5.7 Inverses and radical functions  (Page 3/7)

The original function $f(x)={(x-4)}^2$ is not one-to-one, but the function is restricted to a domain of $x\ge 4$ or $x\le 4$ on which it is one-to-one. See [link] .

To find the inverse, start by replacing $f(x)$ with the simple variable $y.$

This is not a function as written. We need to examine the restrictions on the domain of the original function to determine the inverse. Since we reversed the roles of $x$ and $y$ for the original $f(x),$ we looked at the domain: the values $x$ could assume. When we reversed the roles of $x$ and $y,$ this gave us the values $y$ could assume. For this function, $x\ge 4,$ so for the inverse, we should have $y\ge 4,$ which is what our inverse function gives.

1. The domain of the original function was restricted to $x\ge 4,$ so the outputs of the inverse need to be the same, $f\left(x\right)\ge 4,$ and we must use the + case:
   $$f^{-1}(x)=4+\sqrt{x}$$
2. The domain of the original function was restricted to $x\le 4,$ so the outputs of the inverse need to be the same, $f\left(x\right)\le 4,$ and we must use the – case:
   $$f^{-1}(x)=4-\sqrt{x}$$

We can see this is a parabola with vertex at $(2,\mathrm{–3})$ 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 $x\ge 2.$

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 $x.$

Now we need to determine which case to use. Because we restricted our original function to a domain of $x\ge 2,$ the outputs of the inverse should be the same, telling us to utilize the + case

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.