Functions Modeling Change: A Precalculus Course. Marcel B. Finan Arkansas Tech University c All Rights Reserved


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1 Functions Modeling Change: A Precalculus Course Marcel B. Finan Arkansas Tech University c All Rights Reserved 1
2 PREFACE This supplement consists of my lectures of a freshmenlevel mathematics class offered at Arkansas Tech University. The lectures are designed to accompany the textbook Functions Modeling Change:A preperation for Calculus by HughesHallett et al. This book has been written in a way that can be read by students. That is, the text represents a serious effort to produce exposition that is accessible to a student at the freshmen or high school levels. The lectures cover Chapters 15, 8, and 9 of the book followed by a discussion of trigonometry. These chapters are well suited for a 4hour one semester course in Precalculus. Marcel B. Finan April 003
3 Contents 1 Functions and Function Notation 6 The Rate of Change 11 3 Linear Functions 15 4 Formulas for Linear Functions 18 5 Geometric Properties of Linear Functions 1 6 Linear Regression 6 7 Finding Input/Output of a Function 30 8 Domain and Range of a Function 33 9 Piecewise Defined Functions Inverse Functions: A First Look Rate of Change and Concavity 44 1 Quadratic Functions: Zeros and Concavity Exponential Growth and Decay Exponential Functions Versus Linear Functions The Effect of the Parameters a and b Continuous Growth Rate and the Number e Logarithms and Their Properties Logarithmic and Exponential Equations Logarithmic Functions and Their Graphs 73 0 Logarithmic Scales  Fitting Exponential Functions to Data 77 3
4 1 Vertical and Horizontal Shifts 79 Reflections and Symmetry 84 3 Vertical Stretches and Compressions 89 4 Horizontal Stretches and Compressions 94 5 Graphs of Quadratic Functions 98 6 Composition and Decomposition of Functions Inverse Functions Combining Functions Power Functions Polynomial Functions The ShortRun Behavior of Polynomials 17 3 Rational Functions The ShortRun Behavior of Rational Functions Angles and Arcs Circular Functions Graphs of the Sine and Cosine Functions Graphs of the Other Trigonometric Functions Translations of Trigonometric Functions Verifying Trigonometric Identities Sum and Difference Identities The DoubleAngle and HalfAngle Identities 05 4
5 4 Conversion Identities Inverse Trigonometric Functions Trigonometric Equations 33 5
6 1 Functions and Function Notation Functions play a crucial role in mathematics. A function describes how one quantity depends on others. More precisely, when we say that a quantity y is a function of a quantity x we mean a rule that assigns to every possible value of x exactly one value of y. We call x the input and y the output. In function notation we write y = f(x) Since y depends on x it makes sense to call x the independent variable and y the dependent variable. In applications of mathematics, functions are often representations of real world phenomena. Thus, the functions in this case are referred to as mathematical models. If the set of input values is a finite set then the models are known as discrete models. Otherwise, the models are known as continuous models. For example, if H represents the temperature after t hours for a specific day, then H is a discrete model. If A is the area of a circle of radius r then A is a continuous model. There are four common ways in which functions are presented and used: By words, by tables, by graphs, and by formulas. Example 1.1 The sales tax on an item is 6%. So if p denotes the price of the item and C the total cost of buying the item then if the item is sold at $ 1 then the cost is 1 + (0.06)(1) = $1.06 or C(1) = $1.06. If the item is sold at $ then the cost of buying the item is + (0.06)() = $.1, or C() = $.1, and so on. Thus we have a relationship between the quantities C and p such that each value of p determines exactly one value of C. In this case, we say that C is a function of p. Describes this function using words, a table, a graph, and a formula. Words: To find the total cost, multiply the price of the item by 0.06 and add the result to the price. Table: The chart below gives the total cost of buying an item at price p as a function of p for 1 p 6. p C
7 Graph: The graph of the function C is obtained by plotting the data in the above table. See Figure 1. Formula: The formula that describes the relationship between C and p is given by C(p) = 1.06p. Figure 1 Example 1. The income tax T owed in a certain state is a function of the taxable income I, both measured in dollars. The formula is T = 0.11I 500. (a) Express using functional notation the tax owed on a taxable income of $ 13,000, and then calculate that value. (b) Explain the meaning of T (15, 000) and calculate its value. (a) The functional notation is given by T (13, 000) and its value is T (13, 000) = 0.11(13, 000) 500 = $930. (b) T (15, 000) is the tax owed on a taxable income of $15,000. Its value is T (15, 000) = 0.11(15, 000) 500 = $1,
8 Emphasis of the Four Representations A formula has the advantage of being both compact and precise. However, not much insight can be gained from a formula as from a table or a graph. A graph provides an overall view of a function and thus makes it easy to deduce important properties. Tables often clearly show trends that are not easily discerned from formulas, and in many cases tables of values are much easier to obtain than a formula. Remark 1.1 To evaluate a function given by a graph, locate the point of interest on the horizontal axis, move vertically to the graph, and then move horizontally to the vertical axis. The function value is the location on the vertical axis. Now, most of the functions that we will encounter in this course have formulas. For example, the area A of a circle is a function of its radius r. In function notation, we write A(r) = πr. However, there are functions that can not be represented by a formula. For example, the value of Dow Jones Industrial Average at the close of each business day. In this case the value depends on the date, but there is no known formula. Functions of this nature, are mostly represented by either a graph or a table of numerical data. Example 1.3 The table below shows the daily low temperature for a oneweek period in New York City during July. (a) What was the low temperature on July 19? (b) When was the low temperature 73 F? (c) Is the daily low temperature a function of the date?explain. (d) Can you express T as a formula? D T (a) The low temperature on July 19 was 69 F. (b) On July 17 and July 0 the low temperature was 73 F. (c) T is a function of D since each value of D determines exactly one value of T. (d) T can not be expressed by an exact formula. 8
9 So far, we have introduced rules between two quantities that define functions. Unfortunately, it is possible for two quantities to be related and yet for neither quantity to be a function of the other. Example 1.4 Let x and y be two quantities related by the equation (a) Is x a function of y? Explain. (b) Is y a function of x? Explain. x + y = 4. (a) For y = 0 we have two values of x, namely, x = and x =. So x is not a function of y. (b) For x = 0 we have two values of y, namely, y = and y =. So y is not a function of x. Next, suppose that the graph of a relationship between two quantities x and y is given. To say that y is a function of x means that for each value of x there is exactly one value of y. Graphically, this means that each vertical line must intersect the graph at most once. Hence, to determine if a graph represents a function one uses the following test: Vertical Line Test: A graph is a function if and only if every vertical line crosses the graph at most once. According to the vertical line test and the definition of a function, if a vertical line cuts the graph more than once, the graph could not be the graph of a function since we have multiple y values for the same xvalue and this violates the definition of a function. Example 1.5 Which of the graphs (a), (b), (c) in Figure represent y as a function of x? 9
10 Figure By the vertical line test, (b) represents a function whereas (a) and (c) fail to represent functions since one can find a vertical line that intersects the graph more than once. Recommended Problems (pp. 69): 1, 3, 4, 5, 6, 7, 10, 1, 13, 14, 17, 0, 6, 8. 10
11 The Rate of Change Functions given by tables of values have their limitations in that nearly always leave gaps. One way to fill these gaps is by using the average rate of change. For example, Table 1 below gives the population of the United States between the years d(year) N(in millions) Table 1 This table does not give the population in 197. One way to estimate N(197), is to find the average yearly rate of change of N from 1970 to 1980 given by Then, =.35 million people per year. N(197) = N(1970) + (.35) = million. Average rates of change can be calculated not only for functions given by tables but also for functions given by formulas. The average rate of change of a function y = f(x) from x = a to x = b is given by the difference quotient y x = Change in function value Change in x value = f(b) f(a). b a Geometrically, this quantity represents the slope of the secant line going through the points (a, f(a)) and (b, f(b)) on the graph of f(x). See Figure 3. The average rate of change of a function on an interval tells us how much the function changes, on average, per unit change of x within that interval. On some part of the interval, f may be changing rapidly, while on other parts f may be changing slowly. The average rate of change evens out these variations. 11
12 Figure 3 Example.1 Find the average value of the function f(x) = x from x = 3 to x = 5. The average rate of change is y x = f(5) f(3) 5 3 = 5 9 = 8. Example. (Average Speed) During a typical trip to school, your car will undergo a series of changes in its speed. If you were to inspect the speedometer readings at regular intervals, you would notice that it changes often. The speedometer of a car reveals information about the instantaneous speed of your car; that is, it shows your speed at a particular instant in time. The instantaneous speed of an object is not to be confused with the average speed. Average speed is a measure of the distance traveled in a given period of time. That is, Average Speed = Distance traveled Time elapsed. If the trip to school takes 0. hours (i.e. 1 minutes) and the distance traveled is 5 miles then what is the average speed of your car? The average velocity is given by 1
13 Ave. Speed = 5 miles 0. hours = 5miles/hour. This says that on the average, your car was moving with a speed of 5 miles per hour. During your trip, there may have been times that you were stopped and other times that your speedometer was reading 50 miles per hour; yet on the average you were moving with a speed of 5 miles per hour. Average Rate of Change and Increasing/Decreasing Functions Now, we would like to use the concept of the average rate of change to test whether a function is increasing or decreasing on a specific interval. First, we introduce the following definition: We say that a function is increasing if its graph climbs as x moves from left to right. That is, the function values increase as x increases. It is said to be decreasing if its graph falls as x moves from left to right. This means that the function values decrease as x increases. As an application of the average rate of change, we can use such quantity to decide whether a function is increasing or decreasing. If a function f is increasing on an interval I then by taking any two points in the interval I, say a < b, we see that f(a) < f(b) and in this case f(b) f(a) b a > 0. Going backward with this argument we see that if the average rate of change is positive in an interval then the function is increasing in that interval. Similarly, if the average rate of change is negative in an interval I then the function is decreasing there. Example.3 The table below gives values of a function w = f(t). Is this function increasing or decreasing? t w The average of w over the interval [0, 4] is w(4) w(0) 4 0 = = 10.5
14 The average rate of change of the remaining intervals are given in the chart below time interval [0,4] [4,8] [8,1] [1,16] [16, 0] [0,4] Average Since the average rate of change is always negative on [0, 4] then the function is decreasing on that interval. Of Course, you can see from the table that the function is decreasing since the output values are decreasing as x increases. The purpose of this problem is to show you how the average rate of change is used to determine whether a function is increasing or decreasing. Some functions can be increasing on some intervals and decreasing on other intervals. These intervals can often be identified from the graph. Example.4 Determine the intervals where the function is increasing and decreasing. Figure 4 The function is increasing on (, 1) (1, ) and decreasing on the interval ( 1, 1). Recommended Problems (pp ): 1,, 3, 4, 5, 8, 9, 10, 11, 13, 14,
15 3 Linear Functions In the previous section we introduced the average rate of change of a function. In general, the average rate of change of a function is different on different intervals. For example, consider the function f(x) = x. The average rate of change of f(x) on the interval [0, 1] is f(1) f(0) 1 0 = 1. The average rate of change of f(x) on [1, ] is f() f(1) 1 = 3. A linear function is a function with the property that the average rate of change on any interval is the same. We say that y is changing at a constant rate with respect to x. Thus, y changes by the some amount for every unit change in x. Geometrically, the graph is a straight line ( and thus the term linear). Example 3.1 Suppose you pay $ 19 to rent a booth for selling necklaces at an art fair. The necklaces sell for $ 3. Explain why the function that shows your net income (revenue from sales minus rental fees) as a function of the number of necklaces sold is a linear function. Let P (n) denote the net income from selling n necklaces. Each time a necklace is sold, that is, each time n is increased by 1, the net income P is increased by the same constant, $3. Thus the rate of change for P is always the same, and hence P is a linear function. Testing Data for Linearity Next, we will consider the question of recognizing a linear function given by a table. Let f be a linear function given by a table. Then the rate of change is the same for all pairs of points in the table. In particular, when the x values are evenly spaced the change in y is constant. 15
16 Example 3. Which of the following tables could represent a linear function? x f(x) x g(x) Since equal increments in x yield equal increments in y then f(x) is a linear function. On the contrary, since then g(x) is not linear It is possible to have a table of linear data in which neither the xvalues nor the yvalues go up by equal amounts. However, the rate of change of any pairs of points in the table is constant. Example 3.3 The following table contains linear data, but some data points are missing. Find the missing data points. x 5 8 y Consider the points (, 5), (5, a), (b, 17), (8, 3), and (c, 9). Since the data is linear then we must have a 5 = 3 5 a 5. That is, = 3. Cross multiplying to obtain a 5 = 9 or a = 14. It follows that when x in increased by 1, y increases by 3. Hence, b = 6 and c = 10. Now, suppose that f(x) is a linear function of x. Then f changes at a constant rate m. That is, if we pick two points (0, f(0)) and (x, f(x)) then m = f(x) f(0). x 0 That is, f(x) = mx + f(0). This is the function notation of the linear function f(x). Another notation is the equation notation, y = mx + f(0). We will denote the number f(0) by b. In this case, the linear function will be written as f(x) = mx+b or y = mx+b. Since b = f(0) then the point (0, b) is 16
17 the point where the line crosses the vertical line. We call it the yintercept. So the yintercept is the output corresponding to the input x = 0, sometimes known as the initial value of y. If we pick any two points (x 1, y 1 ) and (x, y ) on the graph of f(x) = mx + b then we must have m = y y 1 x x 1. We call m the slope of the line. Example 3.4 The value of a new computer equipment is $0,000 and the value drops at a constant rate so that it is worth $ 0 after five years. Let V (t) be the value of the computer equipment t years after the equipment is purchased. (a) Find the slope m and the yintercept b. (b) Find a formula for V (t). (a) Since V (0) = 0, 000 and V (5) = 0 then the slope of V (t) is m = 0 0, = 4, 000 and the vertical intercept is V (0) = 0, 000. (b) A formula of V (t) is V (t) = 4, 000t + 0, 000. In financial terms, the function V (t) is known as the straightline depreciation function. Recommended Problems (pp. 35): 1, 3, 5, 7, 9, 11, 1, 13, 18, 0, 1, 8. 17
18 4 Formulas for Linear Functions In this section we will discuss ways for finding the formulas for linear functions. Recall that f is linear if and only if f(x) can be written in the form f(x) = mx + b. So the problem of finding the formula of f is equivalent to finding the slope m and the vertical intercept b. Suppose that we know two points on the graph of f(x), say (x 1, f(x 1 )) and (x, f(x )). Since the slope m is just the average rate of change of f(x) on the interval [x 1, x ] then m = f(x ) f(x 1 ) x x 1. To find b, we use one of the points in the formula of f(x); say we use the first point. Then f(x 1 ) = mx 1 + b. Solving for b we find b = f(x 1 ) mx 1. Example 4.1 Let s find the formula of a linear function given by a table of data values. The table below gives data for a linear function. Find the formula. x f(x) We use the first two points to find the value of m : m = f(1.3) f(1.) To find b we can use the first point to obtain = = 1.. Solving for b we find b =.. Thus, = 1.(1.) + b. f(x) = 1.x +. 18
19 Example 4. Suppose that the graph of a linear function is given and two points on the graph are known. For example, Figure 5 is the graph of a linear function going through the points (100, 1) and (160, 6). Find the formula. The slope m is found as follows: Figure 5 m = = To find b we use the first point to obtain 1 = 0.083(100) + b. Solving for b we find b = 7.3. So the formula for the line is f(x) = x. Example 4.3 Sometimes a linear function is given by a verbal description as in the following problem: In a college meal plan you pay a membership fee; then all your meals are at a fixed price per meal. If 30 meals cost $15.50 and 60 meals cost $50 then find the formula for the cost C of a meal plan in terms of the number of meals n. We find m first: m = = $3.5/meal. To find b or the membership fee we use the point (30, 15.50) in the formula C = mn + b to obtain = 3.5(30) + b. Solving for b we find b = $55. Thus, C = 3.5n
20 So far we have represented a linear function by the expression y = mx + b. This is known as the slopeintercept form of the equation of a line. Now, if the slope m of a line is known and one point (x 0, y 0 ) is given then by taking any point (x, y) on the line and using the definition of m we find y y 0 x x 0 = m. Cross multiply to obtain: y y 0 = m(x x 0 ). This is known as the pointslope form of a line. Example 4.4 Find the equation of the line passing through the point (100, 1) and with slope m = Using the above formula we have: y 1 = 0.01(x 100) or y = 0.01x. Note that the form y = mx + b can be rewritten in the form Ax + By + C = 0. (1) where A = m, B = 1, and C = b. The form (1) is known as the standard form of a linear function. Example 4.5 Rewrite in standard form: 3x + y + 40 = x y. Subtracting x y from both sides to obtain x + 3y + 40 = 0. Recommended Problems (pp. 303): 3, 5, 6, 10, 11, 1, 13, 14, 17, 19, 1,, 3, 6, 7, 30, 31, 3,34. 0
21 5 Geometric Properties of Linear Functions In this section we discuss four geometric related questions of linear functions. The first question considers the significance of the parameters m and b in the equation f(x) = mx + b. We have seen that the graph of a linear function f(x) = mx + b is a straight line. But a line can be horizontal, vertical, rising to the right or falling to the right. The slope is the parameter that provides information about the steepness of a straight line. If m = 0 then f(x) = b is a constant function whose graph is a horizontal line at (0, b). For a vertical line, the slope is undefined since any two points on the line have the same xvalue and this leads to a division by zero in the formula for the slope. The equation of a vertical line has the form x = a. Suppose that the line is neither horizontal nor vertical. If m > 0 then by Section 3, f(x) is increasing. That is, the line is rising to the right. If m < 0 then f(x) is decreasing. That is, the line is falling to the right. The slope, m, tells us how fast the line is climbing or falling. The larger the value of m the more the line rises and the smaller the value of m the more the line falls. The parameter b tells us where the line crosses the vertical axis. Example 5.1 Arrange the slopes of the lines in the figure from largest to smallest. Figure 6 1
22 According to Figure 6 we have m G > m F > m D > m A > m E > m B > m C. The second question of this section is the question of finding the point of intersection of two lines. The point of intersection of two lines is basically the solution to a system of two linear equations. This system can be solved by the method of substitution which we describe in the next example. Example 5. Find the point of intersection of the two lines y + x = 3 and (x + y) = 1 y. Solving the first equation for y we obtain y = 3 x. Substituting this expression in the second equation to obtain ( ( x + 3 x )) ( = 1 3 x ). Thus, x + 6 x = + x x + 6 = + x x + 1 = 4 + x x = 16. Using this value of x in the first equation to obtain y = 3 16 = 11. Our third question in this section is the question of determining when two lines are parallel,i.e. they have no points in common. As we noted earlier in this section, the slope of a line determines the direction in which it points. Thus, if two lines have the same slope then the two lines are either parallel (if they have different vertical intercepts) or coincident ( if they have same yintercept). Also, note that any two vertical lines are parallel even though their slopes are undefined. Example 5.3 Line l in Figure 7 is parallel to the line y = x + 1. Find the coordinates of the point P.
23 Figure 7 Since the two lines are parallel then the slope of the line l is. Since the vertical intercept of l is then the equation of l is y = x. The point P is the xintercept of the line l, i.e., P (x, 0). To find x, we set x = 0. Solving for x we find x = 1. Thus, P (1, 0). Example 5.4 Find the equation of the line l passing through the point (6, 5) and parallel to the line y = 3 3 x. The slope of l is m = since the two lines are parallel. Thus, the equation 3 of l is y = x + b. To find the value of b, we use the given point. Replacing 3 y by 5 and x by 6 to obtain, 5 = (6) + b. Solving for b we find b = 9. 3 Hence, y = 9 x. 3 The final question of this section is the question of determining when two lines are perpendicular. It is clear that if one line is horizontal and the second is vertical then the two lines are perpendicular. So we assume that neither of the two lines is horizontal or vertical. Hence, their slopes are defined and nonzero. Let s see how the slopes of lines that are perpendicular compare. Call the two lines l 1 and l and let A be the point where they intersect. From A take a horizontal segment of length 1 and from the rightendpoint C of that segment construct a vertical line that intersect l 1 at B and l at D. See Figure 8. 3
24 Figure 8 It follows from this construction that if m 1 is the slope of l 1 then Similarly, the slope of l is m 1 = CB CA = CB. m = CD CA = CD. Since ABD is a right triangle at A then DAC = 90 BAC. Similarly, ABC = 90 BAC. Thus, DAC = ABC. A similar argument shows that CDA = CAB. Hence, the triangles ACB and DCA are similar. As a consequence of this similarity we can write or CB CA = CA CD m 1 = 1 m. Thus, if two lines are perpendicular, then the slope of one is the negative reciprocal of the slope of the other. Example 5.5 Find the equation of the line l in Figure 9. 4
25 Figure 9 The slope of l 1 is m 1 =. Since the two lines are perpendicular then the 3 slope of l is m = 3. The horizontal intercept of l is 0. Hence, the equation of the line l is y = 3x. Recommended Problems (pp. 394): 1, 3, 4, 5, 6, 7, 9, 10, 11, 13, 15, 19, 1,, 3, 5, 6, 8. 5
26 6 Linear Regression In general, data obtained from real life events, do not match perfectly simple functions. Very often, scientists, engineers, mathematicians and business experts can model the data obtained from their studies, with simple linear functions. Even if the function does not reproduce the data exactly, it is possible to use this modeling for further analysis and predictions. This makes the linear modeling extremely valuable. Let s try to fit a set of data points from a crankcase motor oil producing company. They want to study the correlation between the number of minutes of TV advertisement per day for their product, and the total number of oil cases sold per month for each of the different advertising campaigns. The information is given in the following table : x:tv ads(min/day) y:units sold(in millions) Using TI83 we obtain the scatter plot of this given data (See Figure 10.) The steps of getting the graph are discussed later in this section. Figure 10 Figure 11 below shows the plot and the optimum linear function that describes the data. That line is called the best fitting line and has been derived with a very commonly used statistical technique called the method of least squares. The line shown was chosen to minimize the sum of the squares of the vertical distances between the data points and the line. 6
27 Figure 11 The measure of how well this linear function fits the experimental points, is called regression analysis. Graphic calculators, such as the TI83, have built in programs which allow us to find the slope and the y intercept of the best fitting line to a set of data points. That is, the equation of the best linear fit. The calculators also give as a result of their procedure, a very important value called the correlation coefficient. This value is in general represented by the letter r and it is a measure of how well the best fitting line fits the data points. Its value varies from  1 to 1. The TI83 prompts the correlation coefficient r as a result of the linear regression. If it is negative, it is telling us that the line obtained has negative slope. Positive values of r indicate a positive slope in the best fitting line. If r is close to 0 then the data may be completely scattered, or there may be a nonlinear relationship between the variables. The square value of the correlation coefficient r is generally used to determine if the best fitting line can be used as a model for the data. For that reason, r is called the coefficient of determination. In most cases, a function is accepted as the model of the data, if this coefficient of determination is greater than 0.5. A coefficient of determination tells us which percent of the variation on the real data is explained by the best fitting line. An r = 0.9 means that 9% of the variation on the data points is described by the best fitting line. The closer the coefficient of determination is to 1 the better the fit. The following are the steps required to find the best linear fit using a TI83 7
28 graphing calculator. 1. Enter the data into two lists L1 and L. a. Push the STAT key and select the Edit option. b. Up arrow to move to the Use the top of the list L1. c. Clear the list by hitting CLEAR ENTER. d. Type in the x values of the data. Type in the number and hit enter. e. Move to the list L, clear it and enter the y data in this list.. Graph the data as a scatterplot. a. Hit nd STAT PLOT. (upper left) b. Move to plot 1 and hit enter. c. Turn the plot on by hitting enter on the ON option. d. Move to the TYPE option and select the dot graph type. Hit enter to select it. e. Move to the Xlist and enter in nd L1. f. Move to the Ylist and enter in nd L. g. Move to Mark and select the small box option. h. Hit ZOOM and select ZOOMSTAT. 3. Fit a line to the data. a. Turn on the option to display the correlation coefficient, r. This is accomplished by hitting nd Catalog (lower left). Scroll down the list until you find Diagnostic On, hit enter for this option and hit enter a second time to activate this option. The correlation coefficient will be displayed when you do the linear regression. b. Hit STAT, CALC, and select option 8 LinREg. c. Enter nd 1, nd, with a comma in between. d. Press ENTER. The equation for a line through the data is shown. The slope is b, the intercept is a, the correlation coefficient is r, and the coefficient of determination is r. 4. Graph the best fit line with the data. a. Press Y=, then press VARS to open the Variables window. 8
29 b. Arrow down to select 5: Statistics... then press ENTER. c. Right arrow over to select EQ and press ENTER. This places the formula for the regression equation into the Y= window. d. Press GRAPH to graph the equation. Your window should now show the graph of the regression equation as well as each of the data points. Recommended Problems (pp ): 1, 3, 5, 7. 9
30 7 Finding Input/Output of a Function In this section we discuss ways for finding the input or the output of a function defined by a formula, table, or a graph. Finding the Input and the Output Values from a Formula By evaluating a function, we mean figuring out the output value corresponding to a given input value. Thus, notation like f(10) = 4 means that the function s output, corresponding to the input 10, is equal to 4. If the function is given by a formula, say of the form y = f(x), then to find the output value corresponding to an input value a we replace the letter x in the formula of f by the input a and then perform the necessary algebraic operations to find the output value. Example 7.1 Let g(x) = x +1. Evaluate the following expressions: 5+x (a) g() (b) g(a) (c) g(a) (d) g(a) g(). (a) g() = +1 = (b) g(a) = a +1 5+a (c) g(a) = a +1 5+a = a a 9 5+a 5+a 5+a (d) g(a) g() = a +1 5 = 7(a +1) 5+a a = 7a 5a 18. 7(5+a) 7 5+a 7a+35 Now, finding the input value of a given output is equivalent to solving an algebraic equation. Example 7. Consider the function y = 1 x 4. (a) Find an xvalue that results in y =. (b) Is there an xvalue that results in y =? Explain. 1 (a) Letting y = to obtain x 4 = or x 4 = 0.5. Squaring both sides to obtain x 4 = 0.5 and adding 4 to otbain x = 4.5. (b) Since y = 1 x 4 then the righthand side is always positive so that y > 0. The equation y = has no solutions. 30
31 Finding Output and Input Values from Tables Next, suppose that a function is given by a table of numeric data. For example, the table below shows the daily low temperature T for a oneweek period in New York City during July. D T Then T (18) = 77 F. This means, that the low temperature on July 18, was 77 F. Remark 7.1 Note that, from the above table one can find the value of an input value given an output value listed in the table. For example, there are two values of D such that T (D) = 75, namely, D = 1 and D =. Finding Output and Input Values from Graphs Finally, to evaluate the output(resp. input) value of a function from its graph, we locate the input (resp. output) value on the horizontal (resp. vertical) line and then we draw a line perpendicular to the xaxis (resp. yaxis) at the input (resp. output) value. This line will cross the graph of the function at a point whose yvalue (resp. xvalue) is the function s output (resp. input) value. Example 7.3 (a) Using Figure 1, evaluate f(.5). (b) For what value of x, f(x) =? Figure 1 31
32 (a) f(.5) 1.5 (b) f( ) =. Recommended Problems (pp ): 5, 6, 9, 10, 11, 13, 16, 17, 18, 19, 0, 3, 5. 3
33 8 Domain and Range of a Function If we try to find the possible input values that can be used in the function y = x we see that we must restrict x to the interval [, ), that is x. Similarly, the function y = 1 takes only certain values for the output, namely, y > 0. Thus, a function is often defined for certain values of x x and the dependent variable often takes certain values. The above discussion leads to the following definitions: By the domain of a function we mean all possible input values that yield an input value. Graphically, the domain is part of the horizontal axis. The range of a function is the collection of all possible output values. The range is part of the vertical axis. The domain and range of a function can be found either algebraically or graphically. Finding the Domain and the Range Algebraically When finding the domain of a function, ask yourself what values can t be used. Your domain is everything else. There are simple basic rules to consider: The domain of all polynomial functions, i.e. functions of the form f(x) = a n x n + a n 1 x n a 1 x + a 0, where n is nonnegative integer, is the Real numbers R. Square root functions can not contain a negative underneath the radical. Set the expression under the radical greater than or equal to zero and solve for the variable. This will be your domain. Fractional functions, i.e. ratios of two functions, determine for which input values the numerator and denominator are not defined and the domain is everything else. For example, make sure not to divide by zero! Example 8.1 Find, algebraically, the domain and the range of each of the following functions. Write your answers in interval notation: (a) y = πx (b) y = 1 x 4 (c) y = + 1 x. (a) Since the function is a polynomial then its domain is the interval (, ). 33
34 To find the range, solve the given equation for x in terms of y obtaining x = ± y. Thus, x exists for y 0. So the range is the interval [0, ). π (b) The domain of y = 1 x 4 consists of all numbers x such that x 4 > 0 or x > 4. That is, the interval (4, ). To find the range, we solve for x in terms of y > 0 obtaining x = x exists for all y > 0. Thus, the range is the y interval (0, ). (c) The domain of y = + 1 is the interval (, 0) (0, ). To find x the range, write x in terms of y to obtain x = 1. The values of y for y which this later formula is defined is the range of the given function, that is, (, ) (, ). Remark 8.1 Note that the domain of the function y = πx of the previous problem consists of all real numbers. If this function is used to model a realworld situation, that is, if the x stands for the radius of a circle and y is the corresponding area then the domain of y in this case consists of all numbers x 0. In general, for a word problem the domain is the set of all x values such that the problem makes sense. Finding the Domain and the Range Graphically We often use a graphing calculator to find the domain and range of functions. In general, the domain will be the set of all x values that has corresponding points on the graph. We note that if there is an asymptote (shown as a vertical line on the TI series) we do not include that x value in the domain. To find the range, we seek the top and bottom of the graph. The range will be all points from the top to the bottom (minus the breaks in the graph). Example 8. Use a graphing calculator to find the domain and the range of each of the following functions. Write your answers in interval notation: (a) y = πx (b) y = 1 x 4 (c) y = + 1 x. (a) The graph of y = πx is given in Figure
35 Figure 13 The domain is the set (, ) and the range is [0, ). (b) The graph of y = 1 x 4 is given in Figure 14 Figure 14 The domain is the set (4, ) and the range is (0, ). (c) The graph of y = + 1 is given in Figure 15. x Figure 15 35
36 The domain is the set (, 0) (0, ) and the range is (, ) (, ). Recommended Problems (pp. 701):, 4, 5, 8, 11, 13, 15, 0,, 7, 8, 3. 36
37 9 Piecewise Defined Functions Piecewisedefined functions are functions defined by different formulas for different intervals of the independent variable. Example 9.1 (The Absolute Value Function) (a) Show that the function f(x) = x is a piecewise defined function. (b) Graph f(x). (a) The absolute value function x is a piecewise defined function since { x for x 0 x = x for x < 0. (b) The graph is given in Figure 16. Figure 16 Example 9. (The Ceiling Function) The Ceiling function f(x) = x is the piecewise defined function given by x = smallest integer greater than x. Sketch the graph of f(x) on the interval [ 3, 3]. The graph is given in Figure 17. An open circle represents a point which is 37
38 not included. Figure 17 Example 9.3 (The Floor Function) The Floor function f(x) = x is the piecewise defined function given by x = smallest integer less than or equal to x. Sketch the graph of f(x) on the interval [ 3, 3]. The graph is given in Figure 18. Figure 18 38
39 Example 9.4 Sketch the graph of the piecewise defined function given by x + 4 for x f(x) = for < x < 4 x for x. The following table gives values of f(x). x f(x) 1 1 The graph of the function is given in Figure 19. Figure 19 We conclude this section with the following realworld situation: Example 9.5 The charge for a taxi ride is $1.50 for the first 1 of a mile, and $0.5 for each 5 additional 1 of a mile (rounded up to the nearest 1 mile). 5 5 (a) Sketch a graph of the cost function C as a function of the distance traveled x, assuming that 0 x 1. (b) Find a formula for C in terms of x on the interval [0, 1]. (c) What is the cost for a 4 mile ride? 5 (a) The graph is given in Figure 0. 39
40 (b) A formula of C(x) is C(x) = Figure 0 (c) The cost for a 4 5 ride is C( 4 5 ) = $ if 0 x if 1 5 < x 5.00 if 5 < x if 3 5 < x if 4 5 < x 1. Recommended Problems (pp ): 1, 3, 4, 5, 7, 8, 11, 1, 14,
41 10 Inverse Functions: A First Look We have seen that when every vertical line crosses a curve at most once then the curve is the graph of a function f. We called this procedure the vertical line test. Now, if every horizontal line crosses the graph at most once then the function can be used to build a new function, called the inverse function and is denoted by f 1, such that if f takes an input x to an output y then f 1 takes y as its input and x as its output. That is f(x) = y if and only if f 1 (y) = x. When a function has an inverse then we say that the function is invertible. Remark 10.1 The test used to identify invertible functions which we discussed above is referred to as the horizontal line test. Example 10.1 Use a graphing calculator to decide whether or not the function is invertible, that is, has an inverse function: (a) f(x) = x (b) g(x) = x. (a) Using a graphing calculator, the graph of f(x) is given in Figure 1. We see that every horizontal line crosses the graph once so the function is invertible. 41
42 (b) The graph of g(x) = x (See Figure 16, Section 9) shows that there are horizontal lines that cross the graph twice so that g is not invertible. Remark 10. It is important not to confuse between f 1 (x) and (f(x)) 1. The later is just the reciprocal of f(x), that is, (f(x)) 1 = 1 whereas the former is how the f(x) inverse function is represented. Domain and Range of an Inverse Function Figure shows the relationship between f and f 1. Figure This figure shows that we get the inverse of a function by simply reversing the direction of the arrows. That is, the outputs of f are the inputs of f 1 and the outputs of f 1 are the inputs of f. It follows that Domain of f 1 = Range of f and Range of f 1 = Domain of f. Example 10. Consider the function f(x) = x 4. (a) Find the domain and the range of f(x). (b) Use the horizontal line test to show that f(x) has an inverse. (c) What are the domain and range of f 1? (a) The function f(x) is defined for all x 4. The range is the interval [0, ). (b) Graphing f(x) we see that f(x) satisfies the horizontal line test and so f has an inverse. See Figure 3. 4
43 (c) The domain of f 1 is the range of f, i.e. the interval [0, ). The range of f 1 is the domain of f, that is, the interval [4, ). Figure 3 Finding a Formula for the Inverse Function How do you find the formula for f 1 from the formula of f? The procedure consists of the following steps: 1. Replace f(x) with y.. Interchange the letters x and y. 3. Solve for y in terms of x. 4. Replace y with f 1 (x). Example 10.3 Find the formula for the inverse function of f(x) = x As seen in Example 10.1, f(x) is invertible. We find its inverse as follows: 1. Replace f(x) with y to obtain y = x Interchange x and y to obtain x = y Solve for y to obtain y 3 = x 7 or y = 3 x Replace y with f 1 (x) to obtain f 1 (x) = 3 x 7. Remark 10.3 More discussion of inverse functions will be covered in Section 7. Recommended Problems (pp ): 1,, 3, 4, 6, 7, 11, 13, 14, 15, 17,, 3. 43
44 11 Rate of Change and Concavity We have seen that when the rate of change of a function is constant then its graph is a straight line. However, not all graphs are straight lines; they may bend up or down as shown in the following two examples. Example 11.1 Consider the following two graphs in Figure 4. Figure 4 (a) What do the graphs above have in common? (b) How are they different? Specifically, look at the rate of change of each. (a) Both graphs represent increasing functions. (b) The rate of change of f(x) is more and more positive so the graph bends up whereas the rate of change of g(x) is less and less positive and so it bends down. The following example deals with version of the previous example for decreasing functions. Example 11. Consider the following two graphs given in Figure 5. 44
45 Figure 5 (a) What do the graphs above have in common? (b) How are they different? Specifically, look at the rate of change of each. (a) Both functions are decreasing. (b) The rate of change of f(x) is more and more negative so the graph bends down, whereas the rate of change of g(x) is less and less negative so the graph bends up. Conclusions: When the rate of change of a function is increasing then the function is concave up. That is, the graph bends upward. When the rate of change of a function is decreasing then the function is concave down. That is, the graph bends downward. The following example discusses the concavity of a function given by a table. Example 11.3 Given below is the table for the function H(x). Calculate the rate of change for successive pairs of points. Decide whether you expect the graph of H(x) to concave up or concave down? x H(x) H(15) H(1) = H(18) H(15) = H(1) H(18) =
46 Since the rate of change of H(x) is increasing then the function is concave up. Remark 11.1 Since the graph of a linear function is a straight line, that is its rate of change is constant, then it is neither concave up nor concave down. Recommended Problems (pp. 834): 1, 3, 5, 6, 7, 9, 10, 11, 13, 15,
47 1 Quadratic Functions: Zeros and Concavity You recall that a linear function is a function that involves a first power of x. A function of the form f(x) = ax + bx + c, a 0 is called a quadratic function. The word quadratus is the latin word for a square. Quadratic functions are useful in many applications in mathematics when a linear function is not sufficient. For example, the motion of an object thrown either upward or downward is modeled by a quadratic function. The graph of a quadratic function is known as a parabola and has a distinctive shape that occurs in nature. Geometrical discussion of quadratic functions will be covered in Section 5. Finding the Zeros of a Quadratic Function In many applications one is interested in finding the zeros or the xintercepts of a quadratic function. This means we wish to find all possible values of x for which ax + bx + c = 0. For example, if v(t) = t 4t + 4 is the velocity of an object in meters per second then one may be interested in finding the time when the object is not moving. Finding the zeros of a quadratic function can be accomplished in two ways: By Factoring: To factor ax + bx + c 1. find two integers that have a product equal to ac and a sum equal to b,. replace bx by two terms using the two new integers as coefficients, 3. then factor the resulting fourterm polynomial by grouping. Thus, obtaining a(x r)(x s) = 0. But we know that if the product of two numbers is zero uv = 0 then either u = 0 or v = 0. Thus, eiher x = r or x = s. Example 1.1 Find the zeros of f(x) = x x 8. 47
48 We need two numbers whose product is 8 and sum is. Such two integers are 4 and. Thus, Thus, either x = or x = 4. x x 8 = x + x 4x 8 = x(x + ) 4(x + ) = (x + )(x 4) = 0. Example 1. Find the zeros of f(x) = x + 9x + 4. We need two integers whose product is ac = 8 and sum equals to b = 9. Such two integers are 1 and 8. Thus, x + 9x + 4 = x + x + 8x + 4 = x(x + 1) + 4(x + 1) = (x + 1)(x + 4) Hence, the zeros are x = 1 and x = 4. By Using the Quadratic Formula: Many quadratic functions are not easily factored. For example, the function f(x) = 3x 7x 7. However, the zeros can be found by using the quadratic formula which we derive next: ax + bx + c = 0 (subtract c from both sides) ax + bx = c (multiply both sides by 4a) 4a x + 4abx = 4ac (add b to both sides) 4a x + 4abx + b = b 4ac (ax + b) = b 4ac ax + b = ± b 4ac b± x = b 4ac a provided that b 4ac 0. This last formula is known as the quadratic formula. Note that if b 4ac < 0 then the equation ax + bx + c = 0 has no solutions. That is, the graph of f(x) = ax + bx + c does not cross the xaxis. 48
49 Example 1.3 Find the zeros of f(x) = 3x 7x 7. Letting a = 3, b = 7 and c = 7 in the quadratic formula we have x = 7 ± Example 1.4 Find the zeros of the function f(x) = 6x x + 5. Letting a = 6, b =, and c = 5 in the quadratic formula we obtain x = ± 1 But 0 is not a real number. Hence, the function has no zeros. Its graph does not cross the xaxis. Concavity of Quadratic Functions Graphs of quadratic functions are called parabolas. They are either always concave up (when a > 0) or always concave down (when a < 0). Example 1.5 Determine the concavity of f(x) = x +4 from x = 1 to x = 5 using rates of change over intervals of length. Graph f(x). Calculating the rates of change we find f(1) f( 1) = 0 1 ( 1) f(3) f(1) = f(5) f(3) = Since the rates of change are getting more and more negative then the graph is concave down from x = 1 to x = 5. See Figure 6. 49
50 Figure 6 Recommended Problems (pp. 889): 1,, 3, 5, 7, 9, 11, 1, 14, 15, 16,
51 13 Exponential Growth and Decay Exponential functions appear in many applications such as population growth, radioactive decay, and interest on bank loans. Recall that linear functions are functions that change at a constant rate. For example, if f(x) = mx + b then f(x + 1) = m(x + 1) + b = f(x) + m. So when x increases by 1, the y value increases by m. In contrast, an exponential function with base a is one that changes by constant multiples of a. That is, f(x + 1) = af(x). Writing a = 1 + r we obtain f(x + 1) = f(x) + rf(x). Thus, an exponential function is a function that changes at a constant percent rate. Exponential functions are used to model increasing quantities such as population growth problems. Example 13.1 Suppose that you are observing the behavior of cell duplication in a lab. In one experiment, you started with one cell and the cells doubled every minute. That is, the population cell is increasing at the constant rate of 100%. Write an equation to determine the number (population) of cells after one hour. Table below shows the number of cells for the first 5 minutes. Let P (t) be the number of cells after t minutes. t P(t) Table At time 0, i.e t=0, the number of cells is 1 or 0 = 1. After 1 minute, when t = 1, there are two cells or 1 =. After minutes, when t =, there are 4 cells or = 4. Therefore, one formula to estimate the number of cells (size of population) after t minutes is the equation (model) f(t) = t. 51
52 It follows that f(t) is an increasing function. Computing the rates of change to obtain f(1) f(0) = f() f(1) 1 = f(3) f() 3 = 4 f(4) f(3) 4 3 = 8 f(5) f(4) 5 4 = 16. Thus, the graph of f(t) is concave up. See Figure 7. Figure 7 Now, to determine the number of cells after one hour we convert to minutes to obtain t = 60 minutes so that f(60) = 60 = cells. Exponential functions can also model decreasing quantities known as decay models. Example 13. If you start a biology experiment with 5,000,000 cells and 45% of the cells are dying every minute, how long will it take to have less than 50,000 cells? Let P (t) be the number of cells after t minutes. Then P (t + 1) = P (t) 45%P (t) or P (t + 1) = 0.55P (t). By constructing a table of data we find 5
53 t P(t) 0 5,000,000 1,750,000 1,51, , , , , , , So it takes 8 minutes for the population to reduce to less than 50,000 cells. A formula of P (t) is P (t) = 5, 000, 000(0.55) t. The graph of P (t) is given in Figure 8. Figure 8 From the previous two examples, we see that an exponential function has the general form P (t) = b a t, a > 0 a 1. Since b = P (0) then we call b the initial value. We call a the base of P (t). If a > 1, then P (t) shows exponential growth with growth factor a. The graph of P will be similar in shape to that in Figure 7. If 0 < a < 1, then P shows exponential decay with decay factor a. The graph of P will be similar in shape to that in Figure 8. Since P (t + 1) = ap (t) then P (t + 1) = P (t) + rp (t) where r = a 1. We call r the percent growth rate. 53
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