5.2 Function Arithmetic

As we mentioned in Section 5.1, in this chapter, we are studying functions in a more abstract and general setting. In this section, we begin our study of what can be considered as the algebra of functions by defining function arithmetic.

Given two real numbers, we have four primary arithmetic operations available to us: addition, subtraction, multiplication, and division (provided we don't divide by $0$ .) Since the functions we study in this text have ranges which are sets of real numbers, it makes sense we can extend these arithmetic notions to functions.

For example, to add two functions means we add their outputs; to subtract two functions, we subtract their outputs, and so on and so forth. More formally, given two functions $f$ and $g$ , we define a new function $f + g$ whose rule is determined by adding the outputs of $f$ and $g$ . That is $(f + g) (x) = f (x) + g (x)$ . While this looks suspiciously like some kind of distributive property, it is nothing of the sort. The ` $+$ ' sign in the expression ` $f + g$ ' is part of the name of the function we are defining,¹ whereas the plus sign ` $+$ ' sign in the expression $f (x) + g (x)$ represents real number addition: we are adding the output from $f$ , $f (x)$ with the output from $g$ , $g (x)$ to determine the output from the sum function, $(f + g) (x)$ .

Of course, in order to define $(f + g) (x)$ by the formula $(f + g) (x) = f (x) + g (x)$ , both $f (x)$ and $g (x)$ need to be defined in the first place; that is, $x$ must be in the domain of $f$ and the domain of $g$ . You'll recall² this means $x$ must be in the intersection of the domains of $f$ and $g$ . We define the following.

Definition

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

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 $g ⁢ open parenthesis x close parenthesis not equal to 0$ .

We put these definitions to work for us in the next example.

Example 1

Consider the following functions:

$f (x) = 6 x^{2} - 2 x$
$g (t) = 3 - \frac{1}{t}$ , $t greater than 0$

$h = {(- 3, 2), (- 2, 0.4), (0, \sqrt{2}), (3, - 6)}$
$s$ whose graph is given below:

Figure 5.19

y = s (t)

Find and simplify the following function values:
1. $open parenthesis f plus g close parenthesis ⁢ open parenthesis 1 close parenthesis$
2. $(s - f) (- 1)$
3. $open parenthesis f ⁢ g close parenthesis ⁢ open parenthesis 2 close parenthesis$
4. $(\frac{s}{h}) (0)$
1. $((s + g) + h) (3)$
2. $(s + (g + h)) (3)$
3. $(\frac{f + h}{s}) (3)$
4. $(f (g - h)) (- 2)$
Find the domain of each of the following functions:
1. $h g$
2. $\frac{f}{s}$
Find expressions for the functions below. State the domain for each.
1. $(f g) (x)$
2. $(\frac{g}{f}) (t)$

1. By definition, $(f + g) (1) = f (1) + g (1)$ . We find $f (1) = 6 {(1)}^{2} - 2 (1) = 4$ and $g (1) = 3 - \frac{1}{1} = 2$ . So we get $(f + g) (1) = 4 + 2 = 6$ .
2. To find $(s - f) (- 1) = s (- 1) - f (- 1)$ , we need both $s ⁢ open parenthesis minus 1 close parenthesis$ and $f ⁢ open parenthesis minus 1 close parenthesis$ . To get $s ⁢ open parenthesis minus 1 close parenthesis$ , we look to the graph of $y = s (t)$ and look for the $y$ -coordinate of the point on the graph with the $t$ -coordinate of $minus 1$ . While not labeled directly, we infer the point $(- 1, - 2)$ is on the graph which means $s (- 1) = - 2$ . For $f ⁢ open parenthesis minus 1 close parenthesis$ , we compute: $f (- 1) = 6 {(- 1)}^{2} - 2 (- 1) = 8$ . Putting it all together, we get $(s - f) (- 1) = (- 2) - (8) = - 10$ .
3. Since $open parenthesis f ⁢ g close parenthesis ⁢ open parenthesis 2 close parenthesis equals f ⁢ open parenthesis 2 close parenthesis ⁢ g ⁢ open parenthesis 2 close parenthesis$ , we first compute $f ⁢ open parenthesis 2 close parenthesis$ and $g ⁢ open parenthesis 2 close parenthesis$ . We find $f (2) = 6 {(2)}^{2} - 2 (2) = 20$ and $g (2) = 2 + \frac{1}{2} = \frac{5}{2}$ , so $(f g) (2) = f (2) g (2) = (20) (\frac{5}{2}) = 50$ .
4. By definition, $(\frac{s}{h}) (0) = \frac{s (0)}{h (0)}$ . Since $open parenthesis 0, minus 2 close parenthesis$ is on the graph of $y = s (t)$ , so we know $s (0) = - 2$ . Likewise, the ordered pair $open parenthesis 0, the square root of 2 close parenthesis ∈ h$ , so $h (0) = \sqrt{2}$ . We get $(\frac{s}{h}) (0) = \frac{s (0)}{h (0)} = \frac{- 2}{\sqrt{2}} = - \sqrt{2}$ .
5. The expression $((s + g) + h) (3)$ involves three functions. Fortunately, they are grouped so that we can apply Definition 5.1 by first considering the sum of the two functions $(s + g)$ and $h$ , then to the sum of the two functions $s$ and $g$ : $((s + g) + h) (3) = (s + g) (3) + h (3) = (s (3) + g (3)) + h (3)$ . To get $s ⁢ open parenthesis 3 close parenthesis$ , we look to the graph of $y = s (t)$ . We infer the point $open parenthesis 3, 2 close parenthesis$ is on the graph of $s$ , so $s ⁢ open parenthesis 3 close parenthesis equals 2$ . We compute $g (3) = 3 - \frac{1}{3} = \frac{8}{3}$ . To find $h ⁢ open parenthesis 3 close parenthesis$ , we note $open parenthesis 3, minus 6 close parenthesis ∈ h$ , so $h (3) = - 6$ . Hence, $((s + g) + h) (3) = (s + g) (3) + h (3) = (s (3) + g (3)) + h (3) = (2 + \frac{8}{3}) + (- 6) = - \frac{4}{3}$ .
6. The expression $(s + (g + h)) (3)$ is very similar to the previous problem, $((s + g) + h) (3)$ except that the $g$ and $h$ are grouped together here instead of the $s$ and $g$ . We proceed as above applying Definition 5.1 twice and find $(s + (g + h)) (3) = s (3) + (g + h) (3) = s (3) + (g (3) + h (3))$ . Substituting the values for $s ⁢ open parenthesis 3 close parenthesis$ , $g ⁢ open parenthesis 3 close parenthesis$ and $h ⁢ open parenthesis 3 close parenthesis$ , we get $(s + (g + h)) (3) = 2 + (\frac{8}{3} + (- 6)) = - \frac{4}{3}$ , which, not surprisingly, matches our answer to the previous problem.
7. Once again, we find the expression $(\frac{f + h}{s}) (3)$ has more than two functions involved. As with all fractions, we treat `' as a grouping symbol and interpret $(\frac{f + h}{s}) (3) = \frac{(f + h) (3)}{s (3)} = \frac{f (3) + h (3)}{s (3)}$ . We compute $f (3) = 6 {(3)}^{2} - 2 (3) = 48$ and have $h (3) = - 6$ and $s ⁢ open parenthesis 3 close parenthesis equals 2$ from above. Hence, $(\frac{f + h}{s}) (3) = \frac{f (3) + h (3)}{s (3)} = \frac{48 + (- 6)}{2} = 21$ .
8. We need to need to exercise caution in parsing $(f (g - h)) (- 2)$ . In this context, $f$ , $g$ , and $h$ are all functions, so we interpret $(f (g - h))$ as the function and $minus 2$ as the argument. We view the function $f (g - h)$ as the product of $f$ and the function $g - h$ . Hence, $(f (g - h)) (- 2) = f (- 2) [(g - h) (- 2)] = f (- 2) [g (- 2) - h (- 2)]$ . We compute $f (- 2) = 6 {(- 2)}^{2} - 2 (- 2) = 28$ , and $g (- 2) = 3 - \frac{1}{- 2} = 3 + \frac{1}{2} = \frac{7}{2} = 3.5$ . Since $open parenthesis minus 2, 0.4 close parenthesis ∈ h$ , $h (- 2) = 0.4$ . Putting this altogether, we get $(f (g - h)) (- 2) = f (- 2) [(g - h) (- 2)] = f (- 2) [g (- 2) - h (- 2)] = 28 (3.5 - 0.4) = 28 (3.1) = 86.8$ .
1. To find the domain of $h g$ , we need to find the real numbers in both the domain of $h$ and the domain of $g$ . The domain of $h$ is ${- 3, - 2, 0, 3}$ and the domain of $g$ is $open brace t ∈ ℝ vertical bar t greater than 0 close brace$ so the only real number in common here is $3$ . Hence, the domain of $h g$ is $open brace 3 close brace$ , which may be small, but it's better than nothing.³
2. To find the domain of $\frac{f}{s}$ , we first note the domain of $f$ is all real numbers, but that the domain of $s$ , based on the graph, is just $open bracket minus 2, ∞ close parenthesis$ . Moreover, $s ⁢ open parenthesis t close parenthesis equals 0$ when $t equals 1$ , so we must exclude this value from the domain of $\frac{f}{s}$ . Hence, we are left with $open bracket minus 2, 1 close parenthesis union open parenthesis 1, ∞ close parenthesis$ .

By definition, $(f g) (x) = f (x) g (x)$ . We are given $f (x) = 6 x^{2} - 2 x$ and $g (t) = 3 - \frac{1}{t}$ so $g (x) = 3 - \frac{1}{x}$ . Hence,
$\begin{matrix} (f g) (x) & = & f (x) g (x) \\ = & (6 x^{2} - 2 x) (3 - \frac{1}{x}) \\ = & 18 x^{2} - 6 x^{2} (\frac{1}{x}) - 2 x (3) + 2 x (\frac{1}{x}) & distribute \\ = & 18 x^{2} - 6 x - 6 x + 2 \\ = & 18 x^{2} - 12 x + 2 \end{matrix}$
To find the domain of $f g$ , we note the domain of $f$ is all real numbers, $open parenthesis minus ∞, ∞ close parenthesis$ whereas the domain of $g$ is restricted to $open brace t ∈ ℝ vertical bar t greater than 0 close brace equals open parenthesis 0, ∞ close parenthesis$ . Hence, the domain of $f g$ is likewise restricted to $open parenthesis 0, ∞ close parenthesis$ . Note if we relied solely on the simplified formula for $(f g) (x) = 18 x^{2} - 12 x + 2$ , we would have obtained the incorrect answer for the domains of $f g$ .

To find an expression for $(\frac{g}{f}) (t) = \frac{f (t)}{g (t)}$ we first note $f (t) = 6 t^{2} - 2 t$ and $g (t) = 3 - \frac{1}{t}$ . Hence:

Table 5.1

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

=

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

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

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

simplify compound fractions

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

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

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

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

factor and cancel

\frac{1}{2 t^{2}}

Hence, $(\frac{g}{f}) (t) = \frac{1}{2 t^{2}} = \frac{1}{2} t^{- 2}$ . To find the domain of $\frac{g}{f}$ , a real number must be both in the domain of $g$ , $open parenthesis 0, ∞ close parenthesis$ , and the domain of $f$ , $open parenthesis minus ∞, ∞ close parenthesis$ so we start with the set $open parenthesis 0, ∞ close parenthesis$ . Additionally, we require $f ⁢ open parenthesis t close parenthesis not equal to 0$ . Solving $f ⁢ open parenthesis t close parenthesis equals 0$ amounts to solving $6 t^{2} - 2 t = 0$ or $2 t (3 t - 1) = 0$ . We find $t equals 0$ or $t = \frac{1}{3}$ which means we need to exclude these values from the domain. Hence, our final answer for the domain of $\frac{g}{f}$ is $(0, \frac{1}{3}) \cup (\frac{1}{3}, \infty)$ . Note that, once again, using the simplified formula for $(\frac{g}{f}) (t)$ to determine the domain of $\frac{g}{f}$ , would have produced erroneous results.

A few remarks are in order. First, in number parts through, we first encountered combinations of three functions despite Definition 5.1 only addressing combinations of two functions at a time. It turns out that function arithmetic inherits many of the same properties of real number arithmetic. For example, we showed above that $((s + g) + h) (3) = (s + (g + h)) (3)$ . In general, given any three functions $f$ , $g$ , and $h$ , $(f + g) + h = f + (g + h)$ that is, function addition is assocative. To see this, choose an element $x$ common to the domains of $f$ , $g$ , and $h$ . Then

$\begin{matrix} ((f + g) + h) (x) & = & (f + g) (x) + h (x) & definiton of ((f + g) + h) (x) \\ = & (f (x) + g (x)) + h (x) & definition of (f + g) (x) \\ = & f (x) + (g (x) + h (x)) & associative property of real number addition \\ = & f (x) + (g + h) (x) & definition of (g + h) (x) \\ = & (f + (g + h)) (x) & definition of (f + (g + h)) (x) \end{matrix}$

The key step to the argument is that $(f (x) + g (x)) + h (x) = f (x) + (g (x) + h (x))$ which is true courtesy of the associative property of real number addition. And just like with real number addition, because function addition is associative, we may write $f + g + h$ instead of $(f + g) + h$ or $f + (g + h)$ even though, when it comes down to computations, we can only add two things together at a time.⁴

For completeness, we summarize the properties of function arithmetic in the theorem below. The proofs of the properties all follow along the same lines as the proof of the associative property and are left to the reader. We investigate some additional properties in the exercises.

Theorem

Suppose $f$ , $g$ and $h$ are functions.

Commutative Law of Addition: $f + g = g + f$
Associative Law of Addition: $(f + g) + h = f + (g + h)$
Additive Identity: The function $Z ⁢ open parenthesis x close parenthesis equals 0$ satisfies: $f + Z = Z + f = f$ for all functions $f$ .
Additive Inverse: The function $F (x) = - f (x)$ for all $x$ in the domain of $f$ satisfies:
$f + F = F + f = Z .$
Commutative Law of Multiplication: $f g = g f$
Associative Law of Multiplication: $(f g) h = f (g h)$
Multiplicative Identity: The function $I ⁢ open parenthesis x close parenthesis equals 1$ satisfies: $f I = I f = f$ for all functions $f$ .
Multiplicative Inverse: If $f ⁢ open parenthesis x close parenthesis not equal to 0$ for all $x$ in the domain of $f$ , then $F (x) = \frac{1}{f (x)}$ satisfies:
$f F = F f = I$
Distributive Law of Multiplication over Addition: $f (g + h) = f g + f h$

In the next example, we decompose given functions into sums, differences, products and/or quotients of other functions. Note that there are infinitely many different ways to do this, including some trivial ones. For example, suppose we were instructed to decompose $f (x) = x + 2$ into a sum or difference of functions. We could write $f = g + h$ where $g (x) = x$ and $h ⁢ open parenthesis x close parenthesis equals 2$ or we could choose $g (x) = 2 x + 3$ and $h (x) = - x - 1$ . More simply, we could write $f = g + h$ where $g (x) = x + 2$ and $h ⁢ open parenthesis x close parenthesis equals 0$ . We'll call this last decomposition a `trivial' decomposition. Likewise, if we ask for a decomposition of $f ⁢ open parenthesis x close parenthesis equals 2 ⁢ x$ as a product, a nontrivial solution would be $f = g h$ where $g ⁢ open parenthesis x close parenthesis equals 2$ and $h (x) = x$ whereas a trivial solution would be $g ⁢ open parenthesis x close parenthesis equals 2 ⁢ x$ and $h ⁢ open parenthesis x close parenthesis equals 1$ . In general, non-trivial solutions to decomposition problems avoid using the additive identity, $0$ , for sums and differences and the multiplicative identity, $1$ , for products and quotients.

Example 2

For $f (x) = x^{2} - 2 x$ , find functions $g$ , $h$ and $k$ to decompose $f$ nontrivially as:
1. $f = g - h$
2. $f = g + h$
3. $f = g h$
4. $f = g (h - k)$
For $F (t) = \frac{2 t + 1}{\sqrt{t^{2} - 1}}$ , find functions $G$ , $H$ and $K$ to decompost $F$ nontrivially as:
1. $F = \frac{G}{H}$
2. $F = G H$
3. $F = G + H$
4. $F = \frac{G + H}{K}$

1. To decompose $f = g - h$ , we need functions $g$ and $h$ so $f (x) = (g - h) (x) = g (x) - h (x)$ . Given $f (x) = x^{2} - 2 x$ , one option is to let $g (x) = x^{2}$ and $h ⁢ open parenthesis x close parenthesis equals 2 ⁢ x$ . To check, we find $(g - h) (x) = g (x) - h (x) = x^{2} - 2 x = f (x)$ as required. In addition to checking the formulas match up, we also need to check domains. There isn't much work here since the domains of $g$ and $h$ are all real numbers which combine to give the domain of $f$ which is all real numbers.
2. In order to write $f = g + h$ , we need $f (x) = (g + h) (x) = g (x) + h (x)$ . One way to accomplish this is to write $f (x) = x^{2} - 2 x = x^{2} + (- 2 x)$ and identify $g (x) = x^{2}$ and $h (x) = - 2 x$ . To check, $(g + h) (x) = g (x) + h (x) = x^{2} - 2 x = f (x)$ . Again, the domains for both $g$ and $h$ are all real numbers which combine to give $f$ its domain of all real numbers.
3. To write $f = g h$ , we require $f (x) = (g h) (x) = g (x) h (x)$ . In other words, we need to factor $f (x)$ . We find $f (x) = x^{2} - 2 x = x (x - 2)$ , so one choice is to select $g (x) = x$ and $h (x) = x - 2$ . Then $(g h) (x) = g (x) h (x) = x (x - 2) = x^{2} - 2 x = f (x)$ , as required. As above, the domains of $g$ and $h$ are all real numbers which combine to give $f$ the correct domain of $open parenthesis minus ∞, ∞ close parenthesis$ .
4. We need to be careful here interpreting the equation $f = g (h - k)$ . What we have is an equality of functions so the parentheses here do not represent function notation here, but, rather function multiplication. The way to parse $g (h - k)$ , then, is the function $g$ times the function $h - k$ . Hence, we seek functions $g$ , $h$ , and $k$ so that $f (x) = [g (h - k)] (x) = g (x) [(h - k) (x)] = g (x) (h (x) - k (x))$ . From the previous example, we know we can rewrite $f (x) = x (x - 2)$ , so one option is to set $g (x) = h (x) = x$ and $k ⁢ open parenthesis x close parenthesis equals 2$ so that $[g (h - k)] (x) = g (x) [(h - k) (x)] = g (x) (h (x) - k (x)) = x (x - 2) = x^{2} - 2 x = f (x)$ , as required. As above, the domain of all constituent functions is $open parenthesis minus ∞, ∞ close parenthesis$ which matches the domain of $f$ .
1. To write $F = \frac{G}{H}$ , we need $G (t)$ and $H (t)$ so $F (t) = (\frac{G}{H}) (t) = \frac{G (t)}{H (t)}$ . We choose $G (t) = 2 t + 1$ and $H (t) = \sqrt{t^{2} - 1}$ . Sure enough, $(\frac{G}{H}) (t) = \frac{G (t)}{H (t)} = \frac{2 t + 1}{\sqrt{t^{2} - 1}} = F (t)$ as required. When it comes to the domain of $F$ , owing to the square root, we require $t squared minus 1 greater than or equal to 0$ . Since we have a denominator as well, we require $\sqrt{t^{2} - 1} \neq 0$ . The former requirement is the same restriction on $H$ , and the latter requirement comes from Definition 5.1. Starting with the domain of $G$ , all real numbers, and working through the details, we arrive at the correct domain of $F$ , $(- \infty, - 1) \cup (1, \infty)$ .
2. Next, we are asked to find functions $G$ and $H$ so $F (t) = (G H) (t) = G (t) H (t)$ . This means we need to rewrite the expression for $F (t)$ as a product. One way to do this is to convert radical notation to exponent notation:
  $F (t) = \frac{2 t + 1}{\sqrt{t^{2} - 1}} = \frac{2 t + 1}{{(t^{2} - 1)}^{\frac{1}{2}}} = (2 t + 1) {(t^{2} - 1)}^{- \frac{1}{2}} .$
  Choosing $G (t) = 2 t + 1$ and $H (t) = {(t^{2} - 1)}^{- \frac{1}{2}}$ , we see $(G H) (t) = G (t) H (t) = (2 t + 1) {(t^{2} - 1)}^{- \frac{1}{2}}$ as required. The domain restrictions on $F$ stem from the presence of the square root in the denominator - both are addressed when finding the domain of $H$ . Hence, we obtain the correct domain of $F$ .
3. To express $F$ as a sum of functions $G$ and $H$ , we could rewrite
  $F (t) = \frac{2 t + 1}{\sqrt{t^{2} - 1}} = \frac{2 t}{\sqrt{t^{2} - 1}} + \frac{1}{\sqrt{t^{2} - 1}},$
  so that $G (t) = \frac{2 t}{\sqrt{t^{2} - 1}}$ and $H (t) = \frac{1}{\sqrt{t^{2} - 1}}$ . Indeed, $(G + H) (t) = G (t) + H (t) = \frac{2 t}{\sqrt{t^{2} - 1}} + \frac{1}{\sqrt{t^{2} - 1}} = \frac{2 t + 1}{\sqrt{t^{2} - 1}} = F (t)$ , as required. Moreover, the domain restrictions for $F$ are the same for both $G$ and $H$ , so we get agreement on the domain, as required.
4. Last, but not least, to write $F = \frac{G + H}{K}$ , we require $F (t) = (\frac{G + H}{K}) (t) = \frac{(G + H) (t)}{K (t)} = \frac{G (t) + H (t)}{K (t)}$ . Identifying $G ⁢ open parenthesis t close parenthesis equals 2 ⁢ t$ , $H ⁢ open parenthesis t close parenthesis equals 1$ , and $K (t) = \sqrt{t^{2} - 1}$ , we get
  $(\frac{G + H}{K}) (t) = \frac{(G + H) (t)}{K (t)} = \frac{G (t) + H (t)}{K (t)} = \frac{2 t + 1}{\sqrt{t^{2} - 1}} = F (t) .$
  Concerning domains, the domain of both $G$ and $H$ are all real numbers, but the domain of $K$ is restricted to $t squared minus 1 greater than or equal to 0$ . Coupled with the restriction stated in Definition 5.1 that $K ⁢ open parenthesis t close parenthesis not equal to 0$ , we recover the domain of $F$ , $(- \infty, - 1) \cup (1, \infty)$ .

The Arithmetic of Change

Recall the average rate of change of a function over the interval $[a, b]$ is the slope of the line connecting the two points $(a, f (a))$ and $(b, f (b))$ and is given by

$\frac{Δ [f (x)]}{Δ x} = \frac{f (b) - f (a)}{b - a} .$

For the purposes of this section, consider a function $f$ defined over an interval containing $x$ and $x + Δ x$ where $Δ ⁢ x not equal to 0$ . The average rate of change of $f$ over the interval $[x, x + Δ x]$ is thus given by the formula:⁵

$\frac{Δ [f (x)]}{Δ x} = \frac{f (x + Δ x) - f (x)}{Δ x}, Δ x \neq 0 .$

Our aim in this section is to develop formulas which relate the rate of change of arithmetic combinations of functions to the rates of change of the component functions. Our first step is to study the difference operator ` $Δ$ ' and how it works with the standard arithmetic operations.

In general, if $u$ is some quantity which assumes two values in a particular order, say $u sub 1$ (the `first' or `initial' value) and $u sub 2$ (the `second' or `final' value), then $Δ u = u_{2} - u_{1}$ . For example, if $u$ represents the temperature of an object before ( $u sub 1$ ) heat is applied and after ( $u sub 2$ ) heat is applied, $Δ u = u_{2} - u_{1}$ represents the increase in temperature of the object.

In the context of functions and rates of change, $u$ is the function $f$ defined on the interval $[x, x + Δ x]$ with $u sub 1 equals f ⁢ open parenthesis x close parenthesis$ and $u_{2} = f (x + Δ x)$ . Here, $Δ u = u_{2} - u_{1} = f (x + Δ x) - f (x) = Δ [f (x)]$ .

Suppose we have two quantities, $u$ and $v$ with $Δ u = u_{2} - u_{1}$ and $Δ v = v_{2} - v_{1}$ . What do we mean by $Δ [u + v]$ ? The initial value of the sum $u + v$ would be the sum of the initial values $u sub 1 plus v sub 1$ . Likewise, the final value of the sum would be the sum of the final values $u sub 2 plus v sub 2$ . Hence:

$\begin{matrix} Δ [u + v] & = & (u_{2} + v_{2}) - (u_{1} + v_{1}) \\ = & u_{2} + v_{2} - u_{1} - v_{1} \\ = & (u_{2} - u_{1}) + (v_{2} - v_{1}) \\ = & Δ u + Δ v \end{matrix}$

A similar calculation gives $Δ [u - v] = Δ u - Δ v$ .

Let's turn our attention to products. We have $Δ [u v] = u_{2} v_{2} - u_{1} v_{1}$ . We'd like to express $Δ [u v]$ in terms of $Δ u$ and $Δ v$ and there seems to be no way obvious way to do that. We take to some geometric reasoning for inspiration. Let's assume the all the quantities we're working with are positive.

We imagine the product $u sub 1 ⁢ v sub 1$ as being the area of a rectangle with width $u sub 1$ and length $v sub 1$ . Likewise, the product $u sub 2 ⁢ v sub 2$ is the area of a (larger) rectangle with width $u sub 2$ and length $v sub 2$ .

Coordinate-plane figure. — **Figure 5.20**

From $Δ u = u_{2} - u_{1}$ , we get $u_{2} = u_{1} + Δ u$ and, likewise, $v_{2} = v_{1} + Δ v$ . Doing so allows us to decompose the larger rectangle into four smaller rectangles.

Using this schematic, we see the area $u sub 2 ⁢ v sub 2$ is the sum of the areas of four smaller rectangles:

$u_{2} v_{2} = u_{1} v_{1} + v_{1} Δ u + u_{1} Δ v + Δ u Δ v$

Hence, $Δ [u v] = u_{2} v_{2} - u_{1} v_{1} = v_{1} Δ u + u_{1} Δ v + Δ u Δ v$ .

To prove this formula holds in general, we can substitute $u_{2} = u_{1} + Δ u$ and $v_{2} = v_{1} + Δ v$ into $Δ [u v] = u_{2} v_{2} - u_{1} v_{1}$ and simplify. We leave the details to the reader.⁶

Next, we turn our attention to quotients. We begin with: $Δ [\frac{u}{v}] = \frac{u_{2}}{v_{2}} - \frac{u_{1}}{v_{1}}$ .

Instead of appealing to geometric reasoning here,⁷ we take a cue from the previous discussion and substitute $u_{2} = u_{1} + Δ u$ and $v_{2} = v_{1} + Δ v$ and set about getting a common denominator:

$\begin{matrix} Δ [\frac{u}{v}] & = & \frac{u_{2}}{v_{2}} - \frac{u_{1}}{v_{1}} \\ = & \frac{u_{1} + Δ u}{v_{1} + Δ v} - \frac{u_{1}}{v_{1}} \\ = & \frac{(u_{1} + Δ u) v_{1}}{(v_{1} + Δ v) v_{1}} - \frac{u_{1} (v_{1} + Δ v)}{v_{1} (v_{1} + Δ v)} \\ = & \frac{u_{1} v_{1} + v_{1} Δ u - u_{1} v_{1} - u_{1} Δ v}{v_{1} (v_{1} + Δ v)} \\ = & \frac{v_{1} Δ u - u_{1} Δ v}{v_{1} (v_{1} + Δ v_{1})} \end{matrix}$

We summarize these results in the following theorem.

Theorem

Suppose $Δ u = u_{2} - u_{1}$ and $Δ v = v_{2} - v_{1}$ :

The Sum Rule for Change: $Δ [u + v] = Δ u + Δ v$
The Difference Rule for Change: $Δ [u - v] = Δ u - Δ v$
The Product Rule for Change: $Δ [u v] = v_{1} Δ u + u_{1} Δ v + Δ u Δ v$
- The Constant Multiple Rule for Change: If $c$ is a constant, then $Δ [c u] = c Δ u$ .
The Quotient Rule for Change: $Δ [\frac{u}{v}] = \frac{v_{1} Δ u - u_{1} Δ v}{v_{1} (v_{1} + Δ v_{1})}$

You'll note we've called out a special case for the Product Rule, the Constant Multiple Rule. If one of the factors is a constant, $c$ , then $Δ ⁢ c equals 0$ (since constants don't change.)

In the following example, we use the Quotient Rule for Change to help approximate the propagated error when using measured quantities (with associated uncertainties) in calculations.⁸

Example 3

The density of a substance, $ρ$ , is calculated by dividing its mass, $m$ , by its volume, $V$ : $ρ = \frac{m}{V}$ . A scientist collects $5$ milliliters (mL) of a substance and determines its mass to be $68.2$ grams (g). She computes the density as: $ρ = \frac{68.2 g}{5 mL} = 13.64 \frac{g}{mL}$ .

Since every measurement in the lab has an associated uncertainty, she notes the pipet she used to measure the volume has an uncertainty of $\pm 0.125$ mL and the balance she used to mass the substance has an uncertainty of $\pm 0.01$ g. This means the actual volume measurement can be anywhere from as low as $5 - 0.125 = 4.875$ mL and as high as $5 + 0.125 = 5.125$ mL. Likewise, the actual mass of the substance can be anywhere from $68.2 - 0.01 = 68.19$ g to $68.2 + 0.01 = 68.21$ g. Our goal is to help estimate the associated uncertainty for the density, $ρ$ .

Use Theorem 5.2 to find an expression for the uncertainty in the volume $Δ ρ$ produced as a result in the uncertainties in the measurements of mass, $Δ m$ , and volume, $Δ V$ .
Calculate $\frac{Δ ρ}{ρ}$ and interpret your answer.

In this scenario, we have two quantities, the volume, $V$ and the mass, $m$ . We'll take $V sub 1$ and $m sub 1$ to be the measured values of volume and mass, respectively, and use the uncertainties in each of the respective measurements as $Δ V$ and $Δ m$ . Since $Δ ρ = Δ [\frac{m}{V}]$ , using the Quotient Rule from Theorem 5.2 gives:
$Δ ρ = \frac{V_{1} Δ m - m_{1} Δ V}{V_{1} (V_{1} + Δ V)} .$
Substituting $V sub 1 equals 5$ mL, $m sub 1 equals 68.2$ g, $Δ V = \pm 0.125$ mL and $Δ m = \pm 0.01$ g gives:
$Δ ρ = \frac{(\pm 0.01 g) (5 mL) - (68.2 g) (\pm 0.125 mL)}{(5 mL) (5 mL \pm 0.125 mL)} .$
Since we have no idea the exact value of each uncertainty, we need to make a judgement call as to which of the sign values, ` $\pm$ ', to use. To get the largest (most conservative) answer for $Δ ρ$ , we select the $\pm$ which generate the largest numerator and smallest denominator:
$Δ ρ = \frac{(+ 0.01 g) (5 mL) - (68.2 g) (- 0.125 mL)}{(5 mL) (5 mL - 0.125 mL)} = \frac{343 g mL}{975 {mL}^{2}} \approx 0.3517 \frac{g}{mL}$
Hence, $\frac{Δ ρ}{ρ} \approx \frac{0.3517 \frac{g}{mL}}{13.64 \frac{g}{mL}} \approx 0.0258 = 2.58 %$ .
We may interpret this as the uncertainties in the measurements for mass and volume in this situation could produce up to a $2.58 %$ error in the calculated density,

In order to establish formulas for the average rate of change for functions, we substitute $f (x)$ for $u sub 1$ and $g (x)$ for $v sub 1$ and divide each of the expressions in Theorem 5.2 by $Δ x$ . For example, if we to find an expression for the average rate of change of $f g$ in terms of $f$ , $g$ , and their respective average rates of change:

$\frac{Δ [(f g) (x)]}{Δ x} = \frac{Δ [f (x)] g (x) + f (x) Δ [g (x)] + Δ [f (x)] Δ [g (x)]}{Δ x} = \frac{Δ [f (x)]}{Δ x} g (x) + f (x) \frac{Δ [g (x)]}{Δ x} + \frac{Δ [f (x)] Δ [g (x)]}{Δ x} .$

Note that with the last term, we may associate the ` $Δ x$ ' with either of the factors in the numerator:

$\frac{Δ [f (x)] Δ [g (x)]}{Δ x} = \frac{Δ [f (x)]}{Δ x} Δ [g (x)] = Δ [f (x)] \frac{Δ [g (x)]}{Δ x} .$

Either way, we've managed to express the average rate of change of the function $f g$ in terms of the changes and rates of change of $f$ and $g$ .

In the result below, we abbreviate the average rate of change as `ARoC' for convenience.

Theorem

Suppose $f$ and $g$ are functions defined on an interval containing $x$ and $x + Δ x$ , $Δ ⁢ x not equal to 0$ :

The Sum Rule for ARoC: $ARoC [(f + g) (x)] = ARoC [f (x)] + ARoC [g (x)]$
The Difference Rule for ARoC: $ARoC [(f - g) (x)] = ARoC [f (x)] - ARoC [g (x)]$
The Product Rule for ARoC:
$\begin{matrix} ARoC [(f g) (x)] & = & ARoC [f (x)] g (x) + f (x) ARoC [g (x)] + ARoC [f (x)] g (x) \\ = & ARoC [f (x)] g (x) + f (x) ARoC [g (x)] + f (x) ARoC [g (x)] \end{matrix}$
- The Constant Multiple Rule for ARoC: If $c$ is a constant, then $ARoC [c f (x)] = c ARoC [f (x)]$ .
The Quotient Rule for ARoC:
$ARoC [(\frac{f}{g}) (x)] = \frac{ARoC [f (x)] g (x) - f (x) ARoC [g (x)]}{g (x) (g (x) + Δ [g (x)])}$

Our final example revisits the scenario in Exercise in Section 2.1.

Example 4

The cost, in dollars, $C (x)$ , to produce $x$ `PortaBoy' handheld game systems is given by: $C (x) = .03 x^{3} - 4.5 x^{2} + 225 x + 250$ . The revenue generated by selling $x$ of the systems, $R (x)$ , also in dollars, is given by $R (x) = - 1.5 x^{2} + 250 x$ .

Find and interpret the average rate of change of $C$ and $R$ over the interval $open bracket 70, 71 close bracket$ .
Use Theorem 5.3 to determine the average rate of change of the profit function, $P$ over the interval $open bracket 70, 71 close bracket$ using your answers to part. What does your answer suggest?

We find $ARoC [C (x)] = \frac{C (71) - C (70)}{71 - 70} = \frac{4277.83 - 4240}{1} = 37.83$ . This means that as we move from producing $70$ to $71$ PortaBoy systems the cost will increase by $$ 37.83$ per system.
For revenue, we find $ARoC [R (x)] = \frac{R (71) - R (70)}{71 - 70} = \frac{10188.5 - 10150}{1} = 38.5$ . This means that as we move from selling $70$ to $71$ PortaBoy systems, the revenue generated will increase by $$ 38.5$ per system.
Since $P (x) = R (x) - C (x)$ , the Difference Rule of Theorem 5.3 gives $ARoC [P (x)] = ARoC [R (x)] - ARoC [C (x)]$ . In this case, we'd get $ARoC [P (x)] = 38.5 - 37.83 = 0.67$ . This means as we move from producing and selling $70$ to $71$ PortaBoy systems, the profit generated will increase by just $67$ cents per system. At this point, the increase in revenue is nearly balanced out by the increase in cost. Since costs typically continue to rise as the number of items is produced while the revenue falls as we try to sell more items,⁹ we are likely near a maximum point with the profit. A quick check of the graph on Desmos confirms our suspicions.
Figure 5.23

Note that in Example 5.2.4, since $Δ ⁢ x equals 1$ , the average rate of change for the cost moving from producing $70$ to $71$ systems, $\frac{C (71) - C (70)}{71 - 70}$ is the same numerical value as the additional cost incurred by producing the $71$ st system, $C ⁢ open parenthesis 71 close parenthesis minus C ⁢ open parenthesis 70 close parenthesis$ . The same goes for the revenue and profit calculations. This is the concept of marginal analysis is studied at length in Economics and Business Calculus classes.¹⁰ For us, it's time for some Exercises.

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

Exercises - refer to the functions $f$ and $g$ whose graphs are below.

$(f + g) (- 4)$
$open parenthesis f plus g close parenthesis ⁢ open parenthesis 0 close parenthesis$
$open parenthesis f minus g close parenthesis ⁢ open parenthesis 4 close parenthesis$
$open parenthesis f ⁢ g close parenthesis ⁢ open parenthesis minus 4 close parenthesis$
$open parenthesis f ⁢ g close parenthesis ⁢ open parenthesis minus 2 close parenthesis$
$open parenthesis f ⁢ g close parenthesis ⁢ open parenthesis 4 close parenthesis$
$(\frac{f}{g}) (0)$
$(\frac{f}{g}) (2)$
$(\frac{g}{f}) (- 1)$
Find the domains of $f + g$ , $f - g$ , $f g$ , $\frac{f}{g}$ and $\frac{g}{f}$ .

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 by

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

Compute the indicated value if it exists.

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

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 -, write the given function as a nontrivial decomposition of functions as directed.

For $p (z) = 4 z - z^{3}$ , find functions $f$ and $g$ so that $p = f - g$ .
For $p (z) = 4 z - z^{3}$ , find functions $f$ and $g$ so that $p = f + g$ .
For $g (t) = 3 t | 2 t - 1 |$ , find functions $f$ and $h$ so that $g = f h$ .
For $r (x) = \frac{3 - x}{x + 1}$ , find functions $f$ and $g$ so $r = \frac{f}{g}$ .
For $r (x) = \frac{3 - x}{x + 1}$ , find functions $f$ and $g$ so $r = f g$ .
Can $f (x) = x$ be decomposed as $f = g - h$ where $g (x) = x + \frac{1}{x}$ and $h (x) = \frac{1}{x}$ ?
Discuss with your classmates how to phrase the quantities revenue and profit in Definition 1.12 terms of function arithmetic as defined in Definition 5.1.
In this exercise, we explore decomposing a function into its positive and negative parts. Given a function $f$ , we define the positive part of $f$ , denoted $f_{+}$ and negative part of $f$ , denoted $f_{-}$ by:
$f_{+} (x) = \frac{f (x) + | f (x) |}{2}, and f_{-} (x) = \frac{f (x) - | f (x) |}{2} .$
1. Using a graphing utility, graph each of the functions $f$ below along with $f_{+}$ and $f_{-}$ .
  - $f (x) = x - 3$
  - $f (x) = x^{2} - x - 6$
  - $f (x) = 4 x - x^{3}$
  Why is $f_{+}$ called the `positive part' of $f$ and $f_{-}$ called the `negative part' of $f$ ?
2. Show that $f = f_{+} + f_{-}$ .
3. Use Definition 1.9 to rewrite the expressions for $f_{+} (x)$ and $f_{-} (x)$ as piecewise defined functions.
Let $U$ be the unit step function defined in Exercise in Section 1.2. For each function $f (t)$ below:
- Write $(U f) (t)$ as a piecewise-defined function.
- Graph $y = f (t)$ and $y = (U f) (t)$ .
1. $f (t) = t - 3$
2. $f (t) = | t + 2 |$
3. $f (t) = {(t - 1)}^{2}$
1. $f (t) = {(t + 1)}^{- 1}$
2. $f (t) = \sqrt[3]{t - 1}$
3. $f (t) = {(t - 2)}^{\frac{2}{3}}$
1. Write a general formula for $(U f) (t)$ for a function $f$ . (Assume the domain of $f$ is $open parenthesis minus ∞, ∞ close parenthesis$ .)
2. Explain how to obtain the graph of $y = (U f) (t)$ from $y = f (t)$ .
3. The function $U (t)$ is used to model a change in state from `off' to `on' (like flipping a light switch.) How does this relate to your observations?
4. Use the graph of $y = f (t)$ below to graph $y = (U f) (t)$ .
  Figure 5.26 $y = f (t)$
Use Example 5.2.3 as a guide to help find the following uncertainties.
1. A chemist combines the solutions from two graduated cylinders into a beaker. The volume of the first solution, $A$ , an acid, is read as $A_{1} = 101 \pm 0.5$ milliliters (mL). The volume of the second solution, a base, $B$ , is measured to be $B_{1} = 16 \pm 0.5$ mL. Estimate the percent propagated error in calculating the volume of the combined solution as $V = A_{1} + B_{1} = 101 + 16 = 117$ mL.
2. A student measures the length, $ℓ$ , and width, $w$ , of a piece of paper. They find $ℓ_{1} = 280 \pm 0.5$ millimeters (mm) $w_{1} = 216 \pm 0.5$ mm. Estimate the percent propagated error in calculating the area of the piece of paper as $A = ℓ_{1} w_{1} = 280 \times 216 = 60480 {mm}^{2}$ .
3. An airplane passenger observers a car travel a distance $d_{1} = 1320 \pm 2$ feet (ft) in time $t_{1} = 15 \pm 0.5$ seconds (s). Estimate the percent propagated error in calculating the speed of the car as $v = \frac{d_{1}}{t_{1}} = \frac{1320}{15} = 88 \frac{ft}{s}$ .
Let us return to Example 5.2.4 where $C (x) = .03 x^{3} - 4.5 x^{2} + 225 x + 250$ denotes the cost, in dollars, of producing $x$ PortaBoy game systems. Recall the average cost ¹¹ is defined as $\bar{C} (x) = \frac{C (x)}{x}$ , $x greater than 0$ , is the cost per item.
1. Find and interpret $C bar ⁢ open parenthesis 75 close parenthesis$ .
2. Define the marginal cost $M C (x) = C (x + 1) - C (x)$ . Find and interpret $M ⁢ C ⁢ open parenthesis 75 close parenthesis$ .
3. How do your answers to parts and compare?
4. Graph $y = \bar{C} (x)$ with help from a graphing utility. What is happening graphically near $x equals 75$ ?
5. Use Theorem 5.3 to show that, in general, $ARoC ⁢ open bracket C bar ⁢ open parenthesis x close parenthesis close bracket equals 0$ when $M C (x) = \bar{C} (x)$ .
  HINT: Note that, by definition, $M C (x) = C (x + 1) - C (x) = Δ [C (x)]$ when $Δ ⁢ x equals 1$ .
  Hence, $ARoC [C (x)] = \frac{Δ [C (x)]}{Δ x} = \frac{Δ [C (x)]}{1} = Δ [C (x)] = M C (x)$ in this case …

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 $g (x) = \frac{1}{x^{2} + 1}$
- $(f + g) (2) = \frac{26}{5}$
- $(f - g) (- 1) = \frac{3}{2}$
- $(g - f) (1) = - \frac{3}{2}$
- $(f g) (\frac{1}{2}) = 1$
- $(\frac{f}{g}) (0) = 1$
- $(\frac{g}{f}) (- 2) = \frac{1}{25}$
$(f + g) (- 4) = - 5$
$(f + g) (0) = 5$
$(f - g) (4) = - 5$
$(f g) (- 4) = 6$
$(f g) (- 2) = 0$
$(f g) (4) = - 6$
$(\frac{f}{g}) (0) = \frac{3}{2}$
$(\frac{f}{g}) (2) = 0$
$(\frac{g}{f}) (- 1) = 0$
The domains of $f + g$ , $f - g$ and $f g$ are all $open bracket minus 4, 4 close bracket$ . The domain of $\frac{f}{g}$ is $[- 4, - 1) \cup (- 1, 4]$ and the domain of $\frac{g}{f}$ is $[- 4, - 2) \cup (- 2, 2) \cup (2, 4]$ .
$(f + g) (- 3) = 2$
$(f - g) (2) = 3$
$(f g) (- 1) = 0$
$(g + f) (1) = 0$
$(g - f) (3) = 3$
$(g f) (- 3) = - 8$
$(\frac{f}{g}) (- 2)$ does not exist
$(\frac{f}{g}) (- 1) = 0$
$(\frac{f}{g}) (2) = 4$
$(\frac{g}{f}) (- 1)$ does not exist
$(\frac{g}{f}) (3) = - 2$
$(\frac{g}{f}) (- 3) = - \frac{1}{2}$
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: $open parenthesis 5, ∞ close parenthesis$
One solution is $f ⁢ open parenthesis z close parenthesis equals 4 ⁢ z$ and $g (z) = z^{3}$ .
One solution is $f ⁢ open parenthesis z close parenthesis equals 4 ⁢ z$ and $g (z) = - z^{3}$ .
One solution is $f ⁢ open parenthesis t close parenthesis equals 3 ⁢ t$ and $h (t) = | 2 t - 1 |$
One solution is $f (x) = 3 - x$ and $g (x) = x + 1$ .
One solution is $f (x) = 3 - x$ and $g (x) = {(x + 1)}^{- 1}$ .
No. The equivalence does not hold when $x equals 0$ .
1. $(f_{+} + f_{-}) (x) = f_{+} (x) + f_{-} (x) = \frac{f (x) + | f (x) |}{2} + \frac{f (x) - | f (x) |}{2} = \frac{2 f (x)}{2} = f (x)$ .
1. $(U f) (t) = {\begin{matrix} 0 & if t < 0, \\ t - 3 & if t \geq 0 . \end{matrix}$
  Figure 5.27 $y = f (t)$
  Figure 5.28 $y = (U f) (t)$
2. $(U f) (t) = {\begin{matrix} 0 & if t < 0, \\ | t + 2 | = t + 2 & if t \geq 0 . \end{matrix}$
  Figure 5.29 $y = f (t)$
  Figure 5.30 $y = (U f) (t)$
3. $(U f) (t) = {\begin{matrix} 0 & if t < 0, \\ {(t - 1)}^{2} & if t \geq 0 . \end{matrix}$
  Figure 5.31 $y = f (t)$
  Figure 5.32 $y = (U f) (t)$
1. $(U f) (t) = {\begin{matrix} 0 & if t < 0, t \neq - 1 \\ {(t + 1)}^{- 1} & if t \geq 0 . \end{matrix}$
  Figure 5.33 $y = f (t)$
  Figure 5.34 $y = (U f) (t)$ - note the hole at $open parenthesis minus 1, 0 close parenthesis$
2. $(U f) (t) = {\begin{matrix} 0 & if t < 0, \\ \sqrt[3]{t - 1} & if t \geq 0 . \end{matrix}$
  Figure 5.35 $y = f (t)$
  Figure 5.36 $y = (U f) (t)$
3. $(U f) (t) = {\begin{matrix} 0 & if t < 0, \\ {(t - 2)}^{\frac{2}{3}} & if t \geq 0 . \end{matrix}$
  Figure 5.37 $y = f (t)$
  Figure 5.38 $y = (U f) (t)$
1. $open parenthesis U ⁢ f close parenthesis ⁢ open parenthesis t close parenthesis equals open brace 0, if t less than 0,; f ⁢ open parenthesis t close parenthesis, if t greater than or equal to 0$
2. The graph of $(U f) (t)$ is $y equals 0$ for $t less than 0$ and $y = f (t)$ for $t greater than or equal to 0$ .
3. The unit step function keeps the function `off' until $t equals 0$ then turns the function `on' for $t greater than or equal to 0$ .
4. Figure 5.39 $y = (U f) (t)$
1. $Δ V = Δ [A + B] = Δ A + Δ B = \pm 0.5 mL + \pm 0.5 mL = \pm 1 mL$ .
  $\frac{Δ V}{V} = \pm \frac{1}{117} \approx 0.85 %$ .
2. $Δ A = Δ [ℓ w] = w_{1} Δ ℓ + ℓ_{1} Δ w + Δ ℓ Δ w = (216 mm) (\pm 0.5 mm) + (280 mm) (\pm 0.5 mm) + (\pm 0.5 mm) (\pm 0.5 mm) = \pm 248.25 {mm}^{2}$ .
  $\frac{Δ A}{A} = \pm \frac{248.25}{60480} \approx 0.41 %$
3. $Δ v = Δ [\frac{d}{t}] = \frac{t_{1} Δ d - d_{1} Δ t}{t_{1} (t_{1} + Δ t)} = \frac{(15 s) (\pm 2 ft) - (1320 ft) (\pm 0.5 s)}{(15 s) (15 \pm 0.5 s)} = \pm \frac{92}{29} \frac{ft}{s} \approx 3.17 \frac{ft}{s}$ .
  $\frac{Δ v}{v} \approx \pm \frac{3.17}{88} \approx 3.60 %$
1. $C bar ⁢ open parenthesis 75 close parenthesis approximately equals 59.58$ . When making $75$ systems, the cost per system is approximately $$ 59.58$ .
2. $M C (75) = C (76) - C (75) = 58.53$ . It costs an additional $$ 58.53$ to make the $76$ th system.
3. $C bar ⁢ open parenthesis 75 close parenthesis$ and $M ⁢ C ⁢ open parenthesis 75 close parenthesis$ appear to be `pretty close.'
4. The graph $y = \bar{C} (x)$ has a local (absolute) minimum right near $x equals 75$ .
5. Per Theorem 5.3, since
  $\begin{matrix} ARoC [\bar{C} (x)] = ARoC [\frac{C (x)}{x}] & = & \frac{ARoC [C (x)] x - C (x) ARoC [x]}{x (x + Δ x)} \\ = & \frac{ARoC [C (x)] x - C (x) (1)}{x (x + Δ x)} & Since ARoC [x] = \frac{Δ x}{Δ x} = 1 \end{matrix}$
  If $ARoC ⁢ open bracket C bar ⁢ open parenthesis x close parenthesis close bracket equals 0$ , then the numerator, $ARoC [C (x)] x - C (x) = 0$ . Solving for $ARoC [C (x)]$ , we get $ARoC [C (x)] = \frac{C (x)}{x} = \bar{C} (x)$ . If we are working with a whole number of items, the smallest meaningful value of $Δ x$ is $1$ , in which case $ARoC [C (x)] = M C (x)$ . Hence, $ARoC ⁢ open bracket C bar ⁢ open parenthesis x close parenthesis close bracket equals 0$ when $M C (x) = \bar{C} (x)$ , that is, when the marginal cost and average cost are the same. At this point, the graph of $y = \bar{C} (x)$ levels off (at a minumum.)¹² Can you reason why this creates a minimum?

We could have just as easily called this new function $S (x)$ for `sum' of $f$ and $g$ and defined $S$ by $S (x) = f (x) + g (x)$ . ↩
see Section A.1. ↩
Since $(h g) (3) = h (3) g (3) = (- 6) (\frac{8}{3}) = - 16$ , we can write $h g = {(3, - 16)}$ . ↩
Addition is a `binary' operation - meaning it is defined only on two objects at once. Even though we write $1 + 2 + 3 = 6$ , mentally, we add just two of numbers together at any given time to get our answer: for example, $1 + 2 + 3 = (1 + 2) + 3 = 3 + 3 = 6$ . ↩
assuming $Δ ⁢ x greater than 0$ ; otherwise, the interval is $[x + Δ x, x]$ . We get the same formula for the difference quotient either way. ↩
Why not do this from the start, then? Carl trained as a geometric topologist. ↩
If you come up with or know of a nice geometric argument and don't mind sharing, feel free to contact Carl . ↩
The adjective `propagated' here means than when we use measured quantities in calculations, the uncertainties in the measured quantities will produce, or `propagate' uncertainty in the calculated quantity. ↩
we've seen this before: to sell more, we lower the price which, in turn, lowers revenue … ↩
We'll do some more exploration as well. See, for example, Exercise . ↩
First mentioned in Definition 3.8 in Section 3.1. ↩
We'll have more to say on this after Section 6.3 in Exercise . ↩

Adapted from Precalculus, Preliminary 4th Edition (integrated calculus), 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.