Question

How would you solve the inverse of the function `f(x)= absx + 1`?

Answer 1

Note that the inverse of `f(x)=abs(x)+1` is not a function.
However we could claim `+-sqrt(x^2-2x+1` is its inverse mapping.

Explanation:

Replacing `f(x)` with `y` for simplicity.
graph{abs(x)+1 [-5.78, 6.704, -0.68, 5.56]}
Note that for every value `y > 1` there are 2 values of `x`
and therefore the inverse of `f(x)` can not be a function .

However, given `y = abs(x)+1`
if we exchange the `x` and `y` variables to get an inverse mapping:
`color(white)("XXX")x=abs(y)+1`

`rarrcolor(white)("XX")abs(y)= x-1`

`rarrcolor(white)("XX")y^2 =(x-1)^2 = x^2-2x+1`

`rarrcolor(white)("XX")y = +-sqrt(x^2-2x+1)`