Section 2.2 Brief Intro to Composite and Inverse Functions

Supplemental Videos.

The main topics of this section are also presented in the following videos:

Subsection 2.2.1 Composition of Functions

We encounter composite functions in the real world every day. As an example, suppose you and some friends are driving from Lincoln, NE to Omaha, NE. Once you arrive in Omaha you drive around the city to see all the beautiful sites. We know that Omaha is 60 miles away from Lincoln, and for each hour spent in Omaha, you drive an additional 5 miles. Then the function

\begin{equation*} m(x)=60+5x \end{equation*}

represents the miles traveled if you and your friends spend $x$ hours in Omaha.

Now, suppose gas for your vehicle costs $0.15 per mile. We can represent this knowledge with the function

\begin{equation*} c(m)=.15m \end{equation*}

where $m$ is in miles. If we want to know how much you and your friends will spend on gas during your trip if you spend $x$ hours in Omaha, we can now use these two functions we created. If you spend $x$ hours in Omaha, you will have traveled $60+5x$ miles. The cost of gas is $.15m$ where $m$ is miles. Then we need only replace $m$ by $60+5x$ to get

\begin{equation*} .15(60+5x) \end{equation*}

as the cost of gas if you spend $x$ hours in Omaha.

In function notation, we write this as follows:

\begin{equation*} c(f(x))=.15f(x)=.15(60+5x). \end{equation*}

Here we are evaluating the function $c(m)$ at the value $f(x)$ since $f(x)$ is the number of miles traveled. The function $c(f(x))$ is called a composite function , or a composition of functions.

Example 2.11.

Given the functions $f(x)=x^2-2x $ and $g(x)=.5 x\text{,}$ identify the function $f(g(x))$ and find the value of $f(g(1))\text{.}$

Solution.

First, let’s find the function $f(g(x))\text{.}$ To identify this function, we are plugging in $.5x$ for $x$ in the function $f(x)=x^2-2x\text{.}$ We get the function

\begin{equation*} f(g(x))=f(.5x)=(.5x)^2-2(.5x)=.25x^2-x. \end{equation*}

Now, to find the value $f(g(1))$ we simply evaluate the function $f(g(x))$ at $1\text{.}$ We have

\begin{equation*} f(g(1))=.25(1)^2-1=.25-1=-.75. \end{equation*}

Looking back at our first example, from the functions $m(x)=60+5x$ and $c(m)=.15m$ we created the new function

\begin{equation*} c(m(x))=.15(60+5x). \end{equation*}

Notice that this composition of functions, $c(m(x))\text{,}$ has time in hours as its input and cost as its output.

In general, given two functions $f(x)$ and $g(x)\text{,}$ the composition $f(g(x))$ has the same input as $g \text{,}$ but has the same output as $f \text{.}$

Subsection 2.2.2 Inverse Functions

When you buy a house, your monthly mortgage payment is a function of the size of the loan. Table 2.12 shows mortgage payments on $30$-year loans of various sizes at $6$% interest.

Table 2.12.

Loan amount, $L$	$150,000$	$175,000$	$200,000$	$225,000$	$250,000$
Mortgage payment, $M$	$899.33$	$1049.21$	$1199.10$	$1348.99$	$1498.88$

For the function $M = f (L)\text{,}$ the input value is the amount of the loan, and the output is the mortgage payment.

However, when you are shopping for a house, you may think of the mortgage payment as the input variable: If you can afford a certain monthly mortgage payment, how large a loan can you finance? Now the mortgage payment is the input value, and the loan amount is the output. By interchanging the inputs and outputs, we define a new function, $L = g(M)\text{,}$ shown in Table 2.13.

Table 2.13.

Mortgage payment, $M$	$899.33$	$1049.21$	$1199.10$	$1348.99$	$1498.88$
Loan amount, $L$	$150,000$	$175,000$	$200,000$	$225,000$	$250,000$

This new function gives the same information as the original function, $f\text{,}$ but from a different point of view. We call the function $g$ the inverse function for $f\text{.}$ The elements of the range of $f$ are used as the input values for $g\text{,}$ and the output values of $g$ are the corresponding domain elements of $f\text{.}$ For example, from the tables you can verify that $f (200,000) = 1199.10\text{,}$ and $g(1199.10) = 200,000\text{.}$ In fact, this property defines the inverse function.

Inverse Functions.

Suppose $g$ is the inverse function for $f\text{.}$ Then

\begin{equation*} g(b) = a \text{ if and only if }f(a) = b. \end{equation*}

Example 2.14.

Suppose $g$ is the inverse function for $f\text{,}$ and we know the following function values for $f\text{:}$

\begin{equation*} f (-3) = 5, ~~ f (2) = 1, ~~ f (5) = 0. \end{equation*}

Find $g(5)$ and $g(0)\text{.}$

Solution.

We know that $g(5) = -3$ because $f (-3) = 5\text{,}$ and $g(0) = 5$ because $f (5) = 0\text{.}$ Tables may be helpful in visualizing the two functions, as shown below.

$y=f(x)$
$x$	$y$
$-3$	$5$
$2$	$1$
$5$	$0$

→ Interchange the columns →

$x=g(y)$
$y$	$x$
$5$	$-3$
$1$	$2$
$0$	$5$

For the function $f\text{,}$ the input variable is $x$ and the output variable is $y\text{.}$ For the inverse function $g\text{,}$ the roles of the variables are interchanged: $y$ is now the input and $x$ is the output.

Checkpoint 2.15.

Suppose $g$ is the inverse function for $f\text{,}$ and suppose we know the following function values for f :

\begin{equation*} f (-1) = 0,~~ f (0) = 1, ~~ f (1) = 2. \end{equation*}

Find $g(0)$ and $g(1)\text{.}$

Subsection 2.2.3 Finding a Formula for the Inverse Function

If a function is given by a table of values, we can interchange the columns (or rows) of the table to obtain the inverse function. Swapping the columns works because we are really interchanging the input and output variables. If a function is defined by an equation, we can find a formula for its inverse function in the same way: Interchange the roles of the variables in the equation so that the old output variable becomes the new input variable.

Example 2.16.

The function $H = f (t) = 6 + 2t$ gives the height of corn seedlings, in inches, $t$ days after they are planted. Find a formula for the inverse function and explain its meaning in this context.
Make a table of values for $f (t)$ and a table for its inverse function.

Solution.

Write the equation for $f$ in the form

\begin{equation*} H = 6 + 2t. \end{equation*}

In this equation, $t$ is the input and $H$ is the output. We interchange the roles of the variables by solving for $t$ to obtain

\begin{equation*} t = \frac{H - 6}{2}. \end{equation*}

In this equation, $H$ is the input and $t$ is the output. The formula for the inverse function is $t=g(H) = \dfrac{H - 6}{2}\text{.}$ The function $g$ gives the number of days it will take the corn seedlings to grow to a height of $H$ inches.
To make a table for $f\text{,}$ we choose values for $t$ and evaluate $f (t) = 6 + 2t$ at those $t$-values, as shown in Table 2.17.

$H=f(t)$

$t$ $H$

$0$ $6$

$1$ $8$

$2$ $10$

$3$ $12$

Table 2.17.

$t=g(H)$

$H$ $t$

$6$ $0$

$8$ $1$

$10$ $2$

$12$ $3$

Table 2.18.

To make a table for $g\text{,}$ we could choose values for $H$ and evaluate $\dfrac{H-6}{2}\text{,}$ but because $g$ is the inverse function for $f\text{,}$ we can simply interchange the columns in our table for $f\text{,}$ as shown in Table 2.18.

You can check that the values in the second table do satisfy the formula for the inverse function, $g(H)\text{.}$ Note once again that the two tables show the same relationship between $t$ and $H\text{,}$ but the roles of input and output have been interchanged. The function $f$ tells us the height of the seedlings after $t$ days, and $g$ tells us how long it will take the seedlings to grow to height $H\text{.}$

Example 2.19.

Carol can burn $600$ calories per hour bicycling and $400$ calories per hour swimming. She would like to lose $5$ pounds, which is equivalent to $16,000$ calories.

Write an equation relating the number of hours of cycling, $x\text{,}$ and the number of hours swimming, $y\text{,}$ that Carol must spend to lose $5$ pounds.
Write $y$ as a function of $x\text{,}$ $y = f (x)\text{.}$ What does $f (10)$ tell you?
Find the inverse function, $x = g(y)\text{.}$ What does $g(10)$ tell you?

Solution.

$\displaystyle 600x + 400y = 16,000$
$y = f (x) = 40 - 1.5x\text{;}$ $f (10) = 25\text{;}$ If Carol cycles for $10$ hrs, she must swim for $25$ hrs.
$x = g(y) = 26\frac{2}{3} - \frac{2}{3} y\text{;}$ $g(10) = 20\text{;}$ If Carol swims for $10$ hrs, she must cycle for $20$ hrs.

Subsection 2.2.4 Inverse Function Notation

If the inverse of a function $f$ is also a function, we denote the inverse by the symbol $f^{-1}\text{,}$ read "$f$ inverse." This notation makes it clear that the two functions are related in a special way. For example, the function $f (t) = 6 + 2t$ in Example 2.16 has inverse function $f^{-1}(H) = \dfrac{H -6}{2}\text{.}$

Example 2.20.

If $y = f (x) = x^3 + 2\text{,}$ find $f^{-1}(10)\text{.}$

Solution.

We first find the inverse function for $y = x^3 + 2$ by solving for $x\text{:}$

\begin{align*} x^3 \amp = y - 2 \amp\amp\text{Subtract 2 from both sides.}\\ x \amp = \sqrt[3]{y - 2}\amp\amp\text{Take cube roots.} \end{align*}

The inverse function is $x = f^{-1}(y) = \sqrt[3]{y - 2}\text{.}$ Now we evaluate the inverse function at $y = 10\text{:}$

\begin{equation*} f^{-1}(10) = \sqrt[3]{10 - 2} = 2. \end{equation*}

Caution 2.21.

Although the same symbol, $^{-1}\text{,}$ is used for both reciprocals and inverse functions, the two notions are not equivalent. That is, the inverse of a given function is usually not the same as the reciprocal of that function. In Example 2.16, note that $f^{-1}(y)$ is not the same as the reciprocal of $f (y)\text{,}$ because

\begin{equation*} \frac{1}{f(y)}= \frac{1}{y^3+ 2} ~~\text{ but }~~ f^{-1}(y)= \sqrt[3]{y - 2} \end{equation*}

To avoid confusion, we use the notation $\dfrac{1}{f}$ to refer to the reciprocal of the function $f\text{.}$

Note 2.22.

In Example 2.16, you can check that $f(2) = 10\text{.}$ In fact, the two statements

\begin{equation*} f^{-1}(10) = 2 ~~\text{ and } ~~ f(2) = 10 \end{equation*}

are equivalent; they convey the same information. This fact is a restatement of our earlier observation about inverse functions, this time using inverse function notation.

Inverse Functions.

Suppose the inverse of $f$ is a function, denoted by $f^{-1}\text{.}$ Then

\begin{equation*} f^{-1}(y) = x \text{ if and only if }f(x) = y. \end{equation*}

Example 2.23.

If $z = f (w) = \dfrac{1}{w + 3}\text{,}$ find $f^{-1}(1)\text{.}$
Write two equations about the value of $f^{-1}(1)\text{,}$ one using $f^{-1}$ and one using $f\text{.}$
Show that $f^{-1}(1)$ is not equal to $\dfrac{1}{f(1)}\text{.}$

Solution.

We first find the inverse function for $f$ by solving for $w\text{:}$

\begin{equation*} \begin{aligned} z \amp = \frac{1}{w+3} \\ z(w+3) \amp = 1 \\ w+3 \amp = \frac{1}{z} \\ w \amp = \frac{1}{z}-3 \end{aligned} \end{equation*}

The inverse function is $w=f^{-1}(z)=\frac{1}{z}-3\text{.}$ Now we evaluate the inverse function at $w=1\text{:}$

\begin{equation*} f^{-1}(1)=\frac{1}{1}-3=-2 \end{equation*}
$f^{-1}(1) = -2\text{,}$ $f (-2) = 1$
$f^{-1}(1) = -2\text{,}$ but $\frac{1}{f (1)}= 4$

We can use a graph of a function $y = f (x)$ to find values of the inverse function $x = f^{-1}(y)\text{.}$ Figure 2.24 shows the graph of $f(x) = x^3 + 2\text{.}$ You already know how to evaluate a function from its graph: We start with the horizontal axis. For instance, to evaluate $f(-2)\text{,}$ we find $-2$ on the $x$-axis, move vertically to the point on the graph with $x = -2\text{,}$ in this case $(-2, -6)\text{,}$ and read the $y$-coordinate of the point. We see that $f(-2) = -6\text{.}$

To evaluate the inverse function, we start with the vertical axis. For example, to find $f^{-1}(10)\text{,}$ we find $10$ on the vertical axis and move horizontally to the point on the graph with $y = 10\text{.}$ In this case, the point is $(2, 10)\text{,}$ so $f^{-1}(10) = 2\text{.}$

described in detail following the image — Figure 2.24.

Example 2.25.

The function $C = h(F)$ gives Celsius temperature as a function of Fahrenheit temperature. The graph of the function is shown in Figure 2.26. Use the graph to evaluate $h(68)$ and $h^{-1}(10)\text{,}$ and then explain their meaning in this context.

Solution.

To evaluate $h(68)\text{,}$ we find the input $F = 68$ on the horizontal axis, then find the point on the graph with $F = 68$ and read its vertical coordinate. We see that the point $(68, 20)$ lies on the graph, so $h(68) = 20\text{.}$ When the Fahrenheit temperature is $68\degree\text{,}$ the Celsius temperature is $20\degree\text{.}$

The inverse function reverses the roles of input and output. Because $C = h(F)\text{,}$ $F = h^{-1}(C)\text{,}$ so the inverse function gives us the Fahrenheit temperature if we know the Celsius temperature. In particular, $h^{-1}(10)$ is the Fahrenheit temperature when the Celsius temperature is $10\degree\text{.}$

To use the graph of $h$ to find values of $h^{-1}\text{,}$ we start with the vertical axis and find the point on the graph with $C = 10\text{.}$ This point is $(50, 10)\text{,}$ so $F = 50$ when $C = 10\text{,}$ or $h^{-1}(10) = 50\text{.}$ When the Celsius temperature is $10\degree\text{,}$ the Fahrenheit temperature is $50\degree\text{.}$

Example 2.27.

Use the graph of $h$ in Figure 2.26 to find $h^{-1}(-10)\text{.}$
Does $h^{-1}(-10) = -h^{-1}(10)\text{?}$
Write two equations, one using $h$ and one using $h^{-1}\text{,}$ stating the Fahrenheit temperature when the Celsius temperature is $0\degree\text{.}$

Solution.

$\displaystyle h^{-1}(-10)=14$
No; $h^{-1}(-10)=14$ and $-h^{-1}(10)=-50\text{.}$
$h(32)=0$ and $h^{-1}(0)=32$

Subsection 2.2.5 Domain and Range

When we interchange the input and output variables to obtain the inverse function, we interchange the domain and range of the function. For the functions graphed in Example 4.15, you can see that

\begin{align*} \text{Domain} (f) \amp=[-4, 12] \amp\text{and}\amp\amp \text{Domain} ( f^{-1}) \amp= [0, 8]\\ \text{Range} ( f ) \amp = [0, 8] \amp\amp\amp \text{Range} ( f -1) \amp = [-4, 12]. \end{align*}

This relationship between the domain and range of a function and its inverse holds in general.

Domain and Range of the Inverse Function.

Suppose $f^{-1}$ is the inverse function for $f\text{.}$ Then

\begin{equation*} f^{-1}\left(f(x)\right) = x\text{ and }f\left(f^{-1}( y)\right) = y \end{equation*}

as long as $x$ is in the domain of $f\text{,}$ and $y$ is in the domain of $f^{-1}\text{.}$

Example 2.28.

Graph the function $y = f (x) = \dfrac{1}{x +3}$ in the window
\begin{align*} \text{Xmin} \amp = -6 \amp\amp \text{Xmax} = 3.4\\ \text{Ymin} \amp = -6 \amp\amp \text{Ymax} = 3\text{.} \end{align*}
Graph the inverse function in the same window, along with the line $y = x\text{.}$
State the domain and range of $f\text{,}$ and of $f^{-1}\text{.}$

Solution.

The graph of $f$ is shown in Figure 2.29. It looks like the graph of $y = \dfrac{1}{x}\text{,}$ shifted $3$ units to the left.

GC graph of function and inverse

Figure 2.29.
To find the inverse function, we solve for $x\text{.}$ Take the reciprocal of both sides of the equation.

\begin{align*} \dfrac{1}{y} \amp= x + 3\amp\amp\text{Subtract 3 from both sides.}\\ x \amp = \dfrac{1}{y}- 3 \end{align*}

The inverse function is $x = f^{-1}(y) = \dfrac{1}{y}- 3\text{,}$ or, using $x$ for the input variable, $f^{-1}(x) = \dfrac{1}{x}- 3\text{.}$ The graph of $f^{-1}$ looks like the graph of $y = \dfrac{1}{x}\text{,}$ shifted down $3$ units, as shown in Figure 2.29.
Because $f$ is undefined at $x = -3\text{,}$ the domain of $f$ is all real numbers except $-3\text{.}$ The graph has a horizontal asymptote at $y = 0\text{,}$ so the range is all real numbers except $0\text{.}$

The inverse function $f^{-1}(x) = \dfrac{1}{x}- 3$ is undefined at $x = 0\text{,}$ so its domain is all real numbers except $0\text{.}$ The graph of $f^{-1}$ has a horizontal asymptote at $y = -3\text{,}$ so its range is all real numbers except $-3\text{.}$

Exercises 2.2.6 Exercises

1. Composition of Piecewise Function.

Suppose we’re given the function

\begin{equation*} f(x)=\begin{cases}\displaystyle{x+2}\amp \text{if}\ 0 \le x \le 2\cr \displaystyle{4-x}\amp \text{if}\ 2 \lt x \le 7\end{cases}. \end{equation*}

a) Evaluate the following.

i) $f(1)=$

ii) $f(4)=$

iii) $f(f(4))=$

b) What is the domain of $f(x)\text{?}$

Domain:

Hint.

a) Determine which piece of the function you should use by looking at the input and comparing it to the restrictions given in the function.

For part iii) we want to evaluate $f(f(4))\text{,}$ which means we need to evaluate the function twice. First, let’s evaluate the inner-most value $f(4)\text{.}$ Then evaluate the function again at this new value.

b) For the domain, we use the restrictions given by the pieces of the piecewise function.

2. Evaluating a Function.

Let $g(x)=-x^2-1x+6\text{.}$

Evaluate the following.

a) $g(0)=$

b) $g(3)=$

c) $g(x+1)=$

Hint.

a) We want to plug $x=0$ into our function.

b) We want to plug $x=3$ into our function.

c) This part is a little different in that we aren’t plugging a number into our function. We will still replace "$x$" in the equation with "$x+1$" as we would with any real number input.

You can simplify this expression, but do not need to.

3. Evaluating a Function Composition.

Let $f(x)=5x-1$ and $g(x)=2-x^2\text{.}$

Evaluate the following.

a) $g(f(8))=$

b) $f(g(x))=$

c) $f(f(x))=$

Hint.

a) We want to first evaluate $f(8)$ and then use that value to plug into $g(x)\text{.}$ Thus, we first plug $x=8$ into $f(x)\text{.}$ Then plug that value into $g(x)\text{.}$

b) In this part, we want to plug the function $g(x)=2-x^2$ into the function $f(x)=5x-1\text{.}$ In other words, every place we have "$x$" in $f(x)$ we will replace it with "$2-x^2$".

c) Use a similar process to part b).

4. Application of Function Composition.

A banner printing company charges $C(a)=3a+35$ dollars to print a banner with an area of $a$ square feet. Taylor wants to print a banner that is $2$ feet high. If her banner is $x$ feet long, the area of her banner is given by $A(x)=2x\text{.}$

a) Evaluate $C(A(10))$ and choose the most appropriate interpretation in the context of the problem.

$C(A(10))=$

It gives the price the company charges for a banner with area 10 square feet.
It gives the area of a banner which is 10 feet long.
It gives the price the company charges for a banner that is 10 feet long.
It gives the area of a banner which costs 10 dollars.
It gives the price the company charges for a banner with area 20 square feet, which is 10 feet long.

b) What are the units of the inputs and the outputs of the function $C(A(x))\text{?}$

Input units:

feet (length)
dollars (cost)
square feet (area)

Output units:

feet (length)
dollars (cost)
square feet (area)

Hint.

a) First, we need to evaluate $C(A(10))\text{.}$ This includes a composition of functions, so we should first evaluate $A(10)$ and then evaluate $C(a)$ at that value.

For the interpretation of $C(A(10))\text{,}$ we have to consider what’s the input and what’s the output. What are the units of the value you plug in and what are the units of the final value you get after evaluation? This will help choose the appropriate interpretation.

b) Using similar reasoning to the interpretation of $C(A(10))\text{,}$ the input units of $C(A(x))$ are going to be the units of $x\text{.}$ The output units are the units from the output of $C(a)\text{.}$

5. Application of Function Composition.

The height, $H\text{,}$ in feet, of water in a cylindrical tank is given by $H(t)=0.5t\text{,}$ where $t$ is time in hours. The volume of water in the tank when the height is $H$ feet is $V(H)=25\pi H$ cubic feet.

a) Find a formula for $V=f(t)\text{,}$ the volume of water in the tank after $t$ hours.

$V=f(t)=$

b) What are the units of the inputs and the outputs of the function $V=f(t)\text{?}$

Input units:

hours (time)
feet (height of water)
cubic feet (volume of water)

Output units:

hours (time)
feet (height of water)
cubic feet (volume of water)

Hint.

a) We are given a formula for the volume of water in a tank at a given height $H\text{:}$ $V(H)=25\pi H\text{.}$ Therefore, to write a formula for the volume of the water in a tank after $t$ hours, we need to have a relationship between the height of the water and time in hours. We will use this information to compose functions to get the formula we need.

b) Think about what the units of $t$ and $V(t)$ are.

6. Evaluating Inverses from a Table.

Fill in the blanks using the table below.

$x$	-10	-7	-4	-1	2
$g(x)$	2	14	-1	8	8

Values of $g(x)$

a) $g(-7)=$

b) $g($$)=8$

c) $g^{-1}($$)=-7$

d) $g^{-1}(8)=$

Hint.

a) Using the table, we want to find the output associated with the input value $x=-7\text{.}$

b) In this case, we want to know what input value gives an output value of 8.

c) This part asks the same question as part a) -- what is the output associated with input value $x=-7\text{?}$

d) This part asks the same question as part b) -- what is the input value that gives us an output of 8?

7. Evaluating Inverses from a Graph.

Using the graph of $g(x)$ below, fill in the blanks:

a) $g(1)=$

b) $g($$)=4$

c) $g^{-1}($$)=1$

d) $g^{-1}(4)=$

Hint.

a) Using the graph, we want to find the output associated with the input value $x=1\text{.}$

b) In this case, we want to know what input value gives an output value of 4.

c) This part asks the same question as part a) -- what is the output associated with input value $x=1\text{?}$

d) This part asks the same question as part b) -- what is the input value that gives us an output of 4?

8. Application of Inverse Functions.

UNL student Schuyler is studying abroad in Copenhagen, Denmark.

A) He knows that the temperature in degrees Fahrenheit, $F\text{,}$ is given by the equation

\begin{equation*} F=g(C)=\frac{9}{5}C+32, \end{equation*}

where $C$ is the temperature in degrees Celsius.

I) Find a formula for $g^{-1}(F)\text{.}$ What are the units of the inputs and outputs?

$g^{-1}(F)=$

Input units:

degrees Fahrenheit
degrees Celsius

Output units:

degrees Fahrenheit
degrees Celsius

II) Evaluate and interpret $g(24)$ and $g^{-1}(24)\text{.}$

$g(24)=$

It is the temperature in degrees Fahrenheit when the temperature is 24 degrees Celsius.
It is the temperature in degrees Celsius when the temperature is 24 degrees Fahrenheit.
It means it’s 24 degrees Fahrenheit.
It means it’s 24 degrees Celsius.

$g^{-1}(24)=$

It is the temperature in degrees Fahrenheit when the temperature is 24 degrees Celsius.
It is the temperature in degrees Celsius when the temperature is 24 degrees Fahrenheit.
It means it’s 24 degrees Fahrenheit.
It means it’s 24 degrees Celsius.

B) Schuyler also always needs to keep the currency exchange rate in mind: $D=f(K)=0.15K\text{,}$ where $D$ is the value in US dollars, and $K$ is the value in Danish krones.

I) Find a formula for $f^{-1}(D)\text{,}$ and interpret its meaning in the context of the problem. What are the units of the inputs and outputs?

$f^{-1}(D)=$

Input units:

US dollars
Danish krones

Output units:

US dollars
Danish krones

II) Evaluate and interpret $f(60)$ and $f^{-1}(100)\text{.}$

$f(60)=$

It is the value of 60 Danish krones in US dollars.
It is the value of 60 US dollars in Danish krones.
It means Schuyler has 60 US dollars.
It means Schuyler has 60 Danish krones.

$f^{-1}(100)=$

It is the value of 100 Danish krones in US dollars.
It is the value of 100 US dollars in Danish krones.
It means Schuyler has 100 US dollars.
It means Schuyler has 100 Danish krones.

C) Suppose the monthly cost in Danish krones of heating Schuyler’s apartment to $x$ degrees Celsius is given by $H(x)=32x\text{.}$ Evaluate and interpret $f(H(g^{-1}(70)))\text{.}$

$f(H(g^{-1}(70)))=$

It is the cost to heat Schuyler’s apartment to 70 degrees Fahrenheit in Danish krones.
It is the cost to heat Schuyler’s apartment to 70 degrees Fahrenheit in US dollars.
It is the cost to heat Schuyler’s apartment to 70 degrees Celsius in US dollars.
It is the cost to heat Schuyler’s apartment to 70 degrees Celsius in Danish krones.

Hint.

A) We are given that $F=g(C)=\dfrac{9}{5}C+32\text{.}$

I) To find a formula for $g^{-1}(F)\text{,}$ we need to solve the given equation above for C.

The input and output units are reversed from $F=g(C)\text{.}$

II) To evaluate $g(24)\text{,}$ we will plug $C=24$ into our original equation. For the interpretation, note that $C=24$ means that we have the input is 24 degrees Celsius. When we plug that into our function, it converts to degrees Fahrenheit.

To evaluate $g^{-1}(24)\text{,}$ we want to plug $F=24$ into our inverse function. Note that we could also use the original equation $g(C)$ and set it equal to $F=24$ to solve for $C\text{.}$

For the interpretation, note that the input value of the inverse function is $F=24$ degrees Fahrenheit. The inverse function gives us the conversion of 24 degrees Fahrenheit to degrees Celsius.

B) This part will be very similar to part a) with different units. Here we are given the equation $D=f(K)=0.15 K\text{.}$

I) To find a formula for $f^{-1}(D)\text{,}$ we need to solve the equation given above for $K\text{.}$

The input and output units are reversed from $D=f(K)\text{.}$

II) To evaluate $f(60\text{,}$ we will plug $K=60$ into our original equation. For the interpretation, note that $K=60$ represents 60 Danish krones. When we plug that into our function, it converts Danish krones to US dollars.

To evaluate $f^{-1}(100)\text{,}$ we want to plug $D=100$ into our inverse function. Note that we could also use the original equation $f(K)$ and set it equal to $D=100$ to solve for $K\text{.}$

For the interpretation, note that the input value of the inverse function is $D=100$ US dollars. The inverse function gives us the conversion of 100 US dollars to Danish krones.

C) For this part, we are given the extra information that the monthly cost in Danish krones to heat Schuyler’s apartment to $x$ degrees Celsius is given by $H(x)=32x\text{.}$ We will break down the evaluation into several steps.

We first evaluate the inner-most part: $g^{-1}(70)\text{.}$ Note that this is taking an input value of 70 degrees Fahrenheit and converting it to degrees Celsius.

We will next evaluate $H(g^{-1}(70))$ by using the information above. Note that this will give us the cost in Danish krones of heating Schuyler’s apartment to 70 degrees Fahrenheit.

Finally, we will evaluate $f(H(g^{-1}(70)))$ by using the information above again. Use the interpretations above to figure out what this interpretation is.

9. Application of Function Composition.

A team of American engineers is working with a team in Germany to design an engine. They need to know the volume of the interior of a piston; in this engine, the volume of each piston is given by \begin{align*} V(x) = 10.125\pi x^2 \end{align*} where $x$ is the diameter of the piston in centimeters. However, the American team took all their measurements in inches. There are 2.54 centimeters in each inch.

a) Write a formula that converts inches to centimeters.

$C=f(I)=$

b) Give a formula for the volume of the cylinder which has input in inches.

$V(I)=$

Hint.

a) Since there are 2.54 centimeters in an inch, we take the number of inches given and multiply by 2.54 to convert to centimeters.

b) Now, to find a formula for the volume of the interior of a piston in the engine given the diameter of the piston in inches we need to take $V(f(I))\text{.}$

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Loan amount, \(L\)	\(150,000\)	\(175,000\)	\(200,000\)	\(225,000\)	\(250,000\)
Mortgage payment, \(M\)	\(899.33\)	\(1049.21\)	\(1199.10\)	\(1348.99\)	\(1498.88\)

Mortgage payment, \(M\)	\(899.33\)	\(1049.21\)	\(1199.10\)	\(1348.99\)	\(1498.88\)
Loan amount, \(L\)	\(150,000\)	\(175,000\)	\(200,000\)	\(225,000\)	\(250,000\)

\(y=f(x)\)
\(x\)	\(y\)
\(-3\)	\(5\)
\(2\)	\(1\)
\(5\)	\(0\)

\(x=g(y)\)
\(y\)	\(x\)
\(5\)	\(-3\)
\(1\)	\(2\)
\(0\)	\(5\)

\(H=f(t)\)
\(t\)	\(H\)
\(0\)	\(6\)
\(1\)	\(8\)
\(2\)	\(10\)
\(3\)	\(12\)

\(t=g(H)\)
\(H\)	\(t\)
\(6\)	\(0\)
\(8\)	\(1\)
\(10\)	\(2\)
\(12\)	\(3\)