College Mathematics Chapter 1 Functions and Graphs PDF

Chapter 1 Functions and Graphs Section 1...

Chapter 1 Functions and Graphs Section 1 Functions Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 1  Solutions to Equations in Two Variables 2 Consider the equation in two variables, y  x 5. A solution to an equation in two variables is an ordered pair of numbers that when substituted into the equation result in a true statement. The ordered pairs (x, y) = (2, 9) and (x, y) = (-2, 9) are solutions to the equation. When values in these ordered pairs are substituted into the equation, the result is a true statement. The ordered pair (x, y) = (0, 2) is not a solution to the equation. When values in this ordered pair are substituted into the equation, the resulting statement is not true. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 2  Finding Solutions to Equations in Two Variables Find a solution to the equation, y  x 2 5, when x 4. One process for finding a solution to an equation in two variables is to choose a value for one variable, substitute that value into the equation, then solve for the other variable. Substitute x = 4 into the equation to obtain the equation, y – 16 = 5. It follows that y = 21 and (4, 21) is a solution to the equation. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 3  Point-by-Point Graphing  To sketch the graph an equation in x and y, we find ordered pairs that solve the equation and plot the ordered pairs on a grid.  We must find sufficiently many pairs so that the shape of the graph is apparent.  This process is called point-by-point plotting.  The points corresponding to the solution are then connected with a smooth curve.  The resulting graph is a visual representation of the solution set for the equation. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 4  Point-by-Point Graphing Create a Table of Solutions Create a table of solutions to the equation, y  x 2 5.  We select a collection of values for one of the variables in the equation. In this example, we select values for the variable, x.  For this example, we will select the x-values, -3, -2, -1, 0, 1, 2, and 3. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 5  Point-by-Point Graphing Create a Table of Solutions 2 Create a table of solutions to the equation, y  x 5.  Using the x-values, -3, -2, -1, 0, 1, 2, and 3, we substitute and solve for y. The results are shown in the table. x y Solution, (x, y) -3 y – (-3)2 = 5 gives y = 9 + 5 = 14 (-3, 14) -2 y – (-2)2 = 5 gives y = 4 + 5 = 9 (-2, 9) -1 y – (-1)2 = 5 gives y = 1 + 5 = 6 (-1, 6) 0 y – (0)2 = 5 gives y = 0 + 5 = 5 (0, 5) 1 y – (1)2 = 5 gives y = 1 + 5 = 6 (1, 6) 2 y – (2)2 = 5 gives y = 4 + 5 = 9 (2, 9) 3 y – (3)2 = 5 gives y = 9 + 5 = 14 (3, 14) Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 6  Point-by-Point Graphing Plotting the Points Solution, (x, y) (-3, 14) (-2, 9) (-1, 6) (0, 5) (1, 6) (2, 9) (3, 14) Plot the points in the solution. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 7  Point-by-Point Graphing Connecting the Points Graph the points Connect the points in the solution. with a smooth curve. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 8  Correspondences  An important aspect of any science is the establishment of correspondences among various types of phenomena.  Correspondences are central to the concept of mathematical functions.  Some familiar correspondences include the following: To each item in a store, there corresponds a price. To each university student, there corresponds their grade-point average. To each circle, there corresponds a circumference. To each number, there corresponds its cube. To each day, there corresponds a maximum temperature. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 9  Mathematical Function A function is a correspondence between two sets of elements such that to each element in the first set, there corresponds one and only one element in the second set. The first set in such a correspondence is called the domain of the function. The second set in such a correspondence is called the range of the function. A key concept for functions is the requirement that each domain element corresponds with one and only one range element. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 10  Function or Not? Domain Range Number Cube of number -2 -8 -1 -1 0 0 1 1 2 8 This table represents a function since each domain element corresponds to one and only one range element. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 11  Function or Not? Domain Number Range -2 Square of number -1 0 0 1 1 4 2 This table represents a function since each domain element corresponds to one and only one range element. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 12  Function or Not? Domain Range Number Square root of number 0 0 1 1 4 -1 9 2 -2 3 -3 This table does not represent a function since there are domain values that correspond to more than one range value (for example, the range values 2, and -2 correspond to the domain value 4). Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 13  Functions Specified by Equations If an equation in two variables has the property that each domain value is associated with one and only one range value, such an equation represents a function. If an equation in two variables has the property that there is at least one domain value that is associated with more than one range value, such an equation does not represent a function. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 14  Functions In the equation, y  x 2 5, y is a function of x.  The collection of x values is the domain of the function.  The collection of y values associated with each x value is the range of the function.  Each domain value in this equation corresponds to one and only one range value. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 15  The Graph of a Function We previously graphed and connected some of the points in the solution set for the equation, y  x 2 5. Additional points in the solution set can be graphed to obtain the graph of the function. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 16  Equations that are Not Functions In the equation, y 2  x 5, y is not a function of x. The domain value, x = 4 corresponds to two range values, y = 3 and y = -3. When an equation has at least one domain value that corresponds to more than one range value, the equation does not represent a function. The graph of this specific correspondence is shown below. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 17  Vertical Line Test for a Function THEOREM 1: An equation specifies a function if each vertical line in the coordinate system passes through, at most, one point on the graph of the equation. If any vertical line passes through two or more points on the graph of an equation, then the equation does not specify a function. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 18 Vertical Line Test for a Function This graph is not the graph of a function because is it possible to draw a vertical line which intersects the graph more than one time. This graph represents a function because any vertical line will intersect the graph in at most one point. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 19  Independent and Dependent Variables Input values for functions are domain values and output values for functions are range values. Function equations often use the variable, x as the domain variable and the variable, y as the range variable. The input variable is called an independent variable. The output variable is called a dependent variable. If a function is specified by an equation and the domain is not indicated, we assume that the domain is the set of all real- number replacements of the independent variable that produce real values for the dependent variable. The range of a function is the set of all outputs corresponding to input values. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 20  Function Notation Functions involve two sets, a domain and a range along with a correspondence that assigns to each element in the domain, exactly one element in the range. Function notation gives an alternate method for naming functions so that the correspondence between specific input and output values can be easily shown. Letters can be used as names for functions as well as names for numbers. The letters, f and g may be used to name the functions specified by the equations y 2 x  1 and y  x 2  2 x  3 as follows: f : y 2 x  1 g : y  x 2  2 x  3. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 21  Function Notation Using the functions named, f : y 2 x 1 and g : y x 2  2 x  3, if x represents an element in the domain of the function, f , we use the notation f ( x) 2 x  1 and g ( x) x 2  2 x  3 to specify these functions. The symbol f(x) is used in place of y to designate the number in the range of f to which x is paired using the equation that is named f. f(x) is read “f of x,” or “the value of f at x.” Similarly, g(x) is read “g of x,” or “the value of g at x.” Whenever we use function notation such as f(x) or g(x), we assume that the variable x is an independent variable and both y and f(x) are dependent variables. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 22  Finding a Domain Find the domain of the function specified by the equation y  10  x , assuming that x is the independent variable. For the dependent variable, y to be real, the expression under the radical sign must be non-negative. We solve the inequality, 10 – x ≥ 0 to obtain -x ≥ -10 which gives x ≤ 10. The domain in inequality notation form is x ≤ 10. The domain in interval notation form is (-∞, 10]. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 23  Function Evaluation Use the functions f ( x) 2 x  1 and g ( x)  x 2  2 x  3 to evaluate f (5), g (5), f (  2), and g (  2).  f(5) = 2∙5 + 1 = 10 + 1 = 11. The point (5, 11) is a point on the graph of the function, f.  g(5) = 52 + 2∙5 – 3 = 25 + 10 – 3 = 35 – 3 = 32. The point (5, 32) is a point on the graph of the function, g.  f(–2) = 2∙(–2) + 1 = –4 + 1 = –3. The point (-2, -3) is a point on the graph of the function, f.  g(–2) = (–2)2 + 2∙(–2) – 3 = 4 – 4 – 3 = –3. The point (–2, –3) is a point on the graph of the function, g.  Since the point (–2, –3) is on the graph of each function, it is a point of intersection for the two graphs. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 24 Graphing Functions Using Calculator Technology Use calculator technology to graph the functions f ( x) 2 x  1 and g ( x)  x 2  2 x  3 in a standard viewing window. Input the functions into the graphing calculator as Y1 = f(x), and Y1 = g(x). Press zoom 6:Zstandard to view the graphs in a standard viewing window. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 25  Function Evaluation The graphs of the functions f ( x) 2 x  1 and g ( x) x 2  2 x  3 are shown in a standard viewing window. The two functions intersect at two different points. The point (–2, –3) was found to be a point on the graphs of both functions. What are the coordinates for the other point of intersection? Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 26  The Difference Quotient  In addition to evaluating functions at specific numbers, an important skill is evaluating functions at expressions that involve one or more variables.  A foundational concept in higher level mathematics areas such as calculus is the difference quotient. The difference quotient for a function f ( x) where x and x  h f ( x  h)  f ( x ) are in the domain of f , with h 0 is. h  The difference quotient computes the slope of the line between two points on the graph of a function. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 27  The Difference Quotient f ( x  h)  f ( x ) The difference quotient for a function f ( x) is. h Find the difference quotient for f ( x) 2 x  1.  f(x + h) = 2(x + h) + 1 = 2x + 2h + 1.  f(x) = 2x + 1.  f(x + h) – f(x) = 2x + 2h + 1 – (2x + 1) = 2x + 2h + 1 – 2x – 1 = 2h. f ( x  h)  f ( x ) 2h  2. h h Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 28  The Difference Quotient g ( x  h)  g ( x ) Find the difference quotient h for the function g ( x)  x 2  2 x  3.  g(x + h) = (x + h)2 + 2(x + h) – 3 = x2 + 2xh + h2 + 2x + 2h – 3  g(x) = x2 + 2x – 3  g(x + h) – g(x) = x2 + 2xh + h2 + 2x + 2h – 3 – (x2 + 2x – 3) = x2 + 2xh + h2 + 2x + 2h – 3 – x2 – 2x + 3 = 2xh + h2 + 2h g ( x  h)  g ( x) 2 xh  h 2  2h  2 x  h  2. h h Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 29  Profit-loss Analysis Companies use Profit-loss analysis to support decisions regarding the pricing of products and appropriate levels of production in order to maximize company profit. A manufacturing company has costs, C, which include fixed costs (plant overhead, product design, setup, and promotion) and variable costs (costs that depend on the number of items produced.) The revenue (income) for a company, R, is the amount of money the company receives from selling its product. If R < C, the company loses money. If R = C, the company breaks even. If R > C, the company makes a profit. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 30  Profit-loss Analysis Profit, P, is equal to revenue, R, minus cost, C. P = R – C. When P < 0, the company loses money (cost exceeds revenue). When P = 0, the company breaks even (cost equals revenue). When P > 0, the company makes a profit (revenue exceeds cost). Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 31  Price-demand Analysis Companies use a price-demand function, p(x), often determined using historical data or sampling techniques, that specifies the relationship between the demand for a product, x, and the price of the product, p. A point (x, p) is on the graph of the price-demand function if x items can be sold at a price of $p per item. Generally, a reduction in price results in an increase in the demand, thus the graph of the price-demand function is expected to go downhill as prices increase from left to right. The revenue, R, is equal to the number of items sold multiplied by the price per item, R = x∙p. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 32  A Price-demand Example The price-demand function for a company is given by p ( x) 1000  5 x, 0  x 100 where x represents the number of items and p ( x) represents the price of the item. Determine the revenue function and find the revenue generated if 50 items are sold. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 33  Price-demand Solution Solution: Revenue = number of items sold times price per item. R(x) = x∙p(x) = x(1000 – 5x) When 50 items are sold R(50) = 50(1000 – 5∙50) = 50(1000 – 250) = 50(750) = 37,500 When 50 items are sold, the price is $750 and the revenue from selling the 50 items is $37,500. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 34  Break-Even and Profit-Loss Analysis Cost, revenue, and profit functions are often represented symbolically as C(x), R(x), and P(x) where the independent variable, x, represents the number of items manufactured and sold. These functions often have the following forms, where a, b, m, and n are positive constants determined from the context of the particular manufacturer. Cost function, C(x) = a + b∙x C = fixed costs + variable costs Price-demand function, p(x) = m – n∙x x is the number of items that can be sold at $p per item. Revenue function, R(x) = x∙p = x(m – n∙x) R = number of items sold times the price per item. Profit function, P(x) = R(x) – C(x) = x(m – n∙x) – (a + b∙x). Be careful not to confuse the price demand function, p(x) with the profit function, P(x). The price function always uses the lower case, p. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 35  Profit-Loss Analysis Example A company manufactures notebook computers. Its marketing research department has determined that the data is modeled by the price-demand function p(x) = 2,000 – 60x, when 1 < x < 25, (x is in thousands). Find the expression representing the company’s revenue function and find the domain for this function? Solution: Since Revenue = Price ∙ Quantity, R(x) = x∙p(x) = x∙(2000 – 60x) = 2000x – 60x2 The domain for this function is the same as the domain of the price-demand function, 1 < x < 25, (x is in thousands of items). Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 36  Profit Problem An analyst for the company in the preceding problem found the following cost function that models producing and selling x thousand notebook computers: C(x) = 4,000 + 500x, 1 < x < 25. Write a profit function for producing and selling x thousand notebook computers, give the domain of this function, and graph the function in a window that covers the domain of the function. Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 37  Solution to Profit Problem Profit = Revenue – Cost. The revenue function was found to be R(x) = 2000x – 60x2. P(x) = R(x) – C(x) = 2000x – 60x2 – (4000 + 500x) = –60x2 + 1500x – 4000. The domain of this function is the same as the domain of the original price-demand function, 1< x < 25 (where x represents the number of computers produced and sold in thousands.) Barnett, College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14e, GE Copyright © 2019, 2015, 2011 Pearson Education Ltd. Slide 38