Kenneth Rosen Edition 6th Exercise Question 18 c (Page No. 147)

Question

Question Determine whether the following function from R to R is a bijection
f(x) = $\frac{(x+1)}{(x+2)}$

Solution:
First of all, this is not a function because f(-2) is not defined.

So, if the domain of the function is R - {-2} then this function is one-to-one.

And, if co-domain of this function is R - {1} then this function is onto.

So, the given function is bijection but with two above mentioned conditions. can someone please confirm?

Comments

Yes, I think the above function is bijection with above two conditions.

Suppose  $y = \frac{x+1}{x+2}$

it is clear that x != -2, otherwise it is not a function, so, we have to satisfy the first condition.

Now,

y(x+2) = (x+1)

xy + 2y = x + 1

2y-1 = (1-y)x

$\frac{2y-1}{1-y} = x$

So, y != 1 which is second condition.

So, for getting Bijection function we have to satisfy these two condition.

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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..

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Relation in which for every input there exist a unique output is called function.

f(x) = (x+1) / (x+2) is a function for domain R- {-2}

For no two different x we have same y => it's one-one

if range = co-domain it's onto {range is R-{1} then only it's onto}

No x can give y=1 so if y=1 is eliminated from range then only function is Onto.

one-one and onto => Bijective function.