# 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,
For x=0, f(x)=0
For all x<0,

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

Now,
$=\frac{1+{x}^{2}-2{x}^{2}}{{\left(1+{x}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{1-{x}^{2}}{{\left(1+{x}^{2}\right)}^{2}}$

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 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.

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