Question

What is the inverse function of `f(x)= absx + 1`?

Answer 1

No inverse as `f(x)` is not a one-to-one function. However, it is possible to obtain a multivalued inverse.

Multivalued inverse:
`f^(-1)(x) = +-(x-1), " " x >= 1`

Explanation:

Multivalued inverse:
`f(x) = |x| + 1`
`f(x)-1 = |x|`
`f^(-1)(x) = +-(x-1), " " x >= 1`

`f^(-1)(x)` can also be found by taking y = x as the axis of symmetry.
Note that the range of f(x) becomes the domain of `f^(-1)(x)`.

Answer 2

As a function from `RR` to `RR`, `f(x)` is not one-one.

As a result it has no well defined inverse function.

Explanation:

For example, `f(1) = f(-1) = 2`, so `f^(-1)(2)` is not well defined - it could be `1` or `-1`.

If we restrict the domain of `f(x)` to `[0, oo)` or to `(-oo, 0]`, then the resulting mapping is one-one onto `[1, oo)` and there is a well defined inverse function with domain `[1, oo)`.

`f_{[0, oo)}(x)` has inverse `f_{[0, oo)}^-1(y) = y - 1` where `y in [1, oo)`

`f_{(-oo, 0]}(x)` has inverse `f_{(-oo, 0]}^-1(y) = 1 - y` where `y in [1, oo)`