Given the function `f(x)=9/(x+1)` determine `f^-1(x)` and its domain and range.

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The function `f(x)=9/(x+1)`

To determine `f^-1(x)` , let `f(x) = y = 9/(x +1)` . Express x in terms of y, this gives `x + 1 = 9/y`

=> `x = 9/y - 1`

Interchange x and y

=> `y = 9/x - 1`

The inverse function `f^-1(x) = 9/x - 1` . The domain of `f^-1(x)` is all real values of x `!=` 0 and the range is all real values of x except x = -1.

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`f(x) = 9/(x+1)`

We have to find the inverse function of f(x).

let `f(x) = y,`


`y = 9/(x+1)`

Now rearrange the above expression so that we get an expression for x.

`x+1 = 9/y`

`x = 9/y - 1`

Now interchange x and y, so that we get the inverse function,

`y = 9/x -1`

`y = (9-x)/x` therefore, `f^(-1)(x) = (9-x)/x`


The domain of f(x) is `{x in R, x!=-1}` . The range of f(x) is

`{x in R}.`


The domain of `f^(-1)(x) = {x in R, x !=0}`
, since x cannot be zero (The denominator cannot be zero).

The range of f^(-1)(x) is obviosuly the domain of f(x) which is,

`{x in R, x!=-1}`


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