1.5: Function Arithmetic

Topics Reviewed in a Typical Corequisite Course

The following topics related to the material in this section are assumed prerequisites for this course. They are only covered if you are enrolled in a class with corequisite support:

• Rationalizing numerators and denominators
• Solving quadratic equations using factoring

Topics that must be assumed prerequisites (even for a class with corequisite support) are:

• Adding, subtracting, multiplying, and dividing polynomial and rational functions

Learning Objectives

• Perform functional arithmetic.
• Find the difference quotient for a given function.
• Continue the theme of using your mathematical knowledge by using function notation and combinations of functions in an applied model.

Combinations of Functions

In Section 1.4, we used function notation to make sense of expressions such as \( f(x)+2\) and \(2f (x)\) for a given function \(f\). It would seem natural, then, that functions should have their own arithmetic 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.

Theorem: Functional 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\[ (f+g)(x)=f(x)+g(x). \nonumber \]
• The difference of \(f\) and \(g\), denoted \(f − g\), is the function defined by the formula\[ (f-g)(x)=f(x)-g(x). \nonumber \]
• The product of \(f\) and \(g\), denoted \(fg\), is the function defined by the formula\[ (f g)(x)=f(x) g(x). \nonumber \]
• The quotient of \(f\) and \(g\), denoted \(\dfrac{f}{g}\), is the function defined by the formula\[ \left(\dfrac{f}{g}\right)(x)=\dfrac{f(x)}{g(x)}, \nonumber \]provided \(g(x) \neq 0\).

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 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 \( \PageIndex{1} \)

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

1. Find \((f+g)(-1)\)
2. Find \((f g)(2)\)
3. Find the domain of \(g − f\) then find and simplify a formula for \((g − f)(x)\).
4. Find the domain of \(\left(\frac{g}{f}\right)\) then find and simplify a formula for \(\left(\frac{g}{f}\right)(x)\).

Solutions

1. To find \((f+g)(-1)\) we first find \(f(−1) = 8\) and \(g(−1) = 4\). By definition, we have that \[(f + g)(−1) = f(−1) + g(−1) = 8 + 4 = 12.\nonumber\]
2. To find \((fg)(2)\), we first need \(f(2)\) and \(g(2)\). Since \(f(2) = 20\) and \(g(2)=\frac{5}{2}\), our formula yields \[(f g)(2)=f(2) g(2)=(20)\left(\dfrac{5}{2}\right)=50.\nonumber\]
3. 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 \((-\infty, 0) \cup(0, \infty)\). 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, \((-\infty, 0) \cup(0, \infty)\).
   
   A second method is to analyze the formula for \((g − f)(x)\) before simplifying looking for the usual domain issues. In this case\[ (g-f)(x)=g(x)-f(x)=\left(3-\dfrac{1}{x}\right)-\left(6 x^{2}-2 x\right), \nonumber \]so we find, as before, the domain \((-\infty, 0) \cup(0, \infty)\).
   
   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\[ \begin{array}{rcl}
   (g-f)(x) & = & g(x)-f(x) \\
   \\
   & = & \left(3-\dfrac{1}{x}\right)-\left(6 x^{2}-2 x\right) \\
   \\
   & = & 3-\dfrac{1}{x}-6 x^{2}+2 x \\
   \\
   & = & \dfrac{3 x}{x}-\dfrac{1}{x}-\dfrac{6 x^{3}}{x}+\dfrac{2 x^{2}}{x} \\
   \\
   & = & \dfrac{3 x-1-6 x^{3}-2 x^{2}}{x} \\
   \\
   & = & \dfrac{-6 x^{3}-2 x^{2}+3 x-1}{x} \\
   \end{array} \nonumber \]
4. As in the previous example, we have two ways to approach finding the domain \(\frac{g}{f}\). First, we can find the domain of \(g\) and \(f\) separately, and find the intersection of these two sets. In addition, since \(\left(\frac{g}{f}\right)(x)=\frac{g(x)}{f(x)}\), 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 \((-\infty, 0) \cup(0, \infty)\) and the domain of \(f\) is \((-\infty, \infty)\). Setting \(f(x) = 0\) 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 \(x = 0\) and \(x=\frac{1}{3}\), or \[(-\infty, 0) \cup\left(0, \dfrac{1}{3}\right) \cup\left(\dfrac{1}{3}, \infty\right).\nonumber\]Alternatively, we may proceed as above and analyze the expression \(\left(\frac{g}{f}\right)(x)=\frac{g(x)}{f(x)}\) before simplifying. In this case,\[ \left(\dfrac{g}{f}\right)(x)=\dfrac{g(x)}{f(x)}=\dfrac{3-\frac{1}{x}}{6 x^{2}-2 x}. \nonumber \]We see immediately from the "little" denominator that \(x \neq 0\). To keep the "big" denominator away from \(0\), we solve \(6 x^{2}-2 x=0\) and get \(x=0\) or \(x=\frac{1}{3}\). Hence, as before, we find the domain of \(\frac{g}{f}\) to be \[(-\infty, 0) \cup\left(0, \dfrac{1}{3}\right) \cup\left(\dfrac{1}{3}, \infty\right).\nonumber\]Next, we find and simplify a formula for \(\left(\frac{g}{f}\right)(x)\).\[ \begin{array}{rcl}
   \left(\dfrac{g}{f}\right)(x) &= & \dfrac{g(x)}{f(x)} \\
   \\
   & = & \dfrac{3-\frac{1}{x}}{6 x^{2}-2 x} \\
   \\
   & = & \dfrac{3-\frac{1}{x}}{6 x^{2}-2 x} \cdot \dfrac{x}{x} \\
   \\
   & = & \dfrac{3 x-1}{\left(6 x^{2}-2 x\right) x} \\
   \\
   & = & \dfrac{3 x-1}{2 x^{2}(3 x-1)} \\
   \\
   & = & \dfrac{1 \cancel{(3x - 1)}}{2 x^{2} \cancel{(3 x-1)}} \\
   \\
   & = & \dfrac{1}{2 x^{2}} \\
   \end{array} \nonumber \]

Note the importance of finding the domain of a function before simplifying its expression. In Example \( \PageIndex{1d} \), had we waited to find the domain of \(\frac{g}{f}\) until after simplifying, we'd have the formula \(\frac{1}{2 x^{2}}\) to go by, and we would (incorrectly!) state the domain as \((-\infty, 0) \cup(0, \infty)\), since the other troublesome number, \(x=\frac{1}{3}\), was canceled away. We'll see what this means geometrically later in this text. This is so important that it deserves some highlighting.

Caution: Find Domains Before Simplifying

When finding the domain of a function, do so before simplifying its expression.

Difference Quotients

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

Definition: Difference Quotient

Given a function \(f\), the difference quotient of \(f\) is the expression\[\dfrac{f(x+h)-f(x)}{h}. \label{dq} \]

We will revisit this concept in the future, but for now, we use it as a way to practice function notation and function arithmetic. For reasons that will become clear in Calculus, "simplifying" a difference quotient means rewriting it in a form where the \(h\) in Equation \( \ref{dq} \) cancels from the denominator. Once that happens, we consider our work to be done.

Example \( \PageIndex{2} \)

Find and simplify the difference quotients for the following functions

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

Solutions

1. 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\[ \begin{array}{rcl}
   f(x+h) & = & (x+h)^{2}-(x+h)-2 \\
   & = & x^{2}+2 x h+h^{2}-x-h-2 \\
   \end{array} \nonumber \]So the difference quotient is\[ \begin{array}{rcl}
   \dfrac{f(x+h)-f(x)}{h} & = & \dfrac{\left(x^{2}+2 x h+h^{2}-x-h-2\right)-\left(x^{2}-x-2\right)}{h} \\
   \\
   & = & \dfrac{x^{2}+2 x h+h^{2}-x-h-2-x^{2}+x+2}{h} \\
   \\
   & = & \dfrac{2 x h+h^{2}-h}{h} \\
   \\
   & = & \dfrac{h(2 x+h-1)}{h} \\
   \\
   & = & \dfrac{\cancel{h}(2 x+h-1)}{\cancel{h}} \\
   \\
   & = & 2x + h - 1. \\
   \end{array} \nonumber \]
2. 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\[\begin{array}{rcl}
   g(x+h) &= & \dfrac{3}{2(x+h)+1} \\
   \\
   & = & \dfrac{3}{2 x+2 h+1}, \\
   \end{array} \nonumber \]which yields\[ \begin{array}{rclr}
   \dfrac{g(x+h)-g(x)}{h} & = & \dfrac{\frac{3}{2 x+2 h+1}-\frac{3}{2 x+1}}{h} & \\
   \\
   & = & \dfrac{\frac{3}{2 x+2 h+1}-\frac{3}{2 x+1}}{h} \cdot \dfrac{(2 x+2 h+1)(2 x+1)}{(2 x+2 h+1)(2 x+1)} & \\
   \\
   & = & \dfrac{3(2 x+1)-3(2 x+2 h+1)}{h(2 x+2 h+1)(2 x+1)} & \left( \text{distributing the numerator only - see Caution below} \right) \\
   \\
   & = & \dfrac{6 x+3-6 x-6 h-3}{h(2 x+2 h+1)(2 x+1)} & \\
   \\
   & = & \dfrac{-6 h}{h(2 x+2 h+1)(2 x+1)} & \\
   \\
   & = & \dfrac{-6 \cancel{h}}{\cancel{h}(2 x+2 h+1)(2 x+1)} & \\
   \\
   & = & \dfrac{-6}{(2 x+2 h+1)(2 x+1)} &  \\
   \end{array} \nonumber \]Since we have managed to cancel the original \(h\) from the denominator, we are done.
3. For \(r(x)=\sqrt{x}\), we get \(r(x+h)=\sqrt{x+h}\) so the difference quotient is\[ \dfrac{r(x+h)-r(x)}{h}=\dfrac{\sqrt{x+h}-\sqrt{x}}{h}. \nonumber \]In order to cancel the \(h\) from the denominator, we rationalize the numerator by multiplying both the numerator and denominator by its conjugate.\[ \begin{array}{rclr}
   \dfrac{r(x+h)-r(x)}{h} & = & \dfrac{\sqrt{x+h}-\sqrt{x}}{h} & \\
   \\
   & = & \dfrac{(\sqrt{x+h}-\sqrt{x})}{h} \cdot \dfrac{(\sqrt{x+h}+\sqrt{x})}{(\sqrt{x+h}+\sqrt{x})} & \\
   \\
   & = & \dfrac{(\sqrt{x+h})^{2}-(\sqrt{x})^{2}}{h(\sqrt{x+h}+\sqrt{x})} & \left( \text{distributing the numerator only - see Caution below} \right) \\
   \\
   & = & \dfrac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})} & \\
   \\
   & = & \dfrac{h}{h(\sqrt{x+h}+\sqrt{x})} & \\
   \\
   & = & \dfrac{\cancelto{~}h^1}{\cancel{h}(\sqrt{x+h}+\sqrt{x})} & \\
   \\
   & = & \dfrac{1}{\sqrt{x+h}+\sqrt{x}} & \\
   \end{array} \nonumber \]Since we have removed the original \(h\) from the denominator, we are done.

Caution: Know When Not to Distribute

With difference quotients, you will commonly need to multiply the numerator and denominator of an expression by either

1. the LCD of all fractions within a compound fraction (a.k.a. simplifying compound fractions), or
2. the conjugate of either the numerator or the denominator (a.k.a. rationalizing).

When doing so, knowing what to distribute and what not to distribute is imperative.

Simplifying Compound Fractions

The entire point of multiplying the numerator and denominator of your compound fraction by the LCD of all the minor fractions is to "get rid of" denominators in those minor fractions; however, with difference quotients, it is common that the denominator of the entire (major) fraction is only \( h \). In this case, do not distribute out the denominator! Doing so will complicate the mathematics, and you will lose visibility of factors that cancel.

Rationalizing

In a difference quotient, the entire point of multiplying the numerator and denominator by the conjugate of either the numerator or the denominator is to clear radicals. This will only happen with the conjugate pairs. So, definitely distribute the conjugate pairs, but do not distribute the non-conjugates.

For example, in Example \( \PageIndex{2c} \), we multiplied the numerator and denominator by the conjugate of the numerator. We then distributed the numerator but did not bother with the distribution in the denominator. This is because distribution in the numerator cleared the radicals, but distribution in the denominator would do nothing other than make a mess.

Applications to Economics

As mentioned before, we will revisit difference quotients later in this text and explain them geometrically. For now, we want to move on to some classic applications of function arithmetic from Economics. For that, we need to think like entrepreneurs.

Suppose you are a manufacturer making a certain product. Let \(n\) be the production level, that is, the number of items produced in a given time period. It is customary to let \(C(n)\) denote the function that calculates the total cost of producing the \(n\) items. The quantity \(C(0)\), which represents the cost of producing no items, is the fixed cost and represents the money required to begin production. This typically includes supplies, machinery, and other operational costs needed to start the production of the item being sold. Associated with the total cost, \(C(n)\), is the cost per item, or average cost, denoted \(\overline{C}(n)\) and read "\( C \)-bar" of \(n\). To compute \(\overline{C}(n)\), we take the total cost, \(C(n)\), and divide it by the number of items produced, \(n\), to get\[ \overline{C}(n)=\dfrac{C(n)}{n}. \nonumber \]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. That is, supply is equal to demand. We regard \(n\) as the independent variable and \(p\) as the dependent variable and speak of the price-demand function, \(p(n)\).² Hence, \(p(n)\) returns the price charged per item when \(n\) items are produced and sold.

Our next function to consider is the revenue function, \(R(n)\). The function \(R(n)\) computes the amount of money collected as a result of selling \(n\) items. Since \(p(n)\) is the price charged per item, we have \[R(n) = n \, p(n).\nonumber\]Finally, the profit function, \(P(n)\) calculates how much money is earned after the costs are paid. That is, \[P(n) = (R − C)(n) = R(n) − C(n).\nonumber\]We summarize all of these functions below.

Summary of Common Economic Functions

Suppose \(n\) represents the quantity of items produced and sold.

• The price-demand function, \(p(n)\), calculates the price per item.
• The revenue function, \(R(n)\), calculates the total money collected by selling \(n\) items at a price \(p(n)\), \[R(n) = n \, p(n).\nonumber\]
• The cost function, \(C(n)\), calculates the cost to produce \(n\) items. The value \(C(0)\) is called the fixed cost or start-up cost.
• The average cost function, \[\overline{C}(n)=\dfrac{C(n)}{n},\nonumber\]calculates the cost per item when making \(n\) items. Here, we necessarily assume \(n > 0\).
• The profit function, \(P(n)\), calculates the money earned after costs are paid when \(n\) items are produced and sold, \[P(n) = (R − C)(n) = R(n) − C(n).\nonumber\]

It is high time for an example.

Example \( \PageIndex{3} \)

Let \(n\) represent the number of dOpi media players produced and sold in a typical week. Suppose the cost, in dollars, to make \(n\) dOpi media players is given by \(C(n) = 100n+ 2000\), for \(n \geq 0\), and the price, in dollars per dOpi, is given by \(p(n) = 450 − 15n\) for \(0 \leq n \leq 30\).

1. Find and interpret \(C(0)\).
2. Find and interpret \(\overline{C}(10)\).
3. Find and interpret \(p(0)\) and \(p(20)\).
4. Solve \(p(n) = 0\) and interpret the result.
5. Find and simplify expressions for the revenue function \(R(n)\) and the profit function \(P(n)\).
6. Find and interpret \(R(0)\) and \(P(0)\).
7. Solve \(P(n) = 0\) and interpret the result.

Solutions

1. We substitute \(n = 0\) into the formula for \(C(n)\) and get \[C(0) = 100(0) + 2000 = 2000.\nonumber\]This means to produce \(0\) dOpi media players, 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.
2. Since \(\overline{C}(n)=\frac{C(n)}{n},\)\[\overline{C}(10)=\dfrac{C(10)}{10}=\dfrac{3000}{10}=300.\nonumber\]This means when \(10\) dOpi media players are produced, the cost to manufacture them amounts to $300 per dOpi.
3. Plugging \(n = 0\) into the expression for \(p(n)\) gives \[p(0) = 450 − 15(0) = 450.\nonumber\]This means no dOpi media players are sold if the price is $450 per dOpi. On the other hand, \[p(20) = 450 − 15(20) = 150\nonumber\]means to sell 20 dOpi media players in a typical week, the price should be set at $150 per dOpi.
4. Setting \(p(n) = 0\) gives \(450 − 15n = 0\). Solving gives \(n = 30\). This means that to sell 30 dOpi media players in a typical week, the price must be $0. Moreover, this means that even if dOpi media players were given away for free, the retailer could only move 30 of them.
5. To find the revenue, we compute \[R(n)=n \, p(n)=n(450-15 n)=450 n-15 n^{2}.\nonumber\]Since the formula for \(p(n)\) is valid only for \(0 \leq n \leq 30\), our formula \(R(n)\) is also restricted to \(0 \leq n \leq 30\). Using the given formula for \(C(n)\) and the derived formula for \(R(n)\), we get \[P(n)=\left(450 n-15 n^{2}\right)-(100 n+2000)=-15 n^{2}+350 n-2000.\nonumber\]As before, the validity of this formula is for \(0 \leq n \leq 30\) only.
6. We find \(R(0) = 0\) which means if no dOpi media players are sold, we have no revenue, which makes sense. Turning to profit, \(P(0) = −2000\) since \(P(n) = R(n)−C(n)\) and \(P(0) = R(0)−C(0) = −2000\). This means that if no dOpi media players are sold, more money ($2000 to be exact!) was put into producing the dOpi media players than was recouped in sales. In number 1, we found the fixed costs to be $2000, so it makes sense that if we sell no dOpi media players, we are out those start-up costs.
7. Setting \(P(n) = 0\) gives \(-15 n^{2}+350 n-2000=0\). Factoring gives \(-5(n-10)(3 n-40)=0\) so \(n = 10\) or \(n=\frac{40}{3}\). What do these values mean in the context of the problem?
   
   Since \(P(n) = R(n) − C(n)\), solving \(P(n) = 0\) is the same as solving \(R(n) = C(n)\). This means that the solutions to \(P(n) = 0\) 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 \(n = 10\) means ten dOpi media players should be produced (and sold) during the week to recoup the cost of production. For \(n=\frac{40}{3}=13 . \overline{3}\), things are a bit more complicated. Even though \(n=13 . \overline{3}\) satisfies \(0 \leq n \leq 30\), 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. Evaluating \(P(13) = 15\) and \(P(14) = −40\), we see that producing and selling 13 dOpi media players 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 a later section.

---

Footnotes

¹ Thus, \(x\) is an element of the intersection of the two domains.

² From a retail perspective, it seems natural to think of the number of items sold, \(n\), as a function of the price charged, \(p\). After all, the retailer can easily adjust the price to sell more products. In the language of functions, \(n\) would be the dependent variable, and \(p\) would be the independent variable, or, using function notation, we have a function \(n (p)\). While we will adopt this convention later in the text, we will hold with tradition at this point and consider the price \(p\) as a function of the number of items sold, \(n\).