College Algebra, Chapter 3, 3.7, Section 3.7, Problem 70

The function $f(x) = (x - 1)^2$ is not one-to-one. Restrict its domain so that the resulting function is one-to-one. Find the inverse of the function with the restricted domain.







If we restrict the domain for $x \geq 1$, the function is now one-to-one, to find its inverse, we set $y = f(x)$.


$\begin{equation} \begin{aligned} y =& (x - 1)^2 && \text{Solve for $x$; take the square root} \\ \\ x - 1 =& \pm \sqrt{y} && \text{Add } 1 \\ \\ x =& 1 \pm \sqrt{y} && \text{Interchange $x$ and $y$} \\ \\ y =& 1 \pm \sqrt{x} && \text{Apply restrictions, for } x \geq 1 \\ \\ y =& 1 + \sqrt{x} && \end{aligned} \end{equation}$



Thus, the inverse of $f(x) = (x - 1)^2$ for $x \geq 1$ is $f^{-1} (x) = 1 + \sqrt{x}$.