x)\right |\leq \frac{7}{\sqrt{5}\sqrt{x^2+5}}\leq \frac{7}{5}.$$
The AM-GM plus the triangular inequality work pretty well, too:
$$\left|\frac{3x+2}{x^2+5}\right|\leq \frac{2}{5}+\frac{3|x|}{x^2+5} \leq \frac{2}{5}+\frac{3}{2\sqrt{5}}.$$
A: Let $f(x)=\frac{3x+2}{x^2+5}$. Extend the domain of the function to the extended real line* $\overline{\Bbb R}=\Bbb R\cup\{-\infty,\infty\}$, as follows: Define $f(\infty)=f(-\infty)=0$. You can check that this extended function is still continuous.
Thus, $f$ is continuous on $\overline{\Bbb R}=[-\infty,\infty]$, a compact set, and therefore bounded.

*The extended real line is $\Bbb R$ with the addition of $\infty$ and $-\infty$. While it's possible to define addition, subtraction, multiplication, and division on $\overline{\Bbb R}$ (at least partially), there's no need to do so here.

A: When $|x| \ge 3$ we have $|3x+2|\le 3|x|+2 \le (x.x+2)<(x.x+5)=|x^2+5|\ne 0 $ so  the absolute value of the ratio is less than $1$. When $|x|<3$ we have $|3x+2|\le 3|x|+2<3.3+2=11$ , and $x^2 + 5 \ge 5$ so the absolute value of the ratio is less than  $11/5.$ For this Q any bound less than infinity will do.
