Example 4 from Algebra and Trigonometry, 5.7 Inverses and Radical Functions

Finding the Inverse of a Quadratic Function When the Restriction Is Not Specified

Restrict the domain and then find the inverse of

f (x) = {(x - 2)}^{2} - 3.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
We can see this is a parabola with vertex at $(2, –3)$ that opens upward. Because the graph will be decreasing on one side of the vertex and increasing on the other side, we can restrict this function to a domain on which it will be one-to-one by limiting the domain to $x \geq 2.$

To find the inverse, we will use the vertex form of the quadratic. We start by replacing $f (x)$ with a simple variable, $y,$ then solve for $x .$

$\begin{matrix} y & = & {(x - 2)}^{2} - 3 & Interchange x and y . \\ x & = & {(y - 2)}^{2} - 3 & Add 3 to both sides . \\ x + 3 & = & {(y - 2)}^{2} & Take the square root . \\ \pm \sqrt{x + 3} & = & y - 2 & Add 2 to both sides . \\ 2 \pm \sqrt{x + 3} & = & y & Rename the function . \\ f^{- 1} (x) & = & 2 \pm \sqrt{x + 3} \end{matrix}$
Now we need to determine which case to use. Because we restricted our original function to a domain of $x \geq 2,$ the outputs of the inverse should be the same, telling us to utilize the + case

$f^{- 1} (x) = 2 + \sqrt{x + 3}$
If the quadratic had not been given in vertex form, rewriting it into vertex form would be the first step. This way we may easily observe the coordinates of the vertex to help us restrict the domain.

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Find the inverse of the function $f (x) = x^{2} + 1,$ on the domain $x \geq 0.$

How did it go?