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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
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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

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oh! got it now... thanks
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1 Answer

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To find inverse of a function if inverse exits.. do this

Put.  f(x) = y  ..........(1)

          f^-1(y)=x      .......(2)

In our case f(x) =ax+b =y  ....... (3)

                      Now put , x= f^-1(y)  in (3)

We have  y = a.f^-1(y)+b

                 (y-b)/a =f^-1(y)

Puy x in place of y

f^-1(x)=(x-b)/a   this is the inverse of f(x)

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