Question: > 
> Show that the set $\{\frac{3x+2}{x^2+5}|x\in \mathbb{R}\}$ is bounded.
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A friend suggested that I should you Cauchy inequality to prove this. But I want to find a simpler way. Please help.

Answer: 
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 $$\displaystyle -\frac{1}{2}\leq y\leq \frac{9}{10}$$
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.

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> \*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.
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