F: R --> R
F: A --> B
a function is not defined at x=-2 that's why they are removing it from the domain.
For making a function onto, I m taking counterexample, I mean two elements from set A is having the same image in set B. which means
f(a) = (a+1) / (a+2) ---------(1)
f(b) =(b+1) / (b+2) ----------(2)
from 1 and 2
(a+1) /(a+2) =(b+1) / (b+2)
on solving a=b
which means my assumption is wrong so, for every element, we have a unique image that's why the function is one to one.
for proving a function onto, we have to find out for every c in set Y, there must exist a preimage in set X
(x+1) / (x+2) = c
x + 1 = cx -2c
x - cx = (-1 - 2c)
x(1-c) = (-1 - 2c)
x = (-1 - 2c)/ (1-c )
There will no preimage for element 1.
for every element other then 1 , we can find pre-image...
so, function is onto..