Exercise 21 Page 578

Before we can find the inverse of the given function, we need to replace $f(x)$ with $y.$ Because the range of $f$ is $y\ge 0$, the domain of the inverse has to be restricted to $x\ge 0.$ To algebraically determine the inverse of the given relation, we exchange $x$ and $y$ and solve for $y.$ The result of isolating $y$ in the new equation will be the inverse of the given function. Now we have the inverse of the given function.

Graphing the Function

To graph the given function let's first determine its domain. To do so, recall that the radicand of a square root is always greater than or equal to $0.$ Therefore, the domain of the given function is all real numbers greater than or equal to $\text{-}3.$ With this in mind, we will make a table of values to graph the function.

$x$	$f(x)=\frac{1}{2}\sqrt{2x+6}$	$f(x)$	Point
$x=\text{-}3$	$f(x)=\frac{1}{2}\sqrt{2(\text{-}3)+6}$	$f(x)=0$	$(\text{-}3,0)$
$x=\text{-}1$	$f(x)=\frac{1}{2}\sqrt{2(\text{-}1)+6}$	$f(x)=1$	$(\text{-}1,1)$
$x=5$	$f(x)=\frac{1}{2}\sqrt{2(5)+6}$	$f(x)=2$	$(5,2)$
$x=15$	$f(x)=\frac{1}{2}\sqrt{2(15)+6}$	$f(x)=3$	$(15,3)$

Let's plot the points and connect them with a smooth curve.

Graphing the Inverse of the Function

Finally, we can graph the inverse of the function by reflecting the graph of the given function across $y=x.$ This means that we should interchange the $x$ and $y$ coordinates of the points that are on the graph.

Points	Reflection across $y=x$
$(\text{-}3,0)$	$(0,\text{-}3)$
$(\text{-}1,1)$	$(1,\text{-}1)$
$(5,2)$	$(2,5)$
$(15,3)$	$(3,15)$

Now that we have the points, let's graph the function!