1.5 Function Arithmetic

In the previous section we used the newly defined function notation to make sense of expressions such as ` $f ⁢ open parenthesis x close parenthesis plus 2$ ' and ` $2 ⁢ f ⁢ open parenthesis x close parenthesis$ ' for a given function $f$ . It would seem natural, then, that functions should have their own arithmetic which is consistent with the arithmetic of real numbers. The following definitions allow us to add, subtract, multiply and divide functions using the arithmetic we already know for real numbers.

Two dashed curves, a U-shaped gray one and a straight green one, are joined by a solid navy curve that lies above or below them, and a slider positions a vertical line where three markers sit on the three curves with two stacked segments connecting the horizontal axis to the marker on the navy curve. Moving the slider drags the markers and the stacked segments across the picture, and where the straight dashed line dips below the axis the upper segment reverses direction so the stack ends at the navy marker below the gray one. Adjustable parameter: Input a (a) = 1.5. Viewing window: x from -12.17 to 13.17, y from -6.83 to 8.83. — The two dashed curves are f(x) = x²/4 − 1 and g(x) = 2 − x/2; the solid curve is their sum f + g. At the input a set by the slider, the three marked heights are f(a), g(a) and (f + g)(a), and the two stacked vertical segments show the sum being assembled: the green segment is g(a) placed on top of the gray segment f(a), ending exactly at the solid curve. Slide a across the picture and the stacking is redone at every input, including places where one height is negative and the segments partly cancel, which is all function addition means.

Function Arithmetic

Suppose $f$ and $g$ are functions and $x$ is in both the domain of $f$ and the domain of $g$ .¹

The sum of $f$ and $g$ , denoted $f + g$ , is the function defined by the formula
$(f + g) (x) = f (x) + g (x)$
The difference of $f$ and $g$ , denoted $f - g$ , is the function defined by the formula
$(f - g) (x) = f (x) - g (x)$
The product of $f$ and $g$ , denoted $f g$ , is the function defined by the formula
$(f g) (x) = f (x) g (x)$
The quotient of $f$ and $g$ , denoted $\frac{f}{g}$ , is the function defined by the formula
$(\frac{f}{g}) (x) = \frac{f (x)}{g (x)},$
provided $g ⁢ open parenthesis x close parenthesis not equal to 0$ .

In other words, to add two functions, we add their outputs; to subtract two functions, we subtract their outputs, and so on. Note that while the formula $(f + g) (x) = f (x) + g (x)$ looks suspiciously like some kind of distributive property, it is nothing of the sort; the addition on the left hand side of the equation is function addition, and we are using this equation to define the output of the new function $f + g$ as the sum of the real number outputs from $f$ and $g$ .

Example 1

Let $f (x) = 6 x^{2} - 2 x$ and $g (x) = 3 - \frac{1}{x}$ .

Find $(f + g) (- 1)$
Find $open parenthesis f ⁢ g close parenthesis ⁢ open parenthesis 2 close parenthesis$
Find the domain of $g - f$ then find and simplify a formula for $(g - f) (x)$ .
Find the domain of $(\frac{g}{f})$ then find and simplify a formula for $(\frac{g}{f}) (x)$ .

To find $(f + g) (- 1)$ we first find $f (- 1) = 8$ and $g (- 1) = 4$ . By definition, we have that $(f + g) (- 1) = f (- 1) + g (- 1) = 8 + 4 = 12$ .
To find $open parenthesis f ⁢ g close parenthesis ⁢ open parenthesis 2 close parenthesis$ , we first need $f ⁢ open parenthesis 2 close parenthesis$ and $g ⁢ open parenthesis 2 close parenthesis$ . Since $f ⁢ open parenthesis 2 close parenthesis equals 20$ and $g (2) = \frac{5}{2}$ , our formula yields $(f g) (2) = f (2) g (2) = (20) (\frac{5}{2}) = 50$ .
One method to find the domain of $g - f$ is to find the domain of $g$ and of $f$ separately, then find the intersection of these two sets. Owing to the denominator in the expression $g (x) = 3 - \frac{1}{x}$ , we get that the domain of $g$ is $open parenthesis minus ∞, 0 close parenthesis union open parenthesis 0, ∞ close parenthesis$ . Since $f (x) = 6 x^{2} - 2 x$ is valid for all real numbers, we have no further restrictions. Thus the domain of $g - f$ matches the domain of $g$ , namely, $open parenthesis minus ∞, 0 close parenthesis union open parenthesis 0, ∞ close parenthesis$ .
A second method is to analyze the formula for $(g - f) (x)$ before simplifying and look for the usual domain issues. In this case,
$(g - f) (x) = g (x) - f (x) = (3 - \frac{1}{x}) - (6 x^{2} - 2 x),$
so we find, as before, the domain is $open parenthesis minus ∞, 0 close parenthesis union open parenthesis 0, ∞ close parenthesis$ .
Moving along, we need to simplify a formula for $(g - f) (x)$ . In this case, we get common denominators and attempt to reduce the resulting fraction. Doing so, we get
$\begin{matrix} (g - f) (x) & = & g (x) - f (x) \\ = & (3 - \frac{1}{x}) - (6 x^{2} - 2 x) \\ = & 3 - \frac{1}{x} - 6 x^{2} + 2 x \\ = & \frac{3 x}{x} - \frac{1}{x} - \frac{6 x^{3}}{x} + \frac{2 x^{2}}{x} & get common denominators \\ = & \frac{3 x - 1 - 6 x^{3} + 2 x^{2}}{x} \\ = & \frac{- 6 x^{3} + 2 x^{2} + 3 x - 1}{x} \end{matrix}$

As in the previous example, we have two ways to approach finding the domain of $\frac{g}{f}$ . First, we can find the domain of $g$ and $f$ separately, and find the intersection of these two sets. In addition, since $(\frac{g}{f}) (x) = \frac{g (x)}{f (x)}$ , we are introducing a new denominator, namely $f (x)$ , so we need to guard against this being $0$ as well. Our previous work tells us that the domain of $g$ is $open parenthesis minus ∞, 0 close parenthesis union open parenthesis 0, ∞ close parenthesis$ and the domain of $f$ is $open parenthesis minus ∞, ∞ close parenthesis$ . Setting $f ⁢ open parenthesis x close parenthesis equals 0$ gives $6 x^{2} - 2 x = 0$ or $x = 0, \frac{1}{3}$ . As a result, the domain of $\frac{g}{f}$ is all real numbers except $x equals 0$ and $x = \frac{1}{3}$ , or $(- \infty, 0) \cup (0, \frac{1}{3}) \cup (\frac{1}{3}, \infty)$ .

Alternatively, we may proceed as above and analyze the expression $(\frac{g}{f}) (x) = \frac{g (x)}{f (x)}$ before simplifying. In this case,

$(\frac{g}{f}) (x) = \frac{g (x)}{f (x)} = \frac{3 - \frac{1}{x}}{6 x^{2} - 2 x}$

We see immediately from the `little' denominator that $x not equal to 0$ . To keep the `big' denominator away from $0$ , we solve $6 x^{2} - 2 x = 0$ and get $x equals 0$ or $x = \frac{1}{3}$ . Hence, as before, we find the domain of $\frac{g}{f}$ to be $(- \infty, 0) \cup (0, \frac{1}{3}) \cup (\frac{1}{3}, \infty)$ .

Next, we find and simplify a formula for $(\frac{g}{f}) (x)$ .

Table 1.1

(\frac{g}{f}) (x)

=

\frac{g (x)}{f (x)}

\frac{3 - \frac{1}{x}}{6 x^{2} - 2 x}

\frac{3 - \frac{1}{x}}{6 x^{2} - 2 x} \cdot \frac{x}{x}

simplify compound fractions

\frac{(3 - \frac{1}{x}) x}{(6 x^{2} - 2 x) x}

\frac{3 x - 1}{(6 x^{2} - 2 x) x}

\frac{3 x - 1}{2 x^{2} (3 x - 1)}

factor

\frac{{(3 x - 1)}^{1}}{2 x^{2} (3 x - 1)}

cancel

\frac{1}{2 x^{2}}

Please note the importance of finding the domain of a function before simplifying its expression. In number in Example 1.5.1 above, had we waited to find the domain of $\frac{g}{f}$ until after simplifying, we'd just have the formula $\frac{1}{2 x^{2}}$ to go by, and we would (incorrectly!) state the domain as $open parenthesis minus ∞, 0 close parenthesis union open parenthesis 0, ∞ close parenthesis$ , since the other troublesome number, $x = \frac{1}{3}$ , was canceled away.²

Next, we turn our attention to the difference quotient of a function.

We will revisit this concept in Section 2.1, but for now, we use it as a way to practice function notation and function arithmetic. For reasons which will become clear in Calculus, `simplifying' a difference quotient means rewriting it in a form where the ` $h$ ' in the definition of the difference quotient cancels from the denominator. Once that happens, we consider our work to be done.

Example 2

Find and simplify the difference quotients for the following functions

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

To find $f (x + h)$ , we replace every occurrence of $x$ in the formula $f (x) = x^{2} - x - 2$ with the quantity $(x + h)$ to get

$\begin{matrix} f (x + h) & = & {(x + h)}^{2} - (x + h) - 2 \\ = & x^{2} + 2 x h + h^{2} - x - h - 2 . \end{matrix}$

So the difference quotient is

Table 1.2

\frac{f (x + h) - f (x)}{h}

\frac{(x^{2} + 2 x h + h^{2} - x - h - 2) - (x^{2} - x - 2)}{h}

\frac{x^{2} + 2 x h + h^{2} - x - h - 2 - x^{2} + x + 2}{h}

[8pt]

\frac{2 x h + h^{2} - h}{h}

[8pt]

\frac{h (2 x + h - 1)}{h}

factor

[8pt]

\frac{h (2 x + h - 1)}{h}

cancel

[8pt]

2 x + h - 1

To find $g (x + h)$ , we replace every occurrence of $x$ in the formula $g (x) = \frac{3}{2 x + 1}$ with the quantity $(x + h)$ to get

$\begin{matrix} g (x + h) & = & \frac{3}{2 (x + h) + 1} \\ = & \frac{3}{2 x + 2 h + 1}, \end{matrix}$

which yields

Table 1.3

\frac{g (x + h) - g (x)}{h}

\frac{\frac{3}{2 x + 2 h + 1} - \frac{3}{2 x + 1}}{h}

[3pt]

\frac{\frac{3}{2 x + 2 h + 1} - \frac{3}{2 x + 1}}{h} \cdot \frac{(2 x + 2 h + 1) (2 x + 1)}{(2 x + 2 h + 1) (2 x + 1)}

[3pt]

\frac{3 (2 x + 1) - 3 (2 x + 2 h + 1)}{h (2 x + 2 h + 1) (2 x + 1)}

[3pt]

\frac{6 x + 3 - 6 x - 6 h - 3}{h (2 x + 2 h + 1) (2 x + 1)}

[3pt]

\frac{- 6 h}{h (2 x + 2 h + 1) (2 x + 1)}

[3pt]

\frac{- 6 h}{h (2 x + 2 h + 1) (2 x + 1)}

[3pt]

\frac{- 6}{(2 x + 2 h + 1) (2 x + 1)}

Since we have managed to cancel the original ` $h$ ' from the denominator, we are done.

For $r (x) = \sqrt{x}$ , we get $r (x + h) = \sqrt{x + h}$ so the difference quotient is
$\frac{r (x + h) - r (x)}{h} = \frac{\sqrt{x + h} - \sqrt{x}}{h}$
In order to cancel the ` $h$ ' from the denominator, we rationalize the numerator by multiplying by its conjugate.³
$\begin{matrix} \frac{r (x + h) - r (x)}{h} & = & \frac{\sqrt{x + h} - \sqrt{x}}{h} \\ = & \frac{(\sqrt{x + h} - \sqrt{x})}{h} \cdot \frac{(\sqrt{x + h} + \sqrt{x})}{(\sqrt{x + h} + \sqrt{x})} & Multiply by the conjugate. \\ = & \frac{{(\sqrt{x + h})}^{2} - {(\sqrt{x})}^{2}}{h (\sqrt{x + h} + \sqrt{x})} & Difference of Squares. \\ = & \frac{(x + h) - x}{h (\sqrt{x + h} + \sqrt{x})} \\ = & \frac{h}{h (\sqrt{x + h} + \sqrt{x})} \\ = & \frac{h^{1}}{h (\sqrt{x + h} + \sqrt{x})} \\ = & \frac{1}{\sqrt{x + h} + \sqrt{x}} \end{matrix}$
Since we have removed the original ` $h$ ' from the denominator, we are done.

As mentioned before, we will revisit difference quotients in Section 2.1 where we will explain them geometrically. For now, we want to move on to some classic applications of function arithmetic from Economics and for that, we need to think like an entrepreneur.⁴

Suppose you are a manufacturer making a certain product.⁵ Let $x$ be the production level, that is, the number of items produced in a given time period. It is customary to let $C (x)$ denote the function which calculates the total cost of producing the $x$ items. The quantity $C ⁢ open parenthesis 0 close parenthesis$ , which represents the cost of producing no items, is called the fixed cost, and represents the amount of money required to begin production. Associated with the total cost $C (x)$ is cost per item, or average cost, denoted $\bar{C} (x)$ and read ` $C$ -bar' of $x$ . To compute $\bar{C} (x)$ , we take the total cost $C (x)$ and divide by the number of items produced $x$ to get

$\bar{C} (x) = \frac{C (x)}{x}$

On the retail end, we have the price $p$ charged per item. To simplify the dialog and computations in this text, we assume that the number of items sold equals the number of items produced. From a retail perspective, it seems natural to think of the number of items sold, $x$ , as a function of the price charged, $p$ . After all, the retailer can easily adjust the price to sell more product. In the language of functions, $x$ would be the dependent variable and $p$ would be the independent variable or, using function notation, we have a function $x (p)$ . While we will adopt this convention later in the text,⁶ we will hold with tradition at this point and consider the price $p$ as a function of the number of items sold, $x$ . That is, we regard $x$ as the independent variable and $p$ as the dependent variable and speak of the price-demand function, $p (x)$ . Hence, $p (x)$ returns the price charged per item when $x$ items are produced and sold. Our next function to consider is the revenue function, $R (x)$ . The function $R (x)$ computes the amount of money collected as a result of selling $x$ items. Since $p (x)$ is the price charged per item, we have $R (x) = x p (x)$ . Finally, the profit function, $P (x)$ calculates how much money is earned after the costs are paid. That is, $P (x) = (R - C) (x) = R (x) - C (x)$ . We summarize all of these functions below.

Summary of Common Economic Functions

Suppose $x$ represents the quantity of items produced and sold.

The price-demand function $p (x)$ calculates the price per item.
The revenue function $R (x)$ calculates the total money collected by selling $x$ items at a price $p (x)$ , $R (x) = x p (x)$ .
The cost function $C (x)$ calculates the cost to produce $x$ items. The value $C ⁢ open parenthesis 0 close parenthesis$ is called the fixed cost or start-up cost.
The average cost function $\bar{C} (x) = \frac{C (x)}{x}$ calculates the cost per item when making $x$ items. Here, we necessarily assume $x greater than 0$ .
The profit function $P (x)$ calculates the money earned after costs are paid when $x$ items are produced and sold, $P (x) = (R - C) (x) = R (x) - C (x)$ .

It is high time for an example.

Example 3

Let $x$ represent the number of dOpi media players (`dOpis'⁷) produced and sold in a typical week. Suppose the cost, in dollars, to produce $x$ dOpis is given by $C (x) = 100 x + 2000$ , for $x greater than or equal to 0$ , and the price, in dollars per dOpi, is given by $p (x) = 450 - 15 x$ for $0 less than or equal to x less than or equal to 30$ .

Find and interpret $C ⁢ open parenthesis 0 close parenthesis$ .
Find and interpret $C bar ⁢ open parenthesis 10 close parenthesis$ .
Find and interpret $p ⁢ open parenthesis 0 close parenthesis$ and $p ⁢ open parenthesis 20 close parenthesis$ .
Solve $p ⁢ open parenthesis x close parenthesis equals 0$ and interpret the result.
Find and simplify expressions for the revenue function $R (x)$ and the profit function $P (x)$ .
Find and interpret $R ⁢ open parenthesis 0 close parenthesis$ and $P ⁢ open parenthesis 0 close parenthesis$ .
Solve $P ⁢ open parenthesis x close parenthesis equals 0$ and interpret the result.

We substitute $x equals 0$ into the formula for $C (x)$ and get $C (0) = 100 (0) + 2000 = 2000$ . This means to produce $0$ dOpis, it costs $$ 2000$ . In other words, the fixed (or start-up) costs are $$ 2000$ . The reader is encouraged to contemplate what sorts of expenses these might be.
Since $\bar{C} (x) = \frac{C (x)}{x}$ , $\bar{C} (10) = \frac{C (10)}{10} = \frac{3000}{10} = 300$ . This means when $10$ dOpis are produced, the cost to manufacture them amounts to $$ 300$ per dOpi.
Plugging $x equals 0$ into the expression for $p (x)$ gives $p (0) = 450 - 15 (0) = 450$ . This means no dOpis are sold if the price is $$ 450$ per dOpi. On the other hand, $p (20) = 450 - 15 (20) = 150$ which means to sell $20$ dOpis in a typical week, the price should be set at $$ 150$ per dOpi.
Setting $p ⁢ open parenthesis x close parenthesis equals 0$ gives $450 - 15 x = 0$ . Solving gives $x equals 30$ . This means in order to sell $30$ dOpis in a typical week, the price needs to be set to $$ 0$ . What's more, this means that even if dOpis were given away for free, the retailer would only be able to move $30$ of them.⁸
To find the revenue, we compute $R (x) = x p (x) = x (450 - 15 x) = 450 x - 15 x^{2}$ . Since the formula for $p (x)$ is valid only for $0 less than or equal to x less than or equal to 30$ , our formula $R (x)$ is also restricted to $0 less than or equal to x less than or equal to 30$ . For the profit, $P (x) = (R - C) (x) = R (x) - C (x)$ . Using the given formula for $C (x)$ and the derived formula for $R (x)$ , we get $P (x) = (450 x - 15 x^{2}) - (100 x + 2000) = - 15 x^{2} + 350 x - 2000$ . As before, the validity of this formula is for $0 less than or equal to x less than or equal to 30$ only.
We find $R ⁢ open parenthesis 0 close parenthesis equals 0$ which means if no dOpis are sold, we have no revenue, which makes sense. Turning to profit, $P (0) = - 2000$ since $P (x) = R (x) - C (x)$ and $P (0) = R (0) - C (0) = - 2000$ . This means that if no dOpis are sold, more money ( $$ 2000$ to be exact!) was put into producing the dOpis than was recouped in sales. In number, we found the fixed costs to be $$ 2000$ , so it makes sense that if we sell no dOpis, we are out those start-up costs.
Setting $P ⁢ open parenthesis x close parenthesis equals 0$ gives $- 15 x^{2} + 350 x - 2000 = 0$ . Factoring gives $- 5 (x - 10) (3 x - 40) = 0$ so $x equals 10$ or $x = \frac{40}{3}$ . What do these values mean in the context of the problem? Since $P (x) = R (x) - C (x)$ , solving $P ⁢ open parenthesis x close parenthesis equals 0$ is the same as solving $R (x) = C (x)$ . This means that the solutions to $P ⁢ open parenthesis x close parenthesis equals 0$ are the production (and sales) figures for which the sales revenue exactly balances the total production costs. These are the so-called `break even' points. The solution $x equals 10$ means $10$ dOpis should be produced (and sold) during the week to recoup the cost of production. For $x = \frac{40}{3} = 13 . \bar{3}$ , things are a bit more complicated. Even though $x equals 13.3 bar$ satisfies $0 less than or equal to x less than or equal to 30$ , and hence is in the domain of $P$ , it doesn't make sense in the context of this problem to produce a fractional part of a dOpi.⁹ Evaluating $P ⁢ open parenthesis 13 close parenthesis equals 15$ and $P (14) = - 40$ , we see that producing and selling $13$ dOpis per week makes a (slight) profit, whereas producing just one more puts us back into the red. While breaking even is nice, we ultimately would like to find what production level (and price) will result in the largest profit, and we'll do just that …in Section 2.3.

Exercises

In Exercises -, use the pair of functions $f$ and $g$ to find the following values if they exist.

$open parenthesis f plus g close parenthesis ⁢ open parenthesis 2 close parenthesis$
$(f - g) (- 1)$
$open parenthesis g minus f close parenthesis ⁢ open parenthesis 1 close parenthesis$

$(f g) (\frac{1}{2})$
$(\frac{f}{g}) (0)$
$(\frac{g}{f}) (- 2)$

$f (x) = 3 x + 1$ and $g (x) = 4 - x$
$f (x) = x^{2}$ and $g (x) = - 2 x + 1$
$f (x) = x^{2} - x$ and $g (x) = 12 - x^{2}$
$f ⁢ open parenthesis x close parenthesis equals 2 ⁢ x cubed$ and $g (x) = - x^{2} - 2 x - 3$
$f (x) = \sqrt{x + 3}$ and $g (x) = 2 x - 1$
$f (x) = \sqrt{4 - x}$ and $g (x) = \sqrt{x + 2}$
$f ⁢ open parenthesis x close parenthesis equals 2 ⁢ x$ and $g (x) = \frac{1}{2 x + 1}$
$f (x) = x^{2}$ and $g (x) = \frac{3}{2 x - 3}$
$f (x) = x^{2}$ and $g (x) = \frac{1}{x^{2}}$
$f (x) = x^{2} + 1$ and $g (x) = \frac{1}{x^{2} + 1}$

In Exercises -, use the pair of functions $f$ and $g$ to find the domain of the indicated function then find and simplify an expression for it.

$(f + g) (x)$
$(f - g) (x)$
$(f g) (x)$
$(\frac{f}{g}) (x)$
$f (x) = 2 x + 1$ and $g (x) = x - 2$
$f (x) = 1 - 4 x$ and $g (x) = 2 x - 1$
$f (x) = x^{2}$ and $g (x) = 3 x - 1$
$f (x) = x^{2} - x$ and $g ⁢ open parenthesis x close parenthesis equals 7 ⁢ x$
$f (x) = x^{2} - 4$ and $g (x) = 3 x + 6$
$f (x) = - x^{2} + x + 6$ and $g (x) = x^{2} - 9$
$f (x) = \frac{x}{2}$ and $g (x) = \frac{2}{x}$
$f (x) = x - 1$ and $g (x) = \frac{1}{x - 1}$
$f (x) = x$ and $g (x) = \sqrt{x + 1}$
$f (x) = \sqrt{x - 5}$ and $g (x) = f (x) = \sqrt{x - 5}$

In Exercises -, find and simplify the difference quotient $\frac{f (x + h) - f (x)}{h}$ for the given function.

$f (x) = 2 x - 5$
$f (x) = - 3 x + 5$
$f ⁢ open parenthesis x close parenthesis equals 6$
$f (x) = 3 x^{2} - x$
$f (x) = - x^{2} + 2 x - 1$
$f ⁢ open parenthesis x close parenthesis equals 4 ⁢ x squared$
$f (x) = x - x^{2}$
$f (x) = x^{3} + 1$
$f (x) = m x + b$ where $m not equal to 0$
$f (x) = a x^{2} + b x + c$ where $a not equal to 0$
$f (x) = \frac{2}{x}$
$f (x) = \frac{3}{1 - x}$
$f (x) = \frac{1}{x^{2}}$
$f (x) = \frac{2}{x + 5}$
$f (x) = \frac{1}{4 x - 3}$
$f (x) = \frac{3 x}{x + 1}$
$f (x) = \frac{x}{x - 9}$
$f (x) = \frac{x^{2}}{2 x + 1}$
$f (x) = \sqrt{x - 9}$
$f (x) = \sqrt{2 x + 1}$
$f (x) = \sqrt{- 4 x + 5}$
$f (x) = \sqrt{4 - x}$
$f (x) = \sqrt{a x + b}$ , where $a not equal to 0$ .
$f (x) = x \sqrt{x}$
$f (x) = \sqrt[3]{x}$ . HINT: $(a - b) (a^{2} + a b + b^{2}) = a^{3} - b^{3}$

In Exercises -, $C (x)$ denotes the cost to produce $x$ items and $p (x)$ denotes the price-demand function in the given economic scenario. In each Exercise, do the following:

Find and interpret $C ⁢ open parenthesis 0 close parenthesis$ .
Find and interpret $C bar ⁢ open parenthesis 10 close parenthesis$ .

Find and interpret $p ⁢ open parenthesis 5 close parenthesis$
Find and simplify $R (x)$ .

Find and simplify $P (x)$ .
Solve $P ⁢ open parenthesis x close parenthesis equals 0$ and interpret.
The cost, in dollars, to produce $x$ “I'd rather be a Sasquatch” T-Shirts is $C (x) = 2 x + 26$ , $x greater than or equal to 0$ and the price-demand function, in dollars per shirt, is $p (x) = 30 - 2 x$ , $0 less than or equal to x less than or equal to 15$ .
The cost, in dollars, to produce $x$ bottles of $100 %$ All-Natural Certified Free-Trade Organic Sasquatch Tonic is $C (x) = 10 x + 100$ , $x greater than or equal to 0$ and the price-demand function, in dollars per bottle, is $p (x) = 35 - x$ , $0 less than or equal to x less than or equal to 35$ .
The cost, in cents, to produce $x$ cups of Mountain Thunder Lemonade at Junior's Lemonade Stand is $C (x) = 18 x + 240$ , $x greater than or equal to 0$ and the price-demand function, in cents per cup, is $p (x) = 90 - 3 x$ , $0 less than or equal to x less than or equal to 30$ .
The daily cost, in dollars, to produce $x$ Sasquatch Berry Pies $C (x) = 3 x + 36$ , $x greater than or equal to 0$ and the price-demand function, in dollars per pie, is $p (x) = 12 - 0.5 x$ , $0 less than or equal to x less than or equal to 24$ .
The monthly cost, in hundreds of dollars, to produce $x$ custom built electric scooters is $C (x) = 20 x + 1000$ , $x greater than or equal to 0$ and the price-demand function, in hundreds of dollars per scooter, is $p (x) = 140 - 2 x$ , $0 less than or equal to x less than or equal to 70$ .

In Exercises -, let $f$ be the function defined by

$f = {(- 3, 4), (- 2, 2), (- 1, 0), (0, 1), (1, 3), (2, 4), (3, - 1)}$

and let $g$ be the function defined

$g = {(- 3, - 2), (- 2, 0), (- 1, - 4), (0, 0), (1, - 3), (2, 1), (3, 2)}$

. Compute the indicated value if it exists.

$(f + g) (- 3)$
$open parenthesis f minus g close parenthesis ⁢ open parenthesis 2 close parenthesis$
$open parenthesis f ⁢ g close parenthesis ⁢ open parenthesis minus 1 close parenthesis$
$open parenthesis g plus f close parenthesis ⁢ open parenthesis 1 close parenthesis$
$open parenthesis g minus f close parenthesis ⁢ open parenthesis 3 close parenthesis$
$open parenthesis g ⁢ f close parenthesis ⁢ open parenthesis minus 3 close parenthesis$
$(\frac{f}{g}) (- 2)$
$(\frac{f}{g}) (- 1)$
$(\frac{f}{g}) (2)$
$(\frac{g}{f}) (- 1)$
$(\frac{g}{f}) (3)$
$(\frac{g}{f}) (- 3)$

Answers

For $f (x) = 3 x + 1$ and $g (x) = 4 - x$
- $(f + g) (2) = 9$
- $(f - g) (- 1) = - 7$
- $(g - f) (1) = - 1$
- $(f g) (\frac{1}{2}) = \frac{35}{4}$
- $(\frac{f}{g}) (0) = \frac{1}{4}$
- $(\frac{g}{f}) (- 2) = - \frac{6}{5}$
For $f (x) = x^{2}$ and $g (x) = - 2 x + 1$
- $(f + g) (2) = 1$
- $(f - g) (- 1) = - 2$
- $(g - f) (1) = - 2$
- $(f g) (\frac{1}{2}) = 0$
- $(\frac{f}{g}) (0) = 0$
- $(\frac{g}{f}) (- 2) = \frac{5}{4}$
For $f (x) = x^{2} - x$ and $g (x) = 12 - x^{2}$
- $(f + g) (2) = 10$
- $(f - g) (- 1) = - 9$
- $(g - f) (1) = 11$
- $(f g) (\frac{1}{2}) = - \frac{47}{16}$
- $(\frac{f}{g}) (0) = 0$
- $(\frac{g}{f}) (- 2) = \frac{4}{3}$
For $f ⁢ open parenthesis x close parenthesis equals 2 ⁢ x cubed$ and $g (x) = - x^{2} - 2 x - 3$
- $(f + g) (2) = 5$
- $(f - g) (- 1) = 0$
- $(g - f) (1) = - 8$
- $(f g) (\frac{1}{2}) = - \frac{17}{16}$
- $(\frac{f}{g}) (0) = 0$
- $(\frac{g}{f}) (- 2) = \frac{3}{16}$
For $f (x) = \sqrt{x + 3}$ and $g (x) = 2 x - 1$
- $(f + g) (2) = 3 + \sqrt{5}$
- $(f - g) (- 1) = 3 + \sqrt{2}$
- $(g - f) (1) = - 1$
- $(f g) (\frac{1}{2}) = 0$
- $(\frac{f}{g}) (0) = - \sqrt{3}$
- $(\frac{g}{f}) (- 2) = - 5$
For $f (x) = \sqrt{4 - x}$ and $g (x) = \sqrt{x + 2}$
- $(f + g) (2) = 2 + \sqrt{2}$
- $(f - g) (- 1) = - 1 + \sqrt{5}$
- $(g - f) (1) = 0$
- $(f g) (\frac{1}{2}) = \frac{\sqrt{35}}{2}$
- $(\frac{f}{g}) (0) = \sqrt{2}$
- $(\frac{g}{f}) (- 2) = 0$
For $f ⁢ open parenthesis x close parenthesis equals 2 ⁢ x$ and $g (x) = \frac{1}{2 x + 1}$
- $(f + g) (2) = \frac{21}{5}$
- $(f - g) (- 1) = - 1$
- $(g - f) (1) = - \frac{5}{3}$
- $(f g) (\frac{1}{2}) = \frac{1}{2}$
- $(\frac{f}{g}) (0) = 0$
- $(\frac{g}{f}) (- 2) = \frac{1}{12}$
For $f (x) = x^{2}$ and $g (x) = \frac{3}{2 x - 3}$
- $(f + g) (2) = 7$
- $(f - g) (- 1) = \frac{8}{5}$
- $(g - f) (1) = - 4$
- $(f g) (\frac{1}{2}) = - \frac{3}{8}$
- $(\frac{f}{g}) (0) = 0$
- $(\frac{g}{f}) (- 2) = - \frac{3}{28}$
For $f (x) = x^{2}$ and $g (x) = \frac{1}{x^{2}}$
- $(f + g) (2) = \frac{17}{4}$
- $(f - g) (- 1) = 0$
- $(g - f) (1) = 0$
- $(f g) (\frac{1}{2}) = 1$
- $(\frac{f}{g}) (0)$ is undefined.
- $(\frac{g}{f}) (- 2) = \frac{1}{16}$
For $f (x) = x^{2} + 1$ and $g (x) = \frac{1}{x^{2} + 1}$
- $(f + g) (2) = \frac{26}{5}$
- $(f - g) (- 1) = \frac{3}{2}$
- $(g - f) (1) = - \frac{3}{2}$
- $(f g) (\frac{1}{2}) = 1$
- $(\frac{f}{g}) (0) = 1$
- $(\frac{g}{f}) (- 2) = \frac{1}{25}$
For $f (x) = 2 x + 1$ and $g (x) = x - 2$
- $(f + g) (x) = 3 x - 1$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$
- $(f - g) (x) = x + 3$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$
- $(f g) (x) = 2 x^{2} - 3 x - 2$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$
- $(\frac{f}{g}) (x) = \frac{2 x + 1}{x - 2}$ Domain: $open parenthesis minus ∞, 2 close parenthesis union open parenthesis 2, ∞ close parenthesis$
For $f (x) = 1 - 4 x$ and $g (x) = 2 x - 1$
- $(f + g) (x) = - 2 x$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$
- $(f - g) (x) = 2 - 6 x$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$
- $(f g) (x) = - 8 x^{2} + 6 x - 1$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$
- $(\frac{f}{g}) (x) = \frac{1 - 4 x}{2 x - 1}$ Domain: $(- \infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$
For $f (x) = x^{2}$ and $g (x) = 3 x - 1$
- $(f + g) (x) = x^{2} + 3 x - 1$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$
- $(f - g) (x) = x^{2} - 3 x + 1$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$
- $(f g) (x) = 3 x^{3} - x^{2}$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$
- $(\frac{f}{g}) (x) = \frac{x^{2}}{3 x - 1}$ Domain: $(- \infty, \frac{1}{3}) \cup (\frac{1}{3}, \infty)$
For $f (x) = x^{2} - x$ and $g ⁢ open parenthesis x close parenthesis equals 7 ⁢ x$
- $(f + g) (x) = x^{2} + 6 x$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$
- $(f - g) (x) = x^{2} - 8 x$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$
- $(f g) (x) = 7 x^{3} - 7 x^{2}$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$
- $(\frac{f}{g}) (x) = \frac{x - 1}{7}$ Domain: $open parenthesis minus ∞, 0 close parenthesis union open parenthesis 0, ∞ close parenthesis$
For $f (x) = x^{2} - 4$ and $g (x) = 3 x + 6$
- $(f + g) (x) = x^{2} + 3 x + 2$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$
- $(f - g) (x) = x^{2} - 3 x - 10$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$
- $(f g) (x) = 3 x^{3} + 6 x^{2} - 12 x - 24$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$
- $(\frac{f}{g}) (x) = \frac{x - 2}{3}$ Domain: $(- \infty, - 2) \cup (- 2, \infty)$
For $f (x) = - x^{2} + x + 6$ and $g (x) = x^{2} - 9$
- $(f + g) (x) = x - 3$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$
- $(f - g) (x) = - 2 x^{2} + x + 15$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$
- $(f g) (x) = - x^{4} + x^{3} + 15 x^{2} - 9 x - 54$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$
- $(\frac{f}{g}) (x) = - \frac{x + 2}{x + 3}$ Domain: $(- \infty, - 3) \cup (- 3, 3) \cup (3, \infty)$
For $f (x) = \frac{x}{2}$ and $g (x) = \frac{2}{x}$
- $(f + g) (x) = \frac{x^{2} + 4}{2 x}$ Domain: $open parenthesis minus ∞, 0 close parenthesis union open parenthesis 0, ∞ close parenthesis$
- $(f - g) (x) = \frac{x^{2} - 4}{2 x}$ Domain: $open parenthesis minus ∞, 0 close parenthesis union open parenthesis 0, ∞ close parenthesis$
- $open parenthesis f ⁢ g close parenthesis ⁢ open parenthesis x close parenthesis equals 1$ Domain: $open parenthesis minus ∞, 0 close parenthesis union open parenthesis 0, ∞ close parenthesis$
- $(\frac{f}{g}) (x) = \frac{x^{2}}{4}$ Domain: $open parenthesis minus ∞, 0 close parenthesis union open parenthesis 0, ∞ close parenthesis$
For $f (x) = x - 1$ and $g (x) = \frac{1}{x - 1}$
- $(f + g) (x) = \frac{x^{2} - 2 x + 2}{x - 1}$ Domain: $open parenthesis minus ∞, 1 close parenthesis union open parenthesis 1, ∞ close parenthesis$
- $(f - g) (x) = \frac{x^{2} - 2 x}{x - 1}$ Domain: $open parenthesis minus ∞, 1 close parenthesis union open parenthesis 1, ∞ close parenthesis$
- $open parenthesis f ⁢ g close parenthesis ⁢ open parenthesis x close parenthesis equals 1$ Domain: $open parenthesis minus ∞, 1 close parenthesis union open parenthesis 1, ∞ close parenthesis$
- $(\frac{f}{g}) (x) = x^{2} - 2 x + 1$ Domain: $open parenthesis minus ∞, 1 close parenthesis union open parenthesis 1, ∞ close parenthesis$
For $f (x) = x$ and $g (x) = \sqrt{x + 1}$
- $(f + g) (x) = x + \sqrt{x + 1}$ Domain: $open bracket minus 1, ∞ close parenthesis$
- $(f - g) (x) = x - \sqrt{x + 1}$ Domain: $open bracket minus 1, ∞ close parenthesis$
- $(f g) (x) = x \sqrt{x + 1}$ Domain: $open bracket minus 1, ∞ close parenthesis$
- $(\frac{f}{g}) (x) = \frac{x}{\sqrt{x + 1}}$ Domain: $open parenthesis minus 1, ∞ close parenthesis$
For $f (x) = \sqrt{x - 5}$ and $g (x) = f (x) = \sqrt{x - 5}$
- $(f + g) (x) = 2 \sqrt{x - 5}$ Domain: $open bracket 5, ∞ close parenthesis$
- $(f - g) (x) = 0$ Domain: $open bracket 5, ∞ close parenthesis$
- $(f g) (x) = x - 5$ Domain: $open bracket 5, ∞ close parenthesis$
- $(\frac{f}{g}) (x) = 1$ Domain: $open parenthesis 5, ∞ close parenthesis$
$2$
$minus 3$
$0$
$6 x + 3 h - 1$
$- 2 x - h + 2$
$8 ⁢ x plus 4 ⁢ h$
$- 2 x - h + 1$
$3 x^{2} + 3 x h + h^{2}$
$m$
$2 a x + a h + b$
$\frac{- 2}{x (x + h)}$
$\frac{3}{(1 - x - h) (1 - x)}$
$\frac{- (2 x + h)}{x^{2} {(x + h)}^{2}}$
$\frac{- 2}{(x + 5) (x + h + 5)}$
$\frac{- 4}{(4 x - 3) (4 x + 4 h - 3)}$
$\frac{3}{(x + 1) (x + h + 1)}$
$\frac{- 9}{(x - 9) (x + h - 9)}$
$\frac{2 x^{2} + 2 x h + 2 x + h}{(2 x + 1) (2 x + 2 h + 1)}$
$\frac{1}{\sqrt{x + h - 9} + \sqrt{x - 9}}$
$\frac{2}{\sqrt{2 x + 2 h + 1} + \sqrt{2 x + 1}}$
$\frac{- 4}{\sqrt{- 4 x - 4 h + 5} + \sqrt{- 4 x + 5}}$
$\frac{- 1}{\sqrt{4 - x - h} + \sqrt{4 - x}}$
$\frac{a}{\sqrt{a x + a h + b} + \sqrt{a x + b}}$
$\frac{3 x^{2} + 3 x h + h^{2}}{{(x + h)}^{3 / 2} + x^{3 / 2}}$
$\frac{1}{{(x + h)}^{2 / 3} + {(x + h)}^{1 / 3} x^{1 / 3} + x^{2 / 3}}$
- $C ⁢ open parenthesis 0 close parenthesis equals 26$ , so the fixed costs are $$ 26$ .
- $C bar ⁢ open parenthesis 10 close parenthesis equals 4.6$ , so when 10 shirts are produced, the cost per shirt is $$ 4.60$ .
- $p ⁢ open parenthesis 5 close parenthesis equals 20$ , so to sell $5$ shirts, set the price at $$ 20$ per shirt.
- $R (x) = - 2 x^{2} + 30 x$ , $0 less than or equal to x less than or equal to 15$
- $P (x) = - 2 x^{2} + 28 x - 26$ , $0 less than or equal to x less than or equal to 15$
- $P ⁢ open parenthesis x close parenthesis equals 0$ when $x equals 1$ and $x equals 13$ . These are the `break even' points, so selling $1$ shirt or $13$ shirts will guarantee the revenue earned exactly recoups the cost of production.
- $C ⁢ open parenthesis 0 close parenthesis equals 100$ , so the fixed costs are $$ 100$ .
- $C bar ⁢ open parenthesis 10 close parenthesis equals 20$ , so when 10 bottles of tonic are produced, the cost per bottle is $$ 20$ .
- $p ⁢ open parenthesis 5 close parenthesis equals 30$ , so to sell $5$ bottles of tonic, set the price at $$ 30$ per bottle.
- $R (x) = - x^{2} + 35 x$ , $0 less than or equal to x less than or equal to 35$
- $P (x) = - x^{2} + 25 x - 100$ , $0 less than or equal to x less than or equal to 35$
- $P ⁢ open parenthesis x close parenthesis equals 0$ when $x equals 5$ and $x equals 20$ . These are the `break even' points, so selling $5$ bottles of tonic or $20$ bottles of tonic will guarantee the revenue earned exactly recoups the cost of production.
- $C ⁢ open parenthesis 0 close parenthesis equals 240$ , so the fixed costs are $240$ ¢ or $$ 2.40$ .
- $C bar ⁢ open parenthesis 10 close parenthesis equals 42$ , so when 10 cups of lemonade are made, the cost per cup is $42$ ¢.
- $p ⁢ open parenthesis 5 close parenthesis equals 75$ , so to sell $5$ cups of lemonade, set the price at $75$ ¢ per cup.
- $R (x) = - 3 x^{2} + 90 x$ , $0 less than or equal to x less than or equal to 30$
- $P (x) = - 3 x^{2} + 72 x - 240$ , $0 less than or equal to x less than or equal to 30$
- $P ⁢ open parenthesis x close parenthesis equals 0$ when $x equals 4$ and $x equals 20$ . These are the `break even' points, so selling $4$ cups of lemonade or $20$ cups of lemonade will guarantee the revenue earned exactly recoups the cost of production.
- $C ⁢ open parenthesis 0 close parenthesis equals 36$ , so the daily fixed costs are $$ 36$ .
- $C bar ⁢ open parenthesis 10 close parenthesis equals 6.6$ , so when 10 pies are made, the cost per pie is $$ 6.60$ .
- $p ⁢ open parenthesis 5 close parenthesis equals 9.5$ , so to sell $5$ pies a day, set the price at $$ 9.50$ per pie.
- $R (x) = - 0.5 x^{2} + 12 x$ , $0 less than or equal to x less than or equal to 24$
- $P (x) = - 0.5 x^{2} + 9 x - 36$ , $0 less than or equal to x less than or equal to 24$
- $P ⁢ open parenthesis x close parenthesis equals 0$ when $x equals 6$ and $x equals 12$ . These are the `break even' points, so selling $6$ pies or $12$ pies a day will guarantee the revenue earned exactly recoups the cost of production.
- $C ⁢ open parenthesis 0 close parenthesis equals 1000$ , so the monthly fixed costs are $1000$ hundred dollars, or $$ 100, 000$ .
- $C bar ⁢ open parenthesis 10 close parenthesis equals 120$ , so when 10 scooters are made, the cost per scooter is $120$ hundred dollars, or $$ 12, 000$ .
- $p ⁢ open parenthesis 5 close parenthesis equals 130$ , so to sell $5$ scooters a month, set the price at $130$ hundred dollars, or $$ 13, 000$ per scooter.
- $R (x) = - 2 x^{2} + 140 x$ , $0 less than or equal to x less than or equal to 70$
- $P (x) = - 2 x^{2} + 120 x - 1000$ , $0 less than or equal to x less than or equal to 70$
- $P ⁢ open parenthesis x close parenthesis equals 0$ when $x equals 10$ and $x equals 50$ . These are the `break even' points, so selling $10$ scooters or $50$ scooters a month will guarantee the revenue earned exactly recoups the cost of production.
$(f + g) (- 3) = 2$
$(f - g) (2) = 3$
$(f g) (- 1) = 0$
$(g + f) (1) = 0$
$(g - f) (3) = 3$
$(g f) (- 3) = - 8$
$(\frac{f}{g}) (- 2)$ does not exist
$(\frac{f}{g}) (- 1) = 0$
$(\frac{f}{g}) (2) = 4$
$(\frac{g}{f}) (- 1)$ does not exist
$(\frac{g}{f}) (3) = - 2$
$(\frac{g}{f}) (- 3) = - \frac{1}{2}$

Adapted from Precalculus, 3rd corrected edition, by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.