Show that the function f : R -- R defined by

f(x) =x / x² +1 , ( x belongs R)

is neither one-one nor onto....

$f\left(x\right)=\frac{x}{x^2+1}$

By a little examination, we can see that:-

For all x>0, $f\left(x\right)\in \left(0, 1\right)$
For x=0, f(x)=0
For all x<0, $f\left(x\right)\in \left(-1, 0\right)$

So, clearly f(x) is not onto, as its range belongs to only $\left(-1, 1\right)$, and not (-R, R).

Now, $$\begin{gathered} f \text{'}\left(x\right)= \left(\frac{1}{1+x^2}\right)\frac{d}{dx}\left(x\right) + x\frac{d}{dx}\left(\frac{1}{1+x^2}\right) \\ =\frac{1}{1+x^2}+x\left\{-\frac{2x}{{\left(1+x^2\right)}^2}\right\} \\ =\frac{1}{1+x^2} - \frac{2x^2}{{\left(1+x^2\right)}^2} \end{gathered}$$
                   $$\begin{gathered} =\frac{1+x^2-2x^2}{{\left(1+x^2\right)}^2} \\ =\frac{1-x^2}{{\left(1+x^2\right)}^2} \end{gathered}$$

So, f '(x)=0 at x=1.
This means, that the slope of f(x) changes sign before and after x=1.
That is, there exists some $x_1 \& x_2$for which $f\left(x_1\right)=f\left(x_2\right)$.
So, f(x) is not one-one.

Thus, it has been proved that f(x) is neither one-one nor onto.