Question

How do you find `f^-1(x)` given `f(x)=(x+2)^2+6`?

Answer 1

`f(x)` has no inverse function, but...

Explanation:

Given:

`f(x) = (x+2)^2+6`

We can attempt to derive an inverse function as follows:

Let:

`y = f(x) = (x+2)^2+6`

Subtract `6` from both ends to get:

`y-6 = (x+2)^2`

Take the square root of both sides, allowing for both possible signs:

`+-sqrt(y-6) = x+2`

Transpose and subtract `2` from both sides to get:

`x = -2+-sqrt(y-6)`

Note that for any value of `y` (apart from `6`) in the range `[6, oo)` of `f(x)`, this formula gives us two possible values of `x`.

Since `f(x)` is many to one on its domain `(-oo, oo)`, its inverse is not a function, but a relation. So we cannot define `f^(-1)(y)` unless we restrict the range.

Alternatively, we can define an inverse image function `F^(-1)(y)` from `[6, oo)` to `2^RR` (the set of subsets of `RR`) by:

`F^(-1)(y) = { -2+sqrt(y-6), -2-sqrt(y-6) }`