Question

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

f(x) =x / x2 +1 , ( x belongs R)

is neither one-one nor onto....

Answer 1

Hey Mate
Here's your answer

For one one function for each result of f(x) the preimage should be unique and if you see the denominator term it has x² in it.
As we all know square roots give ±terms , hence for each square root you get 2values one +ve and one -ve hence these are two preimages which violates the one one rule.

For onto all real number values should be satisfied but here in this function we cannot have 0 as the denominator as it will make the function undefined although 0is a part of the real numbers set R. Getting a 0 and denominators less than 0 is never possible in this function as the denominator is always squared and 1 is added so negative R values are absent.

Hence function is neither one-one nor onto


Hope this helps

Be Brainly

Answer 2

Answer:

Step-by-step explanation:

if you take derivative of that function

you will get  x^2+1 - x(2x) / x^2+1

1-x^2 / (1+x^2)^2

here it can take positive as well as negative

so many one

here range is

y=x/x^2+1

x^2y+y=x

yx^2-x+y=0

x=1+root 1-4y^2 / 2y

domain of x is range of y

so range is [-1/2 , 1/2]

range is subset of codomain

so into

so the function is neither one one nor onto