1.1 Introduction to Functions

In the study of business and economics, we are constantly seeking to understand the relationships between variables: how does a change in price affect demand? How does the number of units produced influence total cost? Functions are the fundamental building blocks for creating mathematical models that describe these relationships. This chapter covers the core definition of a function, its essential properties, and its application in representing the connections between key business variables like price, cost, profit, and demand.

1.2 Core Concepts of Functions

Defining a Function

A function is a rule that matches each number x from a set of inputs, called the domain (D), with exactly one number y from a set of outputs, called the range (R).

In our models, x is the independent variable—the input we can control or observe, such as price or advertising spend. The variable y is the dependent variable, as its value depends on the input x. This structure is perfect for modeling the cause-and-effect relationships that are central to business analysis.

Illustrative Examples

Example 1: Evaluating a Polynomial Function Consider the function f(x) = x³ – 2x² + 3x + 100. To find the value of the function when x=2, we substitute 2 for every instance of x:

• f(2) = (2)³ – 2(2)² + 3(2) + 100
• f(2) = 8 – 2(4) + 6 + 100
• f(2) = 8 – 8 + 6 + 100
• f(2) = 106

Example 2: A Piecewise Function for Commissions A real estate broker’s commission depends on the sale price. The commission is 6% for sales up to $300,000, but for sales above $300,000, it is a flat $6,000 plus 4% of the sale price. This two-tiered structure is modeled using a piecewise function, R(x), where x is the sale price.

• Step 1: Define the function. We need two different rules, or “pieces,” depending on the value of x.
  • For 0 ≤ x ≤ 300,000, the rule is R(x) = 0.06x.
  • For x > 300,000, the rule is R(x) = 6,000 + 0.04x.
• This is written formally as:
• Step 2: Calculate the commission for a $200,000 sale. Since $200,000 is less than or equal to $300,000, we use the first rule: R(200,000) = 0.06 * 200,000 = $12,000
• Step 3: Calculate the commission for a $500,000 sale. Since $500,000 is greater than $300,000, we use the second rule: R(500,000) = 6,000 + 0.04 * 500,000 = 6,000 + 20,000 = $26,000

The Domain of a Function

The domain is the set of all possible input values (x) for which the function is defined. In practical terms, the domain represents the set of realistic inputs for a business model. There are often mathematical or logical constraints on the domain.

• Constraint 1: Division by Zero. For the function f(x) = 1 / (x – 3), the denominator cannot be zero. The value x=3 would make the denominator zero, which is mathematically undefined. Therefore, the domain includes all real numbers except 3.
• Constraint 2: Real Number System. For the function g(x) = √(x – 2), the expression inside the square root must be non-negative, as the square root of a negative number is not a real number. This is a common constraint in economic models which operate within the real number system.
  • Constraint: x – 2 ≥ 0
  • Domain: x ≥ 2

The Composition of Functions

A composite function, written as g(h(x)), is created when the output of one function, h(x), becomes the input for another function, g(u). This is incredibly useful for modeling multi-step processes where one variable depends on an intermediate variable, which in turn depends on a third.

Example 4: Algebraic Composition If g(u) = u² + 3u + 1 and h(x) = x + 1, we find the composite function g(h(x)) by substituting the entire expression for h(x) in for u:

• g(h(x)) = (x + 1)² + 3(x + 1) + 1
• g(h(x)) = (x² + 2x + 1) + (3x + 3) + 1
• g(h(x)) = x² + 5x + 5

Example 5: A Model for Pollution Over Time An environmental study suggests the level of carbon monoxide, C, depends on the population, p (in thousands), according to the function: C(p) = 0.5p + 1

It is also estimated that t years from now, the population will be: P(t) = 10 + 0.1t²

• (a) Express carbon monoxide as a function of time. We can create a single, powerful model that directly links time to pollution by composing the two functions. The output of P(t) becomes the input for C(p).
  • C(P(t)) = 0.5 * (10 + 0.1t²) + 1
  • C(P(t)) = 5 + 0.05t² + 1
  • C(P(t)) = 6 + 0.05t² This composite function gives us the carbon monoxide level directly as a function of time.
• (b) When will the carbon monoxide level reach 6.8 parts per million? We set our new function equal to 6.8 and solve for t.
  • 6.8 = 6 + 0.05t²
  • 0.8 = 0.05t²
  • t² = 0.8 / 0.05 = 16
  • t = 4 In 4 years, the carbon monoxide level will reach 6.8 parts per million.

By understanding the properties of functions, we can construct and analyze these models; the next step is to visualize them through graphs.

1.3 Visualizing Functions: Graphs and Models

Graphing a function provides an intuitive, visual understanding of the relationship between variables. A well-constructed graph makes it easy to identify important features like trends, maximums, minimums, and key points of interest such as break-even points or market equilibrium.

Detailing Graphing Methods

A straightforward method for sketching the graph of a function involves three steps:

1. Choose a representative collection of numbers x from the function’s domain.
2. Construct a table of the corresponding function values, y = f(x).
3. Plot these (x, y) points on a coordinate plane and connect them with a smooth curve.

Example 1: Graphing y = x² We create a table of values:

Plotting these points (-2, 4), (-1, 1), etc., and connecting them reveals the characteristic U-shape of a parabola.

Example 2: Graphing a Piecewise Function Consider the function:

{ x²,   if 0 ≤ x < 1

f(x) = { 2/x,  if 1 ≤ x < 4

{ 3,    if x ≥ 4

We must use the appropriate formula for each value of x:

The graph is constructed from three distinct pieces corresponding to the three different formulas.

Analysis of Parabolic Functions

The graph of a quadratic function, y = ax² + bx + c, is a parabola. These are extremely important in business for modeling profit and cost functions.

• Shape: The parabola opens upward if a > 0, indicating a minimum value (e.g., minimum average cost). It opens downward if a < 0, indicating a maximum value (e.g., maximum profit).
• Vertex: The “peak” or “valley” of the parabola is its vertex. The x-coordinate of the vertex gives the location of the maximum or minimum value and is found with the formula: x = -b / 2a

Example 3: Analyzing y = x² – 6x + 4 Here, a=1, b=-6, and c=4.

• a. Find the Vertex: x = -(-6) / (2 * 1) = 6 / 2 = 3 To find the y-coordinate, substitute x=3 back into the equation: y = (3)² – 6(3) + 4 = 9 – 18 + 4 = -5 The vertex is at (3, -5).
• b. Find the minimum value: Since a=1 > 0, the parabola opens upward, and the vertex represents a minimum. The minimum value for y is -5.
• c. Find the x-intercepts: Set y=0 and solve for x using the quadratic formula: x² – 6x + 4 = 0 x = [-(-6) ± √((-6)² – 4*1*4)] / (2*1) x = [6 ± √(36 – 16)] / 2 = [6 ± √20] / 2 = 3 ± √5

Synthesizing a Business Model

Example 4: The Radio Manufacturer’s Profit A manufacturer can produce radios at a cost of $10 apiece. If they are sold for x dollars, consumers will buy approximately 80 – x radios each month. Let’s build a profit function.

• Step 1: State the relationship in words. Profit = (Number of radios sold) * (Profit per radio)
• Step 2: Translate each part into an algebraic expression.
  • Number of radios sold = 80 – x
  • The selling price is x and the cost is $10, so the profit per radio = x – 10.
• Step 3: Derive the final profit function, P(x). P(x) = (80 – x) * (x – 10) P(x) = 80x – 800 – x² + 10x P(x) = -x² + 90x – 800
• Step 4: Analyze the function. This is a quadratic function with a = -1, which is less than 0. Therefore, its graph is a downward-opening parabola. Its vertex will represent the selling price x that yields the greatest possible profit. Using the algebraic vertex formula, x = -b / 2a, we can find this optimal price: x = -90 / (2 * -1) = 45 Thus, the manufacturer’s profit will be greatest when the radios are sold for $45.

While this algebraic method works for simple parabolic functions, we will soon learn a more powerful and universal method using calculus—finding where the rate of change is zero—to solve far more complex optimization problems.

1.4 Linear Functions and Their Applications

Linear functions are used to model situations involving a constant rate of change. This is a common scenario in business for simple cost models, straight-line depreciation, and basic sales projections. The graph of a linear function is a straight line.

Core Concepts of Lines

• Slope of a Line: The slope measures the rate of change. For a line passing through points (x₁, y₁) and (x₂, y₂), the slope m is: m = (y₂ – y₁) / (x₂ – x₁) It represents the change in y for a one-unit increase in x.
• Horizontal and Vertical Lines:
  • A horizontal line has the equation y = b (where b is a constant) and a slope of 0.
  • A vertical line has the equation x = c (where c is a constant) and an undefined slope.
• Slope-Intercept Form: The equation y = mx + b is the slope-intercept form.
  • m is the slope.
  • b is the y-intercept (the value of y when x=0).
  • Example 2: To find the slope and y-intercept of 3y + 2x = 6, we solve for y: 3y = -2x + 6 y = (-2/3)x + 2 The slope m is -2/3, and the y-intercept b is 2.
• Point-Slope Form: The equation y – y₀ = m(x – x₀) is useful when you know the slope m and one point (x₀, y₀) on the line.
  • Example 3: Find the equation for a line with slope 1/2 passing through (5, 1). y – 1 = (1/2)(x – 5)
  • Example 4: Find the equation for a line passing through (3, -2) and (1, 6). First, find the slope: m = (6 – (-2)) / (1 – 3) = 8 / -2 = -4. Then use the point-slope form with either point: y – 6 = -4(x – 1)

Analysis of Applied Linear Models

Example 5: Price of Bread The price of bread has been rising at a constant rate of 2 cents per month. By November 1 (the 10th month after January 1), the price was $1.06 (106 cents). Let’s model this.

• Translate the problem: “Constant rate of 2 cents per month” means the slope is m=2. “November 1” gives us a point: when x=10 months have passed, the price y=106 cents.
• Use the point-slope form: y – y₀ = m(x – x₀) y – 106 = 2(x – 10) y = 2x – 20 + 106 y = 2x + 86
• Solve the question: “What was the price at the beginning of the year?” This corresponds to x=0. y = 2(0) + 86 = 86 The price was 86 cents at the start of the year.

Example 6: SAT Scores SAT scores have been declining at a constant rate. In 1986, the average was 575; in 1991, it was 545. Using a similar method to the bread example, one can derive the function y = -6x + 575, where x is years since 1986.

Example 7: Manufacturer’s Cost A manufacturer’s cost includes a fixed overhead of $200 plus production costs of $50 per unit.

• Components of the linear cost function C(x) = mx + b:
  • The variable cost per unit is the slope: m = 50.
  • The fixed overhead is the y-intercept (the cost when 0 units are produced): b = 200.
• The function: C(x) = 50x + 200.

Mastering linear models provides the foundation, but the real power of mathematical economics comes from combining these simple building blocks to model the complex, non-linear realities of the market.

1.5 Advanced Functional Models in Business and Economics

By combining the concepts of linear, quadratic, and piecewise functions, we can model sophisticated business scenarios like profit optimization, tiered pricing, break-even points, and market equilibrium.

Model a Profit Function

Example 1: Radio Manufacturer with Price Increases A manufacturer can produce radios at a cost of $2 apiece. The radios have been selling for $5 apiece, and at this price, consumers have been buying 4,000 radios a month. The manufacturer is planning to raise the price and estimates that for each $1 increase in price, 400 fewer radios will be sold. Let’s construct the profit function P(x) where x is the new selling price.

• Profit = (Number of Radios Sold) x (Profit per Radio)
• Profit per Radio: This is the selling price minus the cost: x – 2.
• Number of Radios Sold: This starts at 4,000 and decreases by 400 for every dollar the price increases above $5. The number of $1 increases is (x – 5).
  • Number sold = 4000 – 400(x – 5)
  • Number sold = 4000 – 400x + 2000
  • Number sold = 6000 – 400x, which simplifies to 400(15 – x).
• Final Profit Function: P(x) = (400(15 – x))(x – 2)

Model Piecewise Functions for Tiered Pricing

Example 2: Water Shortage Rates To discourage water use, a district implements a tiered pricing structure. Let x be the water used in 100-cubic-feet units.

• Tier 1 (0 ≤ x ≤ 12): The cost is simply $1.22 per unit: C(x) = 1.22x.
• Tier 2 (12 < x ≤ 24): The cost is the sum of two parts: the fixed cost for the first 12 units (1.22 * 12 = $14.64) plus the variable cost for any additional usage. The number of additional units is (x – 12), and these are charged at the higher rate of $10 per unit. So, C(x) = 14.64 + 10(x – 12) = 10x – 105.36.
• Tier 3 (x > 24): The cost is the fixed cost for the first 24 units ($14.64 for the first 12, plus $10 * 12 = $120 for the next 12) plus the variable cost for any additional usage. The number of additional units is (x – 24), charged at $50 per unit. So, C(x) = 14.64 + 120 + 50(x – 24) = 50x – 1065.36.

This results in the three-part piecewise function:

{ 1.22x,           if 0 ≤ x ≤ 12

C(x) = { 10x – 105.36,    if 12 < x ≤ 24

{ 50x – 1065.36,   if x > 24

Analysis of Break-Even Points

Break-even analysis is the process of finding the production level at which total revenue equals total cost.

• Total Cost Function, C(x): The total cost to produce x units.
• Total Revenue Function, R(x): The total revenue from selling x units.
• Break-Even Point: The value of x where R(x) = C(x).
  • If R(x) > C(x), the company is in a profit zone.
  • If R(x) < C(x), the company is in a loss zone.

Example 3: Green-Belt Company Belts cost $2 each to manufacture, plus $300 per day in fixed costs. They sell for $3 each.

• C(x) = 2x + 300
• R(x) = 3x
• Set R(x) = C(x): 3x = 2x + 300 x = 300 The break-even point is 300 belts. The company must sell more than 300 belts to make a profit.

Example 4: A Quadratic Model A company’s cost is C(x) = 500 + 140x. The price per item is P = 200 – x.

• Cost: C(x) = 500 + 140x
• Revenue: R(x) = Price * x = (200 – x)x = 200x – x²
• Profit: P(x) = R(x) – C(x) = (200x – x²) – (500 + 140x) = -x² + 60x – 500
• Break-Even: Set R(x) = C(x): 200x – x² = 500 + 140x 0 = x² – 60x + 500 0 = (x – 10)(x – 50) There are two break-even points: x=10 and x=50. This means that profit occurs between these two production levels.

Analysis of Market Equilibrium

The economic law of supply and demand states that, generally, the quantity of a product manufacturers are willing to supply increases as the price goes up, while the quantity consumers are willing to demand decreases as the price goes up.

• Point of Market Equilibrium: This is the intersection of the supply and demand curves. It represents the price (p) at which the quantity supplied equals the quantity demanded. At this price, there is neither a surplus nor a shortage of the commodity.

Example 5: Finding Equilibrium A commodity’s supply function is S(p) = p² + 3p – 70, and its demand function is D(p) = 410 – p.

• Set Supply = Demand: p² + 3p – 70 = 410 – p
• Solve for p: p² + 4p – 480 = 0 (p – 20)(p + 24) = 0 The solutions are p=20 and p=-24. Since price cannot be negative, we discard p=-24.
• Find Equilibrium Quantity: The equilibrium price is $20. We can plug this into either function to find the quantity. Using the simpler demand function: D(20) = 410 – 20 = 390 At a price of $20, 390 units will be supplied and demanded.

Functions provide a versatile and powerful framework for modeling a wide range of economic phenomena, from simple costs to complex market dynamics.

1.6 Chapter 1 Practice Problems

Problems 1-6 focus on the core definitions and properties of functions. Problems 7-11 and 16-18 challenge you to build functional models from word problems. Problems 20 and 21 focus on break-even analysis.

1. Compute the indicated values of the given function. a. f(x) = 3x² + 5x – 2; f(-1), f(0), f(2) b. g(x) = x + 1/x; g(-1), g(1), g(2) c. h(t) = t² + 2t + 4; h(2), h(0), h(-4) d. f(t) = (2t – 1)⁻³/²; f(1), f(5), f(13) e. f(t) = { 3 if t < -5; t+1 if -5 ≤ t ≤ 5; √t if t > 5 }; f(-6), f(-5), f(16)
2. Specify the domain of the given function. a. g(x) = (x² + 1) / (x + 2) b. y = √(x – 5) c. g(t) = √(t² + 9) d. f(t) = (2t – 4)²/³ e. f(x) = (x – 1) / (√(x² – 9))
3. Suppose the total cost of manufacturing q units of a certain commodity is given by the function C(q) = q³ – 30q² + 400q + 500. a. Compute the cost of manufacturing 20 units. b. Compute the cost of manufacturing the 20th unit.
4. Find the composite function g(h(x)). a. g(u) = u² + 4, h(x) = x – 1 b. g(u) = 3u² + 2u – 6, h(x) = x + 2 c. g(u) = (u – 1)³ + 2u², h(x) = x + 1 d. g(u) = 1/u², h(x) = x – 1 e. g(u) = u² + 1, h(x) = 1/(x-1)
5. Find the indicated composite function. a. f(x+1) where f(x) = x² + 5 b. f(x-2) where f(x) = 2x² – 3x + 1 c. f(x-1) where f(x) = (x+1)⁵ – 3x² d. f(1/x) where f(x) = 3x + 2/x e. f(x²+3x-1) where f(x) = √x f. f(x+1) where f(x) = (x-1)/(x)
6. Find functions h(x) and g(u) such that f(x) = g(h(x)). a. f(x) = (x⁵ – 3x² + 12)³ b. f(x) = √(3x – 5) c. f(x) = (x-1)² + 2(x-1) + 3 d. f(x) = (x/(x+1))² – 4(x/(x+1)) + 4 e. f(x) = √(x+4) – 2/(x+4)³
7. At a certain factory, the total cost of manufacturing q units during the daily production run is C(q) = q² + q + 900 dollars. On a typical workday, q(t) = 25t units are manufactured during the first t hours of a production run. a. Express the total manufacturing cost as a function of t. b. How much will have been spent on production by the end of the 3rd hour? c. When will the total manufacturing cost reach $11,000?
8. In economics, revenue R is defined as R = xp. Suppose that p and x are related by the demand equation p = -1/10x + 20 for 0 ≤ x ≤ 200. Express the revenue R as a function of the number x of units sold.
9. The price p and the quantity x sold of a certain product obey the demand equation p = -1/6x + 100 for 0 ≤ x ≤ 600. a. Express the revenue R as a function of x. b. What is the revenue of the company if 200 units are sold? c. Graph the revenue function. d. What quantity x maximizes revenue? What is the maximum revenue? e. What price should the company charge to maximize revenue?
10. The price p and the quantity x sold of a certain product obey the demand equation x = -5p + 100 for 0 ≤ p ≤ 20. a. Express the revenue R as a function of x. b. What is the revenue of the company if 15 units are sold? c. Graph the revenue function. d. What quantity x maximizes revenue? What is the maximum revenue? e. What price should the company charge to maximize revenue?
11. A manufacturer can produce cassette tape recorders at a cost of $20 apiece. It is estimated that if the tape recorders are sold for x dollars a piece, consumers will buy 120 – x of them a month. Express the manufacturer’s monthly profit as a function of price, graph this function, and use the graph to estimate the optimal selling price.
12. Write the equation for the line with the given properties. a. Through (2,0) with slope 1. b. Through (5,-2) with slope -1/2. c. Through (2,5) and parallel to the x axis. d. Through (1,0) and (0,1). e. Through (2,5) and (1,-2). f. Through (1,5) and (3,5).
13. A manufacturer’s total cost consists of a fixed overhead of $5,000 plus production costs of $60 per unit. Express the total cost as a function of the number of units produced and draw the graph.
14. A doctor owns $1,500 worth of medical books which, for tax purposes, are assumed to depreciate linearly to zero over a 10-year period. Express the value of the books as a function of time and draw the graph.
15. Since the beginning of the month, a local reservoir has been losing water at a constant rate. On the 12th of the month the reservoir held 200 million gallons of water, and on the 21st it held only 164 million gallons. Express the amount of water in the reservoir as a function of time and draw the graph.
16. A manufacturer can produce radios at a cost of $2 apiece. The radios have been selling for $5 apiece, and at this price, consumers have been buying 4,000 radios a month. The manufacturer is planning to raise the price of the radios and estimates that for each $1 increase in the price, 400 fewer radios will be sold each month. Express the number of radios sold as a function of the manufacturer’s selling price.
17. Each unit of a certain commodity costs p = 35x + 15 cents when x units of the commodity are produced. If all x units are sold at this price, express the revenue derived from the sales as a function of x.
18. A manufacturer has been selling lamps at the price of $6 per lamp, and at this price, consumers have been buying 3,000 lamps a month. The manufacturer wishes to raise the price and estimates that for each $1 increase in the price, 1,000 fewer lamps will be sold each month. The manufacturer can produce the lamps at a cost of $4 per lamp. Express the manufacturer’s monthly profit as a function of the price that the lamps are sold, draw the graph, and estimate the optimal selling price.
19. A cable is to be run from a power plant on one side of a river 900 meters wide to a factory on the other side, 3,000 meters downstream. The cable will be run in a straight line from the power plant to some point P on the opposite bank and then along the bank to the factory. The cost of running the cable across the water is $5 per meter, while the cost over land is $4 per meter. Let x be the distance from P to the point directly across the river from the power plant and express the cost of installing the cable as a function of x.
20. A furniture manufacturer can sell dining-room tables for $70 apiece. The manufacturer’s total cost consists of a fixed overhead of $8,000 plus production costs of $30 per table. a. How many tables must the manufacturer sell to break even? b. How many tables must the manufacturer sell to make a profit of $6,000? c. What will the manufacturer’s profit or loss be if 150 tables are sold? d. On the same axes, graph the manufacturer’s total revenue and total cost functions.
21. Since the beginning of the year, the price of gasoline has been increasing at a constant rate of 2 cents per gallon per month. By June first, the price had reached $1.03 per gallon. a. Express the price of the gasoline as a function of time and draw the graph. b. What was the price at the beginning of the year? c. What will the price be on October first?