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Show that the function f(x) = ax + b from R->R is invertible, where a and b are constants, with a$\neq$0, and find the inverse of f

How to check whether this function is onto?

pls give a detailed solution
in Set Theory & Algebra by Active (5k points)
edited by | 48 views
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Inverse of f:A->B exists iff  'f' is a bijection

1) One-One

f(x1) = f(x2)

a(x1) + b = a (x2) + b

x1 = x2

hence it's one-one function

 

2) Onto

ax + b = c

ax = b-c

x = b - c / a

given a !=0

I want to find out for every "C" there exists an element in domain or not

therefore ,we can see that Yes for every value of 'C' we get the pre-image in Domain

hence , it's Onto -function

 

0
it'll be x=y-b /a right?
0

   yeah

but you ask in the question that How to check whether this function is onto?

that'Y i just explain you one-one and  onto 

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@Magma

inverse will be y-b/a right

if take ax+b = y

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@Magma could you pls elaborate the onto part a little more? it's not clear to me

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y = ax+b

x = ay+b

x-b/a = y

f-(x) = x -b /a
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onto mean : for every element in co-domain there should be a pre-image in domain right ??

now assume there's is an element in co-domain 'c'

c is an arbitrary element

now definitely element 'c' have pre-image in domain right ?? 

x = b - c / a

 now a and b are const it's given in the question and a ! =0

for any  value of c we get the pre-image in domain

you can check it now

C can be any real number

for every value in c we get pre-image in domain of the function

0
oh! got it now... thanks

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