Question

How do you find the inverse of `f(x)=sqrt(x-3)`?

Answer 1

`f^-1(x)=x^2+3` where `x>=0`

Explanation:

You can find the inverse function by exchanging `x` and `y` and then solve for `y` in terms of `x`.

`y=sqrt(x-3)` (Original function)
`x=sqrt(y-3)`
`x^2=y-3`
`x^2+3=y`
Therefore
`f^-1(x)=x^2+3`

The domain of the inverse function, `f^-1(x)` is equal to the range of the original function `f(x)`. The range should be `[0, +oo)` if the domain of `f(x)` is `[3, +oo)`.

Answer 2

The inverse is `f^-1(x)=x^2+3`

Explanation:

The domain of `f(x)` is `x in [3,+oo)` and the range is `y in [0,+oo)`

Let `color(blue)(y)=sqrt(color(red)(x)-3)`

Interchange `x` and `y`

`color(red)(x)=sqrt(color(blue)(y)-3)`

Now, we express `y` in terms of `x`

`x^2=y-3`

`y=x^2+3`

Therefore,

`f^-1(x)=x^2+3`

The domain and range of `f^-1(x)` is the range and domain of `f(x)`

The domain of `f^-1(x)` is `x in [0,+oo)` and the range is `y in [3,+oo)`

Verification

`f(f^-1(x))=f(x^2+3)=sqrt(x^2+3-3)=x`