Question: The function $f$ is defined by $f(x)=x/(x^2+1)$, $x$ is an element of a set of real number.
If $a$ is an element of a set of real number and $a$ is not $0$, find the image of $1/a$ under $f$. (This I know how to solve it, but after that... I don't know)

Deduce that $f$ is not one-to-one.(How can we deduce?By diagram?)

Show that if $a,b$ are an element of real number with $a>b≥1$,then $f(b)>f(a)$.

Deduce that, if the domain of $f$ is restricted to the subset of $\mathbb{R}$ given by $\{x:x≥1\}$, then $f$ is one-to-one.

State the range of $f$ in this case.

Can anybody help me to solve this question?

Answer: 
**Hint:** If you are asking about
$$
f(x)=\frac{x}{x^2+1}
$$
then consider the relation between $f(1/x)$ and $f(x)$.
f(0) = 0; f(x) $\leq$ f(1) for all x; limf(x) = 0 when x tends to $\infty$. Hence for all number y of the open intervale (0, 1/2) there are two distinct numbers $x\_1$ and $x\_2$ with f($x\_1$) = f($x\_2$) = y. Thus f is not one-to-one.