## Section5.10Inverse Functions

##### One-to-One Functions.

Suppose that $y$ is a function of $x$ and that the name of the function is $f\text{.}$ We say that $f$ is a one-to-one function if no two ordered pairs in the set that makes up the function share the same $y$-coordinate. Another way of putting it is that a function is one-to-one if and only every value of $y$ in the range of the function corresponds to exactly one value of $x$ in the domain of the function.

###### Example5.10.1.

Determine whether or note the functions described in Figure 5.10.2 and Figure 5.10.3 are one-to-one functions.

Solution

The function $f$ described in Figure 5.10.2 is not one-to-one. There are two different ordered pairs with $y$-coordinates equal to $6\text{.}$ The function $g$ described in Figure 5.10.3 is one-to-one. The $y$-coordinate of each ordered pair in the table is distinct from every other $y$-coordinate in the table.

###### Example5.10.4.

Determine whether or not the functions shown in Figure 5.10.5 and Figure 5.10.6 are one-to-one. Figure 5.10.5. $y=h(x)$ Figure 5.10.6. $y=k(x)$
Solution

The function $h$ shown in Figure 5.10.5 is one-to-one; no two points share the same $y$-coordinate. One way of visualizing this it to imagine passing a horizontal line over the entirety of the graph. At no place on the graph would the horizontal line intersect the function in two or points, so there are no two points that share the same $y$-coordinate.

On the other hand, performing the same horizontal line test on the function $k$ shown in Figure 5.10.6, we see that there are several locations where the horizontal line intersects the function in two points. For example, as shown in, Figure 5.10.7, the line $y=2$ intersects the function in two distinct points which demonstrates two or more points with the same $y$-coordinate, thereby establishing that the function, $k$ is not one-to-one. Figure 5.10.7. $y=k(x)$

In general, linear functions, square root functions, and other radical functions are one-to-one whereas absolute value functions and quadratic functions are not one-to-one (at least not without domain restrictions, which will be discusses later).

###### Example5.10.8.

Determine which of the following functions are one-to-one. If a given function is not one-to-one, state an explicit example of a pair of points that demonstrate it is not one-to-one.

1. $g(x)=6-11x$
2. $m(x)=7-x^2$
3. $f(x)=\abs{x-3}+4$
4. $h(x)=\sqrt{x+9}$
Solution
1. The function $g$ is a one-to-one linear function.

2. The function $m$ is not one-to-one. For example, both the points $(-3,-2)$ and $(3,-2)$ lie on a graph of the function.

3. The function $f$ is not one-to-one. For example, both the points $(-5,12)$ and $(11,12)$ lie on a graph of the function.

4. The function $h$ is a one-to-one radical function.

##### Inverse Functions.

One-to-one functions, and only one-to-one functions, are said to be invertible. Specifically, if the one-to-one function $f$ consists of the ordered pairs $\{(a,b)\}\text{,}$ then the inverse function of $f$ consists of the ordered pairs $\{(b,a)\}\text{.}$The inverse function of $f$ is denoted as $f^{-1}$ which is read aloud as "$f$ inverse." Please note that in this context the $-1$ is in no way acting as an exponent, That is,

\begin{equation*} f^{-1} \neq \frac{1}{f}. \end{equation*}
###### Example5.10.9.

Three function values for the one-to-one function $f$ are $f(9)=2\text{,}$ $f(3)=-5\text{,}$ and $f(1)=0\text{.}$ State three function values for $f^{-1}\text{.}$

Solution

From the given information we know that three ordered pairs in the set that makes up $f$ are $(9,2)\text{,}$ $(3,-5)\text{,}$ and $(1,0)\text{.}$ This implies that three ordered pairs in the set that makes up $f^{-1}$ are $(2,9)\text{,}$ $(-5,3)\text{,}$ and $(0,1)\text{.}$ Thus, three function values for $f^{-1}$ are $f^{-1}(2)=9\text{,}$ $f^{-1}(-5)=3\text{,}$ and $f^{-1}(0)=1\text{.}$

Recall that if $y=f(x)\text{,}$ then the set of all $x$-coordinates in the set that makes up $f$ is called the domain of $f$ and the set of all $y$-coordinates in the set that makes up $f$ is called the range of $f\text{.}$ Assuming that $f$ is a one-to-one function, all of the $x$-coordinates in $f$ become $y$-coordinates in $f^{-1}$ and all of the $y$-coordinates in $f$ become $x$-coordinates in $f^{1}\text{.}$

###### Example5.10.10.

Suppose that $f$ is a one-to-one function. Suppose further that the domain of $f$ is $(-\infty,6]$ and that the range of $f$ is $[-4,11]\text{.}$ What are the domain and range of $f^{-1}\text{?}$

Solution

The domain of $f^{-1}$ is $[-4,11]$ and the range of $f^{-1}$ is $(-\infty,6]\text{.}$

As discussed previously, if $f$ is a one-to-one function and $f(a)=b\text{,}$ it follows that $f^{-1}(b)=a\text{.}$ Putting these two function values together we get the following.

\begin{equation} b\xrightarrow{\,\,\,f^{-1}\,\,\,}a\xrightarrow{\,\,\,\,\,f\,\,\,\,\,}b\label{functions-inverses-figure-4}\tag{5.10.1} \end{equation}
\begin{align*} (f \circ f^{-1})(b)\amp=f(f^{-1}(b))\\ \amp=f(a)\\ \amp=b \end{align*}
\begin{equation} a\xrightarrow{\,\,\,\,\,f\,\,\,\,\,}b\xrightarrow{\,\,\,f^{-1}\,\,\,}a\label{functions-inverses-figure-5}\tag{5.10.2} \end{equation}
\begin{align*} (f^{-1} \circ f)(a)\amp=f^{-1}(f(a))\\ \amp=f^{-1}(b)\\ \amp=a \end{align*}

From these to observations, we infer the following.

The functions $f$ and $g$ are inverse functions if and only if:

\begin{equation*} (f \circ g)(x)=x\,\text{for all values of}\, x\,\text{in the domain of}\, g \end{equation*}

and

\begin{equation*} (g \circ f)(x)=x\,\text{for all values of}\, x\,\text{in the domain of}\, f. \end{equation*}
###### Example5.10.11.

Determine whether or not the functions $f(x)=\frac{1}{3}x-\frac{7}{6}$ and $g(x)=3x+\frac{7}{2}$ are inverses of one another.

Solution

We need to determine whether or not it is always the case that the expressions $(f \circ g)(x)$ and $(g \circ f)(x)$ both simplify to $x\text{.}$

\begin{align*} (f \circ g)(x)\amp=f(g(x))\\ \amp=f\left(3x+\frac{7}{2}\right)\\ \amp=\frac{1}{3}\left(3x+\frac{7}{2}\right)-\frac{7}{6}\\ \amp=x+\frac{7}{6}-\frac{7}{6}\\ \amp=x \end{align*}

and

\begin{align*} (g \circ f)(x)\amp=g(f(x))\\ \amp=g\left(\frac{1}{3}x-\frac{7}{6}\right)\\ \amp=3\left(\frac{1}{3}x-\frac{7}{6}\right)+\frac{7}{2}\\ \amp=x-\frac{7}{2}+\frac{7}{2}\\ \amp=x \end{align*}

Since it is indeed the case that $(f \circ g)(x)=x$ and $(g \circ f)(x)=x$ for all $x\text{,}$ $f$ and $g$ are inverses of one another.

###### Example5.10.12.

Determine whether or not the functions $f(x)=x^2+6$ and $g(x)=\sqrt{x-6}$ are inverses of one another.

Solution

We need to determine whether or not it is always the case that the expressions $(f \circ g)(x)$ and $(g \circ f)(x)$ both simplify to $x\text{.}$

\begin{align*} (f \circ g)(x)\amp=f(g(x))\\ \amp=f\left(\sqrt{x-6}\right)\\ \amp=\left(\sqrt{x-6}\right)^2+6\\ \amp=x-6+6;\,\, x \ge 6\\ \amp=x;\,\, x \ge 6 \end{align*}

and

\begin{align*} (g \circ f)(x)\amp=g(f(x))\\ \amp=g(x^2+6)\\ \amp=\sqrt{x^2+6-6}\\ \amp=\sqrt{x^2}\\ \amp=\abs{x} \end{align*}

Since $(g \circ f)(x)$ does not simplify to $x\text{,}$ $f$ and $g$ are not inverses of one another.

In the last example, we could have simply noted that $f$ is not a one-to-one function (e.g., $f(-1)=f(1))$) and, consequently, it is not the inverse of any function. However, with a modification called a restricted domain, we can make $f$ a one-to-one function and the inverse of $g$ as well. If we restrict the domain of $f$ to $[0,\infty)\text{,}$ then $f$ becomes one-to-one and the evaluation of $(g \circ f)(x)$ has one more line.

\begin{align*} (g \circ f)(x)\amp=g(f(x))\\ \amp=g(x^2+6);\,\,x \ge 0\\ \amp=\sqrt{x^2+6-6};\,\,x \ge 0\\ \amp=\sqrt{x^2};\,\,x \ge 0\\ \amp=\abs{x};\,\,x \ge 0\\ \amp=x;\,\,x \ge 0 \end{align*}

Now, because both $(f \circ g)(x)$ and $(g \circ f)(x)$ simplify to $x\text{,}$ $f$ and $g$ are inverse functions. That is, $f^{-1}(x)=g(x)$ and $g^{-1}(x)=f(x)\text{.}$ We don't need to worry about the fact that the domains are neither all real numbers near even the same. Remember, the domain of $f$ is the range of $f^{-1}$ and the domain of $f^{-1}$ is the range of $f\text{.}$ There's nothing that says the two domains need to be the same.

##### Determining the Formula of an Inverse Function.

We now turn our attention to the determination of the formulas for inverse functions. Let's assume that $y=f(x)$ is a one-to-one function. To determine the formula for the inverse of $f\text{,}$ we go through the following steps.

1. In lieu of writing $y=f(x)\text{,}$ we exchange the placements of $x$ and $y\text{,}$ that is we write $x=f(y)$ using the actual formula for $f\text{.}$

2. We then solve the resultant equation for $y\text{.}$ The formula on the other side of the equal sign from $y$ is the formula for $f^{-1}\text{.}$

###### Example5.10.13.

Determine the formula for $f^{-1}(x)$ given that $f(x)=5x+2\text{.}$

Solution

We begin by noting that the ordered pairs in $f$ satisfy the equation $y=5x+2\text{,}$ so the ordered pairs in $f^{-1}$ satisfy the equation $x=5y+2\text{.}$ We then isolate $y$ in the second equation.

\begin{align*} x\amp=5y+2\\ x\subtractright{2}\amp=5y+2\subtractright{2}\\ x-2\amp=5y\\ \divideunder{x-2}{5}\amp=\divideunder{5y}{5}\\ \frac{1}{5}x-\frac{2}{5}\amp=y \end{align*}

We conclude by stating that

\begin{equation*} f^{-1}(x)=\frac{1}{5}x-\frac{2}{5}\text{.} \end{equation*}

The reader should confirm that both $(f \circ f^{-1})(x)=x$ and $(f^{-1} \circ f)(x)=x\text{.}$

###### Example5.10.14.

Determine the formula for $g^{-1}(x)$ given that $g(x)=\sqrt{x+9}-2\text{.}$

Solution

We being by noting that the ordered pairs in $g$ satisfy the equation $y=\sqrt{x+9}-2\text{,}$ so the ordered pairs in $g^{-1}$ satify the equation $x=\sqrt{y+9}-2\text{.}$ We then isolate $y$ in the second equation.

We conclude by stating that $g^{-1}(x)=(x+2)^3-9\text{.}$ The reader should confirm that both $(g \circ g^{-1})(x)=x$ and $(g^{-1} \circ g)(x)=x\text{.}$

###### Example5.10.15.

Determine the formula for $h^{-1}(x)$ given that $h(x)=\frac{x+7}{x-3}\text{.}$

Solution

We begin by noting that the ordered pairs that make up $h$ all satisfy the equation $y=\frac{x+7}{x-3}\text{,}$ so the ordered pairs that make up $h^{-1}$ all satisfy the equation $x=\frac{y+7}{y-3}\text{.}$ We the isolate $y$ in the second equation.

\begin{align*} x\amp=\frac{y+7}{y-3}\\ \amp\highlight{\text{(we begin by clearing away the fraction)}}\\ x\multiplyright{(y-3)}\amp=\frac{y+7}{y-3}\multiplyright{(y-3)}\\ \amp\highlight{\text{(we go ahead and distribute on the left side of the equation)}}\\ xy-3x\amp=y+7\\ \amp\highlight{(\text{to isolate}\,y\,\text{we need all occurrences of}\,y\,\text{on the same side of}\,=)}\\ xy-3x\subtractright{y}\amp=y+7\subtractright{y}\\ xy-y-3x\amp=7\\ \amp\highlight{(\text{we need terms without factors of}\,y\,\text{on the other side of}\,=)}\\ xy-y-3x\addright{3x}\amp=7\addright{3x}\\ xy-y\amp=3x+7\\ \amp\highlight{(\text{we factor out}\,y\,\text{so that we know what to divide on the next line})}\\ y(x-1)\amp=3x+7\\ \divideunder{y(x-1)}{x-1}\amp=\divideunder{3x+7}{x-1}\\ y\amp=\frac{3x+7}{x-1} \end{align*}

That was pretty intense. Let's go ahead and check together.

\begin{align*} (h \circ h^{-1})(x)\amp=h(h^{-1}(x))\\ \amp=h\left(\frac{3x+7}{x-1}\right)\\ \amp=\frac{\frac{3x+7}{x-1}+7}{\frac{3x+7}{x-1}-3}\\ \amp=\frac{\frac{3x+7}{x-1}+7}{\frac{3x+7}{x-1}-3}\multiplyright{\frac{\frac{x-1}{1}}{\frac{x-1}{1}}}\\ \amp=\frac{3x+7+7(x-1)}{3x+7-3(x-1)}\\ \amp=\frac{3x+7+7x-7}{3x+7-3x+3}\\ \amp=\frac{10x}{10}\\ \amp=x \end{align*}

and

\begin{align*} (h^{-1} \circ h)(x)\amp=h^{-1}(h(x))\\ \amp=h^{-1}\left(\frac{x+7}{x-3}\right)\\ \amp=\frac{3 \cdot \frac{x+7}{x-3}+7}{\frac{x+7}{x-3}-1}\\ \amp=\frac{\frac{3x+21}{x-3}+7}{\frac{x+7}{x-3}-1}\multiplyright{\frac{\frac{x-3}{1}}{\frac{x-3}{1}}}\\ \amp=\frac{3x+21+7(x-3)}{x+7-(x-3)}\\ \amp=\frac{3x+21+7(x-3)}{x+7-(x-3)}\\ \amp=\frac{3x+21+7x-21}{x+7-x+3}\\ \amp=\frac{10x}{10}\\ \amp=x \end{align*}

So we can confidently conclude that $h^{-1}(x)=\frac{3x+7}{x-1}\text{.}$

##### Graphs of Inverse Functions.

When graphed, inverse functions display a special characteristic.

Suppose that two of the points of the function $f$ are $(-5,2)$ and $(3,6)\text{.}$ Then two of the points on the function $f^{-1}$ are $(2,-5)$ and $(6,3)\text{.}$ All four of the points are shown in Figure.

The points with interchanged coordinates have been connected with line segments. Note that both of these line segments are perpendicular to the line $y=x$ and that the endpoints of thee line segment are equidistant from the line $y=x\text{.}$ We refer this phenomenon as symmetry across the line $y=x$. When graphed on the same grid, inverse function are always symmetric across the line $y=x\text{.}$ Figure 5.10.16. Symmetry across $y=x$
###### Example5.10.17.

A function $g$ is shown in Figure. Sketch the graph of $g^{-1}\text{.}$ Figure 5.10.18. $y=g(x)$

Solution

We begin by noting several points on $g$ and the implied points on $g^{-1}\text{.}$ These are shown in Figure and Figure.

We the plot the know points on $g^{-1}$ and the line $y=x\text{.}$ Finally we connect the dots and extend the curve keeping in mind the need to maintain symmetry across the line $y=x\text{.}$ This is all shown in Figure. Figure 5.10.21. $\highlightr{y=g(x)}$ and $\highlight{y=g^{-1}(x)}$

### ExercisesExercises

Verify that each stated pair of functions are in fact inverse functions by showing that both $(f \circ f^{-1})(x)$ and $(f^{-1} \circ f)(x)$ simplify to $x\text{.}$

###### 1.

$f(x)=-\frac{1}{8}x+3$ and $f^{-1}(x)=-8x+24$

Solution
\begin{align*} (f \circ f^{-1})(x)\amp=f(f^{-1}(x))\\ \amp=f(-8x+24)\\ \amp=-\frac{1}{8}(-8x+24)+3\\ \amp=x-3+3\\ \amp=x \end{align*}

and

\begin{align*} (f^{-1} \circ f)(x)\amp=f^{-1}(f(x))\\ \amp=f^{-1}\left(-\frac{1}{8}x+3\right)\\ \amp=-8\left(-\frac{1}{8}x+3\right)+24\\ \amp=x-24+24\\ \amp=x \end{align*}
###### 2.

$f(x)=\frac{3-x}{7}$ and $f^{-1}(x)=3-7x$

Solution
\begin{align*} (f \circ f^{-1})(x)\amp=f(f^{-1}(x))\\ \amp=f(3-7x)\\ \amp=\frac{3-(3-7x)}{7}\\ \amp=\frac{3-3+7x}{7}\\ \amp=\frac{7x}{7}\\ \amp=x \end{align*}

and

\begin{align*} (f^{-1} \circ f)(x)\amp=f^{-1}(f(x))\\ \amp=f^{-1}\left(\frac{3-x}{7}\right)\\ \amp=3-7\left(\frac{3-x}{7}\right)\\ \amp=3-(3-x)\\ \amp=3-3+x\\ \amp=x \end{align*}
###### 3.

$f(x)=9(x-5)^2-8;\,\,x \ge 5$ and $f^{-1}(x)=\frac{1}{3}\sqrt{x+8}+5$

Solution
\begin{align*} (f \circ f^{-1})(x)\amp=f(f^{-1}(x))\\ \amp=f\left(\frac{1}{3}\sqrt{x+8}+5\right)\\ \amp=9\left(\frac{1}{3}\sqrt{x+8}+5-5\right)^2-8\\ \amp=9\left(\frac{1}{3}\sqrt{x+8}\right)^2-8\\ \amp=9 \cdot \frac{1}{9}(x+8)-8;;\,\,x \ge -8\\ \amp=x+8-8;\,\,x \ge -8\\ \amp=x;\,\,x \ge -8 \end{align*}

and

\begin{align*} (f^{-1} \circ f)(x)\amp=f^{-1}(f(x))\\ \amp=f^{-1}\left(9(x-5)^2-8\right);\,\,x \ge 5\\ \amp=\frac{1}{3}\sqrt{9(x-5)^2-8+8}+5;\,\,x \ge 5\\ \amp=\frac{1}{3}\sqrt{9(x-5)^2}+5;\,\,x \ge 5\\ \amp=\frac{1}{3} \cdot 3(x-5)+5;\,\,x \ge 5\\ \amp=x-5+5;\,\,x \ge 5\\ \amp=x;\,\,x \ge 5 \end{align*}
###### 4.

$f(x)=11-\sqrt{6-2x}$ and $f^{-1}(x)=-\frac{1}{2}(11-x)^7+3$

Solution
\begin{align*} (f \circ f^{-1})(x)\amp=f(f^{-1}(x))\\ \amp=f\left(-\frac{1}{2}(11-x)^7+3\right)\\ \amp=11-\sqrt{6-2\left(-\frac{1}{2}(11-x)^7+3\right)}\\ \amp=11-\sqrt{6+(11-x)^7-6}\\ \amp=11-\sqrt{(11-x)^7}\\ \amp=11-(11-x)\\ \amp=11-11+x\\ \amp=x \end{align*}

and

\begin{align*} (f^{-1} \circ f)(x)\amp=f^{-1}(f(x))\\ \amp=f^{-1}\left(11-\sqrt{6-2x}\right)\\ \amp=-\frac{1}{2}\left[11-\left(11-\sqrt{6-2x}\right)\right]^7+3\\ \amp=-\frac{1}{2}\left[11-11+\sqrt{6-2x}\right]^7+3\\ \amp=-\frac{1}{2}\left(\sqrt{6-2x}\right)^7+3\\ \amp=-\frac{1}{2}(6-2x)+3\\ \amp=-3+x+3\\ \amp=x \end{align*}
###### 5.

$f(x)=\frac{x-7}{x}$ and $f^{-1}(x)=\frac{7}{1-x}$

Solution
\begin{align*} (f \circ f^{-1})(x)\amp=f(f^{-1}(x))\\ \amp=f\left(-\frac{7}{1-x}\right)\\ \amp=\frac{\frac{7}{1-x}-7}{\frac{7}{1-x}}\\ \amp=\frac{\frac{7}{1-x}-7}{\frac{7}{1-x}}\multiplyright{\frac{1-x}{1-x}}\\ \amp=\frac{7-7(1-x)}{7}\\ \amp=\frac{7-7+7x}{7}\\ \amp=\frac{7x}{7}\\ \amp=x \end{align*}

and

\begin{align*} (f^{-1} \circ f)(x)\amp=f^{-1}(f(x))\\ \amp=f^{-1}\left(\frac{x-7}{x}\right)\\ \amp=\frac{7}{1-\frac{x-7}{x}}\\ \amp=\frac{7}{1-\frac{x-7}{x}}\multiplyright{\frac{x}{x}}\\ \amp=\frac{7x}{x-(x-7)}\\ \amp=\frac{7x}{x-x+7}\\ \amp=\frac{7x}{7}\\ \amp=x \end{align*}
###### 6.

$f(x)=\frac{x+3}{2-x}$ and $f^{-1}(x)=\frac{2x-3}{x+1}$

Solution
\begin{align*} (f \circ f^{-1})(x)\amp=f(f^{-1}(x))\\ \amp=f\left(\frac{2x-3}{x+1}\right)\\ \amp=\frac{\frac{2x-3}{x+1}+3}{2-\frac{2x-3}{x+1}}\\ \amp=\frac{\frac{2x-3}{x+1}+3}{2-\frac{2x-3}{x+1}}\multiplyright{\frac{x+1}{x+1}}\\ \amp=\frac{2x-3+3(x+1)}{2(x+1)-(2x-3)}\\ \amp=\frac{2x-3+3x+3}{2x+2-2x+3}\\ \amp=\frac{5x}{5}\\ \amp=x \end{align*}

and

\begin{align*} (f^{-1} \circ f)(x)\amp=f^{-1}(f(x))\\ \amp=f^{-1}\left(\frac{x+3}{2-x}\right)\\ \amp=\frac{2 \cdot \frac{x+3}{2-x}-3}{\frac{x+3}{2-x}+1}\\ \amp=\frac{2 \cdot \frac{x+3}{2-x}-3}{\frac{x+3}{2-x}+1}\multiplyright{\frac{2-x}{2-x}}\\ \amp=\frac{2(x+3)-3(2-x)}{x+3+2-x}\\ \amp=\frac{2x+6-6+3x}{5}\\ \amp=\frac{5x}{5}\\ \amp=x \end{align*}

For each stated function, determine the formula for $f^{-1}(x)\text{.}$

###### 7.

$f(x)=12x-8$

Solution

We begin by noting that the ordered pairs in $f$ satisfy the equation $y=12x-8\text{.}$ It follows that the ordered pairs in $f^{-1}$ satisfy the equation $x=12y-8\text{.}$ This latter equation needs to be solved for $y\text{.}$

We conclude that $f^{-1}(x)=\frac{1}{12}x+\frac{2}{3}$ (after checking the answer, of course).

###### 8.

$f(x)=-\frac{2}{9}x+\frac{4}{9}$

Solution

We begin by noting that the ordered pairs in $f$ satisfy the equation $y=-\frac{2}{9}x+\frac{4}{9}\text{.}$ It follows that the ordered pairs in $f^{-1}$ satisfy the equation $x=-\frac{2}{9}y+\frac{4}{9}\text{.}$ This latter equation needs to be solved for $y\text{.}$

\begin{align*} x\amp=-\frac{2}{9}y+\frac{4}{9}\\ x\subtractright{\frac{4}{9}}\amp=-\frac{2}{9}y+\frac{4}{9}\subtractright{\frac{4}{9}}\\ x-\frac{4}{9}\amp=-\frac{2}{9}y\\ \multiplyleft{-\frac{9}{2}}\left(x-\frac{4}{9}\right)\amp=\multiplyleft{-\frac{9}{2}}-\frac{2}{9}y\\ -\frac{9}{2}x+2\amp=y \end{align*}

We conclude that $f^{-1}(x)=-\frac{9}{2}x+2$ (after checking the answer, of course :).

###### 9.

$f(x)=8-\sqrt{2x+8}$

Solution

We begin by noting that the ordered pairs in $f$ satisfy the equation $y=8-\sqrt{2x+8}\text{.}$ It follows that the ordered pairs in $f^{-1}$ satisfy the equation $x=8-\sqrt{2y+8}\text{.}$ This latter equation needs to be solved for $y\text{.}$

\begin{align*} x\amp=8-\sqrt{2y+8}\\ x\subtractright{8}\amp=8-\sqrt{2y+8}\subtractright{8}\\ x-8\amp=-\sqrt{2y+8}\\ \multiplyleft{-1}(x-8)\amp=\multiplyleft{-1}-\sqrt{2y+8}\\ -x+8\amp=\sqrt{2y+8}\\ (-x+8)^3\amp=\left(\sqrt{2y+8}\right)^3\\ (-x+8)^3\amp=2y+8\\ (-x+8)^3\subtractright{8}\amp=2y+8\subtractright{8}\\ (-x+8)^3-8\amp=2y\\ \divideunder{(-x+8)^3-8}{2}\amp=\divideunder{2y}{2}\\ \frac{1}{2}(-x+8)^3-4\amp=y \end{align*}

We conclude that $f^{-1}(x)=\frac{1}{2}(-x+8)^3-4$ (once we've checked, that is).

###### 10.

$f(x)=2\sqrt{11-x}+22$

Solution

We begin by noting that the ordered pairs in $f$ satisfy the equation $y=2\sqrt{11-x}+22\text{.}$ It follows that the ordered pairs in $f^{-1}$ satisfy the equation $x=2\sqrt{11-y}+22\text{.}$ This latter equation needs to be solved for $y\text{.}$

\begin{align*} x\amp=2\sqrt{11-y}+22\\ x\subtractright{22}\amp=2\sqrt{11-y}+22\subtractright{22}\\ x-22\amp=2\sqrt{11-y}\\ \divideunder{x-22}{2}\amp=\divideunder{2\sqrt{11-y}}{2}\\ \frac{1}{2}x-11\amp=\sqrt{11-y}\\ \left(\frac{1}{2}x-11\right)^5\amp=\left(\sqrt{11-y}\right)^5\\ \left(\frac{1}{2}x-11\right)^5\amp=11-y\\ \left(\frac{1}{2}x-11\right)^5\subtractright{11}\amp=11-y\subtractright{11}\\ \left(\frac{1}{2}x-11\right)^5-11\amp=-y\\ \multiplyleft{-1}\left[\left(\frac{1}{2}x-11\right)^5-11\right]\amp=\multiplyleft{-1}-y\\ -\left(\frac{1}{2}x-11\right)^5+11\amp=y \end{align*}

We conclude that $f^{-1}(x)=-\left(\frac{1}{2}x-11\right)^5+11$ (once we've thoroughly checked).

###### 11.

$f(x)=\frac{x}{x-12}$

Solution

We begin by noting that the ordered pairs in $f$ satisfy the equation $y=\frac{x}{x-12}\text{.}$ It follows that the ordered pairs in $f^{-1}$ satisfy the equation $x=\frac{y}{y-12}\text{.}$ This latter equation needs to be solved for $y\text{.}$

\begin{align*} x\amp=\frac{y}{y-12}\\ x\multiplyright{(y-12)}\amp=\frac{y}{y-12}\multiplyright{(y-12)}\\ xy-12x\amp=y\\ xy-12x\subtractright{y}\amp=y\subtractright{y}\\ xy-12x-y\amp=0\\ xy-12x-y\addright{12x}\amp=0\addright{12x}\\ xy-y\amp=12x\\ y(x-1)\amp=12x\\ \divideunder{y(x-1)}{x-1}\amp=\divideunder{12x}{x-1}\\ y\amp=\frac{12x}{x-1} \end{align*}

We conclude that $f^{-1}(x)=\frac{12x}{x-1}$ (upon check, of course.).

###### 12.

$f(x)=\frac{2x+3}{4-x}$

Solution

We begin by noting that the ordered pairs in $f$ satisfy the equation $y=\frac{2x+3}{4-x}\text{.}$ It follows that the ordered pairs in $f^{-1}$ satisfy the equation $x=\frac{2y+3}{4-y}\text{.}$ This latter equation needs to be solved for $y\text{.}$

\begin{align*} x\amp=\frac{2y+3}{4-y}\\ x\multiplyright{(4-y)}\amp=\frac{2y+3}{4-y}\multiplyright{(4-y)}\\ 4x-xy\amp=2y+3\\ 4x-xy\addright{xy}\amp=2y+3\addright{xy}\\ 4x\amp=2y+3+xy\\ 4x\subtractright{3}\amp=2y+3+xy\subtractright{3}\\ 4x-3\amp=2y+xy\\ 4x-3\amp=y(x+2)\\ \divideunder{4x-3}{x+2}\amp=\divideunder{y(x+2)}{x+2}\\ \frac{4x-3}{x+2}\amp=y \end{align*}

We conclude that $f^{-1}(x)=\frac{4x-3}{x+2}$ (oh yeah, I checked, as should you :).

Draw the inverse of each graphed function.

###### 13. Figure 5.10.22. $\highlightr{y=f(x)}$
Solution Figure 5.10.23. $\highlightr{y=f(x)}$ and $\highlight{y=f^{-1}(x)}$
###### 14. Figure 5.10.24. $\highlightr{y=g(x)}$
Solution Figure 5.10.25. $\highlightr{y=g(x)}$ and $\highlight{y=g^{-1}(x)}$
###### 15. Figure 5.10.26. $\highlightr{y=k(x)}$
Solution Figure 5.10.27. $\highlightr{y=k(x)}$ and $\highlight{y=k^{-1}(x)}$