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

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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}`

y=1/(x+1)

**Domain: {all real numbers except -1}** ;since we cannot allow the denominator be equal to zero.

**Range: {all real numbers except zero}**

In solving for the inverse, simply replace x by y and y by x, then solve for y in terms of x.

y=1/(x+1)

x=1/(y+1)

y+1=1/x

**y=(1/x) - 1**

**Domain: {all real numbers except zero}** ;since we cannot allow the denominator be equal to zero.

**Range: {all real numbers except -1}**; since the first term will not be equal to 0.

Note: the domain of a function is equal to the range of its inverse function, and the range of a function is the domain of its inverse function.

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