If a point $\,(a,b)\,$ is on the graph of a onetoone function $\,f\,$,
then the point $\,(b,a)\,$ is on the graph of its inverse $\,f^{1}\,$.
This is because the input/output roles for a function and its inverse are switched—an input to one is an output from the other.
As discussed below, the point $\,(b,a)\,$ is found by reflecting the point $\,(a,b)\,$
about the line $\,y = x\,$,
when graphed in a coordinate system where
‘$\,1\,$’ on the $\,x\,$axis is the same as
‘$\,1\,$’ on the $\,y\,$axis.
From this observation, we will see below
that the graph of an inverse function is very easy to obtain
if we have the
graph of the original function—just reflect the original graph about the line
$\,y = x\,$.

By definition, the graph of a function is the set of all its input/output pairs,
as the inputs vary over the domain.
Thus,
$$\text{the graph of } f^{1} = \{(x,f^{1}(x))\ \ x\in\text{dom}(f^{1})\}\,$$
Since $\,x\,$ is a dummy variable, this set can also be written using the dummy variable $\,y\,$:
$$\text{the graph of } f^{1} = \{(y,f^{1}(y))\ \ y\in\text{dom}(f^{1})\}\,$$
Now, we have:
the graph of $\,f^{1}\,$  
$= \{(y,f^{1}(y))\ \ y\in\text{dom}(f^{1})\}$  definition of the graph of a function 
$= \{(y,f^{1}(y))\ \ y\in\text{ran}(f)\}$  since $\text{ran}(f) = \text{dom}(f^{1})$ 
$= \{(f(x),x)\ \ x\in\text{dom}(f)\}$  $y\in\text{ran}(f)\,$ if and only if there exists $\,x\in\text{dom}(f)\,$ such that $\,f(x) = y\,$; and, $\,f(x) = y\,$ if and only if $x = f^{1}(y)$ 
Thus, the graph of $\,f^{1}\,$ is precisely the set of points that make up the graph of $\,f\,$, but with
the coordinates switched!
We've already seen that ‘switching coordinates’ is accomplished by
reflecting about the line $\,y = x\,$
in a coordinate system where the scales on the horizontal and vertical axes are
identical.
Thus, we have:
Several different onetoone functions and their reflections in the line $\,y = x\,$ are shown below.
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
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