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: 
Being injective (also badly termed one-to-one) for a function $f$ means that any equation $f(x)=c$ has *at most one* solution $x$, for any given value of$~c$. (A proper term would therefore be at-most-one-to-one.)

So just write down the equation $\frac x{x^2+1}=c$ for some fixed$~c$ and solve for $x$. The equation is equivalent to $cx^2-x+c=0$, which you can solve by the formula for the quadratic equation, provided that $c\neq0$ (for the case $c=0$ you have a linear equation, which indeed has a unique solution, so that specific case is OK). Now a quadratic equation rarely has a unique solution; here the discriminant is $1-4c^2$, which can be positive, zero, or negative. It is the positive values that cause a problem: for any $c\in(-\frac12,\frac12)$ with $c\neq0$ the discriminant is positive, and the equation $f(x)=c$ has *two* solutions; even having just one value of $c$ for which there is more than one solution would suffice to show that $f$ is not injective.

To see whether a restriction of $f$ is injective, you must show that for every $c$ there is at most one $x$ *in the restricted domain* that solves $f(x)=c$. For this purpose it is probably useful to actually write down the solution that the quadratic formula gives you, and to show that in case of multiple solutions, at least one solution is eliminated by the restriction.
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.