$f: \mathbb R \to \mathbb R$ by $f(x) = (x-1)^2$, what is $f^{-1}( [0,1] )$?

Question

$f: \mathbb R \to \mathbb R$ by $f(x) = (x-1)^2$, what is $f^{-1}( [0,1] )$?

How would you go about solving the inverse $f^{-1}([0,1])$?

If I took the inverse of the function itself it would be: $\pm\sqrt{x} +1$

The solution for it is $[0,2]$ but after plugging the numbers in, I'm not sure how that is the case.

Answer 1

$x\in f^{-1}[0,1]\iff f(x)\in [0,1]$

$\iff (x-1)^2\in [0,1]\iff 0\le (x-1)^2\le 1 $

$\iff 0\le |x-1|\le 1\iff |x-1|\le 1$

$\iff 0\le x\le 2\iff x\in [0,2] .$