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 $f^{-1} (x) = 1 + \sqrt{x}$.