The inverse function of $f(x) = |x - 1|$

Question

The inverse function of $f(x) = |x - 1|$ would be $y = x + 1$ by isolating the $y$ right?

Answer 1

A function can only have an inverse if it is injective. This means that in order for $f(x)=|x-1|$ to have an inverse, you must restrict its domain to a set $[a,b]$, where either $a\ge 1$ or $b \leq 1$. If for instance you take the set $[1, +\infty[$, $f(x)= x - 1$ and you can obtain the expression of the inverse solving $y = x -1$ with respect to $x$.

Answer 2

You cannot "isolate the $y$", as you claim.

One way of seeing your error is to first view $f(x)$ as $\sqrt{(x-1)^2}$. So starting with $x=|y-1|$ and solving for $y$: \begin{align*} x&=|y-1|\\ x&=\sqrt{(y-1)^2}\\ x^2&=(y-1)^2 \end{align*} ...and we know that we cannot simply "cancel" the exponents. For example, if $x=1$ and $y=0$ then $x^2=(y-1)^2$ but $x\neq y-1$. So there is no way to "isolate the $y$" here. (The problem is that the function $f(x)$ is non-injective, and indeed neither is the quadratic function $g(x):=x^2$ which is why we could not cancel the exponents above.)