Question

How do you find the inverse of `f(x)=(x+3)^(1/2) + 2` and is it a function?

Answer 1

`f^(-1)(y) = (y-2)^2-3`

Explanation:

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

Assuming that we are dealing with Real square roots, the (implicit) domain of this function is `[-3, oo)`.

Let `y = (x+3)^(1/2)+2`

Rearrange this equation so that `x` is isolated on one side.

First subtract `2` from both sides to get:

`y-2 = (x+3)^(1/2)`

Square both sides to get:

`(y-2)^2 = x+3`

Subtract `3` from both sides and transpose to get:

`x = (y-2)^2-3`

For any Real value of `y`, this formula gives us a unique value of `x`. So this is a suitable inverse function:

`f^(-1)(y) = (y-2)^2-3`

The domain of this inverse function is the whole of `RR`, i.e. `(-oo, oo)` and its range is `[-3, oo)`, the domain of `f(x)`.

Footnote

Note that though `f^(-1)(y)` is an inverse function for `f(x)`, `f(x)` is not an inverse function for `f^(-1)(y)`.

The function `f^(-1)(y)` is not one-one and we find (for example):

`f(f^(-1)(color(blue)(1))) = f(-2) = color(blue)(3)`