what is the inverse of y=1/x+1? State the domain and range show complete solution

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`y = 1/(x+1)`

The domain of y is `{x in R; x!=-1}`

The range of y is `{y in R;y!=0}`

To find the inverse function, let's rearrange the avove equation,

`y = 1/(x+1)`

`x+1 = 1/y`

`x = 1/y -1`

 

Therefore, the inverse function is,

`y = 1/x -1`

 

The denominator cannot be zero at anytime.

Domain is the range of previous function, `{x in R; x!=0}`

and the range is the domain of previous function ,` {y in R, y!=-1}`

 

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