So, there is this question in which it needs to proven of the function the following has to be true:
$f: \mathbb{R} \rightarrow \mathbb{R}$, $f(x) = \frac{x}{4+9x^2}$ is $[-\frac{1}{12}, \frac{1}{12}]$
(The function of domain all real numbers projected on the co-domain of all real numbers, of the given function has the restricted domain).
To approach this, each have to be a subset of each other (function and image). Therefore:
Let $y\in f(\mathbb{R}) \Rightarrow$
$y = f(x) = \frac{x}{4+9x^2} \Rightarrow$
$[-\frac{1}{12}] \leq \frac{x}{4+9x^2} \leq [-\frac{1}{12}] \Rightarrow$
Then solve. Geting how to show that:
$f(\mathbb{R}) \subseteq [-\frac{1}{12}, \frac{1}{12}]$
is known, however showing the reverse that:
$[-\frac{1}{12}, \frac{1}{12}] \subseteq f(\mathbb{R})$
is where the problem is. To start:
Let $y\in -[\frac{1}{12}, \frac{1}{12}] \Rightarrow$
$y = f(x) = \frac{x}{4+9x^2} \Rightarrow$
$y = \frac{x}{4+9x^2} \Rightarrow$
$9yx^2+4y -x = 0 \Rightarrow$
(which is now a quadratic).
Now, it says that at least one $x$ needs to be satisfied to prove the subset is equal. But why? Then after you say that the discriminant of this quadratic has to be $\geq 0$ you get:
$1-144y^2 \geq 0 \Rightarrow$
$(1-12y)(1+12y) \geq 0 \Rightarrow$
Now, it says since:
$-1 \leq 12y \leq 1$
That it implies:
$1-12y \geq 0$ and $1+12y \geq 0$
Now, it says, because it has a solution the restricted image is a subset of the function. It also notes after that, because this quadratic can only be true when $x \neq 0$ so then do then consider the following:
For $y = 0$, take $x = 0$ and so, $0\in\mathbb{R}$
But why? Thanks.
Other important points of the domain:
$-\frac{1}{12} \leq y \leq \frac{1}{12} = -1 \leq 12y \leq 1 =-\frac{1}{144} \leq y^2 \leq \frac{1}{144}$
Thanks.
What it means for $a$ to be in the image of $g$ is that we have $g(c)=a$ for some $c$. All we need is one $c$: if $g(3)=17$, then $17$ is in the image of $g$, full stop - maybe $g(5)$ also equals $17$, or maybe not, it doesn't matter.
So as soon as you've found a single $x$ such that $f(x)=y$, you know that $y$ is in the image of $f$. Now since $y$ was an arbitrary element of $[-{1\over 2},{1\over 2}]$, what you've really shown is that $[-{1\over 2},{1\over 2}]\subseteq f(\mathbb{R})$.