College Algebra, Chapter 3, 3.7, Section 3.7, Problem 72

The function $f(x) = |x -3|$ 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.







By using the property of absolute value, $f(x) = |x - 3| \to f(x) = \begin{array}{cc} x - 3 & \text{for } x \geq 3 \\ -x + 3 & \text{for } x < 3 \end{array}$

If we restrict the domain for $x \geq 3$, the function is now one-to-one, to find its inverse, we set $y = f(x)$.


$\begin{equation} \begin{aligned} y =& x - 3 && \text{Solve for $x$; add } 3 \\ \\ x =& y + 3 && \text{Interchange $x$ and $y$} \\ \\ y =& x + 3 && \end{aligned} \end{equation}$



Thus, the inverse of $f(x) = |x - 3|$ for $x \geq 3$ is $f^{-1} (x) = x + 3$.