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 $f(x)$.
If $f$ is not one to one, then if $f(x\_1) = f(x\_2)$, it is possible that $x\_1 \neq x\_2$. 

So, we can start by saying, 

$$ \dfrac{x\_1}{x\_1^2 + 1} = \dfrac{x\_2}{x\_2^2+1}$$
$$ x\_1 x\_2^2 + x\_1 = x\_2 x\_1^2 + x\_2$$
$$x\_1x\_2^2 - x\_2x\_1^2 = x\_2 - x\_1$$
$$ x\_1x\_2(x\_2 - x\_1) = x\_2 - x\_1$$
$$x\_1x\_2 = 1 $$

Hence, this can't be one to one.