Question

Show that function x/x2+1 is neither one one nor onto

Answer 1

Steps and Understanding:
1) Let
$f(x) = \frac{x}{ {x}^{2} + 1 }$
2) By Graphical Approach,

If there exists a line parallel to x-axis which cuts the graph of the function at least two points, then function is many -one otherwise
one-one.

For Onto,
If Co-Domain = Range,
then f(x) is onto function.

Here,
Co-Domain is Range.

3) f'(x) = (1-x^2)/ (x^2+1)^2
Critical Points are x = ±1 .
It is increasing in [-1,+1].

4)
$f(x) \: max \: = f(1) = \frac{1}{2} \\ f(x)min \: = f( - 1) = \frac{ - 1}{2}$

Since,
Range of f(x) = [-1/2 , 1/2]
So, It is not onto function as Co - domain is set of Real Numbers.

5) On making graph by critical points, it's monontonicity.
See pic 2 :
There exits horizontal line parallel to x -axis which cuts the graph at two -points , so
it is many -one function.
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Method - 2 ( For Many -one)
1) For -one one Function.

If a, b belongs to real and f(a) = f(b) if and only if, a = b.

$f(a) = f(b) \\ = > \frac{a}{ {a}^{2} + 1} = \frac{b}{ {b}^{2} + 1} \\ = > a {b}^{2} + a = {a}^{2} b + b \\ = > ab(b - a) - (b - a) = 0 \\ = > (b - a)(ab - 1) = 0 \\ = > b = a \: \: or \: ab = 1$
Here, we got two conditions
a = b or ab= 1
Hence, it is many -one function.

Therefore,
$f(x) = \frac{x}{ {x}^{2} + 1} \: is \: many \: \: one \\ \: and \: into \: \: function \:$