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

To find the inverse, start by replacing [latex]f\left(x\right)[/latex] with the simple variable [latex]y[/latex].

[latex]\begin{align}y&={\left(x - 4\right)}^{2} &&\text{Interchange } x \text{ and }y. \\[1mm] x&={\left(y - 4\right)}^{2} &&\text{Take the square root}. \\[1mm] \pm \sqrt{x}&=y - 4 &&\text{Add } 4 \text{ to both sides}. \\[1mm] 4\pm \sqrt{x}&=y \end{align}[/latex]

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 [latex]x[/latex] and [latex]y[/latex] for the original [latex]f(x)[/latex], we looked at the domain: the values [latex]x[/latex] could assume. When we reversed the roles of [latex]x[/latex] and [latex]y[/latex], this gave us the values [latex]y[/latex] could assume. For this function, [latex]x\ge 4[/latex], so for the inverse, we should have [latex]y\ge 4[/latex], which is what our inverse function gives.

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

Analysis of the Solution

On the graphs below, we see the original function graphed on the same set of axes as its inverse function. Notice that together the graphs show symmetry about the line [latex]y=x[/latex]. The coordinate pair [latex]\left(4, 0\right)[/latex] is on the graph of [latex]f[/latex] and the coordinate pair [latex]\left(0, 4\right)[/latex] is on the graph of [latex]{f}^{-1}[/latex]. For any coordinate pair, if [latex](a,b)[/latex] is on the graph of [latex]f[/latex], then [latex](b,a)[/latex] is on the graph of [latex]{f}^{-1}[/latex]. Finally, observe that the graph of [latex]f[/latex] intersects the graph of [latex]{f}^{-1}[/latex] on the line [latex]y=x[/latex]. Points of intersection for the graphs of [latex]f[/latex] and [latex]{f}^{-1}[/latex] will always lie on the line [latex]y=x[/latex].

We can see this is a parabola with vertex at [latex]\left(2, -3\right)[/latex] 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 [latex]x\ge 2[/latex].

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

[latex]\begin{align}y&={\left(x - 2\right)}^{2}-3 && \text{Interchange } x \text{ and } y. \\[1mm] x&={\left(y - 2\right)}^{2}-3 && \text{Add 3 to both sides}. \\[1mm] x+3&={\left(y - 2\right)}^{2} && \text{Take the square root}. \\[1mm] \pm \sqrt{x+3}&=y - 2 && \text{Add 2 to both sides}. \\[1mm] 2\pm \sqrt{x+3}&=y && \text{Rename the function}. \\[3mm] {f}^{-1}\left(x\right)&=2\pm \sqrt{x+3} \end{align}[/latex]

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

[latex]{f}^{-1}\left(x\right)=2+\sqrt{x+3}[/latex]

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.

Analysis of the Solution

Notice that we arbitrarily decided to restrict the domain on [latex]x\ge 2[/latex]. We could just have easily opted to restrict the domain on [latex]x\le 2[/latex], in which case [latex]{f}^{-1}\left(x\right)=2-\sqrt{x+3}[/latex]. Observe the original function graphed on the same set of axes as its inverse function in the graph below. Notice that both graphs show symmetry about the line [latex]y=x[/latex]. The coordinate pair [latex]\left(2,\text{ }-3\right)[/latex] is on the graph of [latex]f[/latex] and the coordinate pair [latex]\left(-3,\text{ }2\right)[/latex] is on the graph of [latex]{f}^{-1}[/latex]. Observe from the graph of both functions on the same set of axes that

[latex]\text{domain of }f=\text{range of } {f}^{-1}=\left[2,\infty \right)[/latex]

and

[latex]\text{domain of }{f}^{-1}=\text{range of } f=\left[-3,\infty \right)[/latex]

Finally, observe that the graph of [latex]f[/latex] intersects the graph of [latex]{f}^{-1}[/latex] along the line [latex]y=x[/latex].

Note that the original function has range [latex]f\left(x\right)\ge 0[/latex]. Replace [latex]f\left(x\right)[/latex] with [latex]y[/latex], then solve for [latex]x[/latex].

[latex]\begin{align} y&=\sqrt{x - 4} && \text{Replace}f\left(x\right)\text{with }y.\\[1mm] x& =\sqrt{y - 4}&& \text{Interchange }x\text{ and }y. \\[1mm] x& =\sqrt{y - 4}&& \text{Square each side}. \\[1mm] {x}^{2}& =y - 4&& \text{Add 4}. \\[1mm] {x}^{2}+4&=y&& \text{Rename the function }{f}^{-1}\left(x\right). \\[4mm] {f}^{-1}\left(x\right)& ={x}^{2}+4 \end{align}[/latex]

Recall that the domain of this function must be limited to the range of the original function.

[latex]{f}^{-1}\left(x\right)={x}^{2}+4,x\ge 0[/latex]

Analysis of the Solution

Notice in the graph below that the inverse is a reflection of the original function over the line [latex]y=x[/latex]. Because the original function has only positive outputs, the inverse function has only positive inputs.