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A graphical representation of a function––here the number of hours of daylight as a function of the time of year at various latitudes–– is often the most natural and convenient way to represent the function. Functions and Models The fundamental objects that we deal with in calculus are functions. This chapter prepares the way for calculus by discussing the basic ideas concerning functions, their graphs, and ways of transforming and combining them. We stress that a function can be represented in different ways: by an equation, in a table, by a graph, or in words. We look at the main types of functions that occur in calculus and describe the process of using these functions as mathematical models of real-world phenomena. We also discuss the use of graphing calculators and graphing software for computers. |||| 1.1 Four Ways to Represent a Function Functions arise whenever one quantity depends on another. Consider the following four situations. A. The area Aof a circle depends on the radius rof the circle. The rule that connects r and Ais given by the equation A(cid:2)(cid:2)r2. With each positive number rthere is associated one value of A,and we say that Ais a functionof r. B. The human population of the world Pdepends on the time t. The table gives estimates Population of the world population P(cid:2)t(cid:3)at time t,for certain years. For instance, Year (millions) P(cid:2)1950(cid:3)(cid:4)2,560,000,000 1900 1650 1910 1750 But for each value of the time tthere is a corresponding value of P,and we say that 1920 1860 Pis a function of t. 1930 2070 C. The cost Cof mailing a first-class letter depends on the weight wof the letter. 1940 2300 Although there is no simple formula that connects wand C,the post office has a rule 1950 2560 for determining Cwhen wis known. 1960 3040 1970 3710 D. The vertical acceleration aof the ground as measured by a seismograph during an 1980 4450 earthquake is a function of the elapsed time t.Figure 1 shows a graph generated by 1990 5280 seismic activity during the Northridge earthquake that shook Los Angeles in 1994. 2000 6080 For a given value of t,the graph provides a corresponding value of a. a {cm/s@} 100 50 5 10 15 20 25 30 t(seconds) FIGURE 1 _50 Vertical ground acceleration during the Northridge earthquake Calif. Dept. of Mines and Geology ❙❙❙❙ 12 CHAPTER 1 FUNCTIONS AND MODELS Each of these examples describes a rule whereby,given a number (r,t,w,or t),another number (A,P,C,or a) is assigned. In each case we say that the second number is a function of the first number. A function f is a rule that assigns to each element xin a set Aexactly one element,called f(cid:2)x(cid:3),in a set B. We usually consider functions for which the sets Aand Bare sets of real numbers. The set Ais called the domainof the function. The number f(cid:2)x(cid:3)is the value of f at xand is read “f of x.”The rangeof f is the set of all possible values of f(cid:2)x(cid:3)as xvaries throughout the domain. A symbol that represents an arbitrary number in the domainof a function f is called an independent variable. A symbol that represents a number in the rangeof f is called a dependent variable. In Example A,for instance,ris the independent variable and Ais the dependent variable. It’s helpful to think of a function as a machine(see Figure 2). If xis in the domain of x f ƒ the function f,then when xenters the machine,it’s accepted as an input and the machine (input) (output) produces an output f(cid:2)x(cid:3) according to the rule of the function. Thus, we can think of the domain as the set of all possible inputs and the range as the set of all possible outputs. FIGURE 2 The preprogrammed functions in a calculator are good examples of a function as a Machine diagram for a function ƒ machine. For example, the square root key on your calculator computes such a function. You press the key labeled s (or sx)and enter the input x. If x(cid:4)0,then xis not in the domain of this function; that is, xis not an acceptable input,and the calculator will indi- cate an error. If x(cid:3)0,then an approximationto sxwill appear in the display. Thus,the sxkey on your calculator is not quite the same as the exact mathematical function f defined by f(cid:2)x(cid:3)(cid:2)sx. x ƒ Another way to picture a function is by an arrow diagramas in Figure 3. Each arrow connects an element of A to an element of B. The arrow indicates that f(cid:2)x(cid:3) is associated a f(a) with x,f(cid:2)a(cid:3)is associated with a,and so on. The most common method for visualizing a function is its graph. If f is a function with domain A,then its graphis the set of ordered pairs f A B (cid:5)(cid:2)x,f(cid:2)x(cid:3)(cid:3)(cid:6)x(cid:2)A(cid:7) FIGURE 3 (Notice that these are input-output pairs.) In other words, the graph of f consists of all Arrow diagram for ƒ points (cid:2)x,y(cid:3)in the coordinate plane such that y(cid:2)f(cid:2)x(cid:3)and xis in the domain of f. The graph of a function f gives us a useful picture of the behavior or “life history”of a function. Since the y-coordinate of any point (cid:2)x,y(cid:3)on the graph is y(cid:2)f(cid:2)x(cid:3),we can read the value of f(cid:2)x(cid:3) from the graph as being the height of the graph above the point x (see Figure 4). The graph of f also allows us to picture the domain of f on the x-axis and its range on the y-axis as in Figure 5. y {x, ƒ} y range y (cid:2) ƒ(x) ƒ f(2) f(1) 0 1 2 x x 0 x domain FIGURE 4 FIGURE 5 ❙❙❙❙ SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION 13 EXAMPLE 1 The graph of a function f is shown in Figure 6. (a) Find the values of f(cid:2)1(cid:3)and f(cid:2)5(cid:3). (b) What are the domain and range of f? y 1 0 1 x FIGURE 6 SOLUTION (a) We see from Figure 6 that the point (cid:2)1,3(cid:3)lies on the graph of f,so the value of f at 1 is f(cid:2)1(cid:3)(cid:2)3. (In other words,the point on the graph that lies above x(cid:2)1 is 3 units above the x-axis.) When x(cid:2)5,the graph lies about 0.7 unit below the x-axis,so we estimate that f(cid:2)5(cid:3)(cid:4)(cid:5)0.7. |||| The notation for intervals is given in (b) We see that f(cid:2)x(cid:3)is defined when 0(cid:8)x(cid:8)7,so the domain of f is the closed inter- Appendix A. val (cid:8)0,7(cid:9). Notice that f takes on all values from (cid:5)2 to 4,so the range of f is (cid:6) (cid:5)y (cid:5)2(cid:8)y(cid:8)4(cid:7)(cid:2)(cid:8)(cid:5)2,4(cid:9) EXAMPLE 2 Sketch the graph and find the domain and range of each function. (a) f(cid:2)x(cid:3)(cid:2)2x(cid:5)1 (b) t(cid:2)x(cid:3)(cid:2)x2 SOLUTION y (a) The equation of the graph is y(cid:2)2x(cid:5)1,and we recognize this as being the equation of a line with slope 2 and y-intercept (cid:5)1. (Recall the slope-intercept form of the equation of a line:y(cid:2)mx(cid:7)b. See Appendix B.) This enables us to sketch the graph of y=2x-1 f in Figure 7. The expression 2x(cid:5)1is defined for all real numbers,so the domain of f is the set of all real numbers,which we denote by (cid:2). The graph shows that the range is 0 1 x also (cid:2). -1 2 (b) Since t(cid:2)2(cid:3)(cid:2)22(cid:2)4and t(cid:2)(cid:5)1(cid:3)(cid:2)(cid:2)(cid:5)1(cid:3)2(cid:2)1,we could plot the points (cid:2)2,4(cid:3)and (cid:2)(cid:5)1,1(cid:3),together with a few other points on the graph,and join them to produce the graph (Figure 8). The equation of the graph is y(cid:2)x2,which represents a parabola (see FIGURE 7 Appendix C). The domain of tis (cid:2). The range of tconsists of all values of t(cid:2)x(cid:3),that is, all numbers of the form x2. But x2(cid:3)0for all numbers xand any positive number yis a (cid:6) square. So the range of tis (cid:5)y y(cid:3)0(cid:7)(cid:2)(cid:8)0,(cid:6)(cid:3). This can also be seen from Figure 8. y (2, 4) y=≈ (_1, 1) 1 0 1 x FIGURE 8 ❙❙❙❙ 14 CHAPTER 1 FUNCTIONS AND MODELS Representations of Functions There are four possible ways to represent a function: (cid:3)(cid:3) verbally (by a description in words) (cid:3)(cid:3) numerically (by a table of values) (cid:3)(cid:3) visually (by a graph) (cid:3)(cid:3) algebraically (by an explicit formula) If a single function can be represented in all four ways,it is often useful to go from one representation to another to gain additional insight into the function. (In Example 2, for instance, we started with algebraic formulas and then obtained the graphs.) But certain functions are described more naturally by one method than by another. With this in mind, let’s reexamine the four situations that we considered at the beginning of this section. A. The most useful representation of the area of a circle as a function of its radius is probably the algebraic formula A(cid:2)r(cid:3)(cid:2)(cid:2)r2,though it is possible to compile a table of values or to sketch a graph (half a parabola). Because a circle has to have a positive (cid:6) radius,the domain is (cid:5)r r(cid:10)0(cid:7)(cid:2)(cid:2)0,(cid:6)(cid:3),and the range is also (cid:2)0,(cid:6)(cid:3). B. We are given a description of the function in words:P(cid:2)t(cid:3)is the human population of the world at time t. The table of values of world population on page 11 provides a convenient representation of this function. If we plot these values,we get the graph (called a scatter plot) in Figure 9. It too is a useful representation; the graph allows us to absorb all the data at once. What about a formula? Of course,it’s impossible to devise an explicit formula that gives the exact human population P(cid:2)t(cid:3)at any time t. But it is possible to find an expression for a function that approximates P(cid:2)t(cid:3). In fact, using methods explained in Section 1.5,we obtain the approximation P(cid:2)t(cid:3)(cid:4)f(cid:2)t(cid:3)(cid:2)(cid:2)0.008079266(cid:3) (cid:9) (cid:2)1.013731(cid:3)t and Figure 10 shows that it is a reasonably good “fit.”The function f is called a mathematical modelfor population growth. In other words,it is a function with an explicit formula that approximates the behavior of our given function. We will see, however,that the ideas of calculus can be applied to a table of values; an explicit formula is not necessary. P P 6x10' 6x10' 1900 1920 1940 1960 1980 2000 t 1900 1920 1940 1960 1980 2000 t FIGURE 9 FIGURE 10 ❙❙❙❙ SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION 15 The function Pis typical of the functions that arise whenever we attempt to apply calculus to the real world. We start with a verbal description of a function. Then we |||| A function defined by a table of values is may be able to construct a table of values of the function,perhaps from instrument called a tabular function. readings in a scientific experiment. Even though we don’t have complete knowledge of the values of the function,we will see throughout the book that it is still possible to perform the operations of calculus on such a function. w(ounces) C(cid:2)w(cid:3)(dollars) C. Again the function is described in words:C(cid:2)w(cid:3)is the cost of mailing a first-class letter with weight w. The rule that the U.S. Postal Service used as of 2002 is as follows: 0(cid:4)w(cid:8)1 0.37 The cost is 37 cents for up to one ounce,plus 23 cents for each successive ounce up 1(cid:4)w(cid:8)2 0.60 to 11 ounces. The table of values shown in the margin is the most convenient repre- 2(cid:4)w(cid:8)3 0.83 sentation for this function,though it is possible to sketch a graph (see Example 10). 3(cid:4)w(cid:8)4 1.06 D. The graph shown in Figure 1 is the most natural representation of the vertical acceler- 4(cid:4)w(cid:8)5 1.29 ation function a(cid:2)t(cid:3). It’s true that a table of values could be compiled,and it is even (cid:9) (cid:9) (cid:9) (cid:9) possible to devise an approximate formula. But everything a geologist needs to (cid:9) (cid:9) know—amplitudes and patterns—can be seen easily from the graph. (The same is true for the patterns seen in electrocardiograms of heart patients and polygraphs for lie- detection.) Figures 11 and 12 show the graphs of the north-south and east-west accel- erations for the Northridge earthquake; when used in conjunction with Figure 1,they provide a great deal of information about the earthquake. a a {cm/s@} {cm/s@} 400 200 200 100 5 10 15 20 25 30 t 5 10 15 20 25 30 t (seconds) (seconds) _200 _100 _400 _200 Calif. Dept. of Mines and Geology Calif. Dept. of Mines and Geology FIGURE 11 North-south acceleration for the Northridge earthquake FIGURE 12 East-west acceleration for the Northridge earthquake In the next example we sketch the graph of a function that is defined verbally. EXAMPLE 3 When you turn on a hot-water faucet,the temperature T of the water depends on how long the water has been running. Draw a rough graph of T as a function of the T time tthat has elapsed since the faucet was turned on. SOLUTION The initial temperature of the running water is close to room temperature because of the water that has been sitting in the pipes. When the water from the hot- water tank starts coming out,T increases quickly. In the next phase,T is constant 0 t at the temperature of the heated water in the tank. When the tank is drained,T decreases to the temperature of the water supply. This enables us to make the rough sketch of T as FIGURE 13 a function of tin Figure13. ❙❙❙❙ 16 CHAPTER 1 FUNCTIONS AND MODELS A more accurate graph of the function in Example 3 could be obtained by using a ther- mometer to measure the temperature of the water at 10-second intervals. In general, sci- entists collect experimental data and use them to sketch the graphs of functions,as the next example illustrates. EXAMPLE 4 The data shown in the margin come from an experiment on the lactonization t C(cid:2)t(cid:3) of hydroxyvaleric acid at 25(cid:11)C. They give the concentration C(cid:2)t(cid:3)of this acid (in moles 0 0.0800 per liter) after t minutes. Use these data to draw an approximation to the graph of the 2 0.0570 concentration function. Then use this graph to estimate the concentration after 5minutes. 4 0.0408 SOLUTION We plot the five points corresponding to the data from the table in Figure 14. 6 0.0295 The curve-fitting methods of Section 1.2 could be used to choose a model and graph it. 8 0.0210 But the data points in Figure 14 look quite well behaved,so we simply draw a smooth curve through them by hand as in Figure 15. C(t) C(t) 0.08 0.08 0.06 0.06 0.04 0.04 0.02 0.02 0 1 2 3 4 5 6 7 8 t 0 1 2 3 4 5 6 7 8 t FIGURE 14 FIGURE 15 Then we use the graph to estimate that the concentration after 5 minutes is C(cid:2)5(cid:3)(cid:4)0.035mole(cid:10)liter In the following example we start with a verbal description of a function in a physical situation and obtain an explicit algebraic formula. The ability to do this is a useful skill in solving calculus problems that ask for the maximum or minimum values of quantities. EXAMPLE 5 A rectangular storage container with an open top has a volume of 10 m3. The length of its base is twice its width. Material for the base costs $10 per square meter; material for the sides costs $6 per square meter. Express the cost of materials as a function of the width of the base. SOLUTION We draw a diagram as in Figure 16 and introduce notation by lettingwand2w be the width and length of the base,respectively,and hbe the height. The area of the base is (cid:2)2w(cid:3)w(cid:2)2w2,so the cost,in dollars,of the material for the base is 10(cid:2)2w2(cid:3). Two of the sides have area whand the other two have area 2wh,so the h cost of the material for the sides is 6(cid:8)2(cid:2)wh(cid:3)(cid:7)2(cid:2)2wh(cid:3)(cid:9). The total cost is therefore w C(cid:2)10(cid:2)2w2(cid:3)(cid:7)6(cid:8)2(cid:2)wh(cid:3)(cid:7)2(cid:2)2wh(cid:3)(cid:9)(cid:2)20w2(cid:7)36wh 2w To express Cas a function of walone,we need to eliminate hand we do so by using the FIGURE 16 fact that the volume is 10 m3. Thus w(cid:2)2w(cid:3)h(cid:2)10 10 5 which gives h(cid:2) (cid:2) 2w2 w2 ❙❙❙❙ SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION 17 |||| In setting up applied functions as in Substituting this into the expression for C,we have (cid:11) (cid:12) Example5, it may be useful to review the principles of problem solving as discussed on 5 180 page 80, particularly Step 1: Understand the C(cid:2)20w2(cid:7)36w w2 (cid:2)20w2(cid:7) w Problem. Therefore,the equation 180 C(cid:2)w(cid:3)(cid:2)20w2(cid:7) w(cid:10)0 w expresses Cas a function of w. EXAMPLE 6 Find the domain of each function. 1 (a) f(cid:2)x(cid:3)(cid:2)sx(cid:7)2 (b) t(cid:2)x(cid:3)(cid:2) x2(cid:5)x SOLUTION |||| If a function is given by a formula and the (a) Because the square root of a negative number is not defined (as a real number),the domain is not stated explicitly, the convention is domain of f consists of all values of xsuch that x(cid:7)2(cid:3)0.This is equivalent to that the domain is the set of all numbers for x(cid:3)(cid:5)2,so the domain is the interval (cid:8)(cid:5)2,(cid:6)(cid:3). which the formula makes sense and defines a real number. (b) Since 1 1 t(cid:2)x(cid:3)(cid:2) (cid:2) x2(cid:5)x x(cid:2)x(cid:5)1(cid:3) and division by 0is not allowed,we see that t(cid:2)x(cid:3)is not defined when x(cid:2)0or x(cid:2)1. Thus,the domain of tis (cid:6) (cid:5)x x(cid:3)0, x(cid:3)1(cid:7) which could also be written in interval notation as (cid:2)(cid:5)(cid:6), 0(cid:3)(cid:3)(cid:2)0, 1(cid:3)(cid:3)(cid:2)1, (cid:6)(cid:3) The graph of a function is a curve in the xy-plane. But the question arises:Which curves in the xy-plane are graphs of functions? This is answered by the following test. The Vertical Line Test A curve in the xy-plane is the graph of a function of xif and only if no vertical line intersects the curve more than once. The reason for the truth of the Vertical Line Test can be seen in Figure 17. If each vertical line x(cid:2)a intersects a curve only once, at (cid:2)a,b(cid:3), then exactly one functional value is defined by f(cid:2)a(cid:3)(cid:2)b. But if a line x(cid:2)a intersects the curve twice, at (cid:2)a,b(cid:3) and (cid:2)a,c(cid:3), then the curve can’t represent a function because a function can’t assign two different values to a. y y x=a x=a (a, c) (a, b) (a, b) 0 a x 0 a x FIGURE 17 ❙❙❙❙ 18 CHAPTER 1 FUNCTIONS AND MODELS For example,the parabola x(cid:2)y2(cid:5)2shown in Figure 18(a) is not the graph of a function of xbecause,as you can see,there are vertical lines that intersect the parabola twice. The parabola,however,does contain the graphs of twofunctions of x. Notice that the equation x(cid:2)y2(cid:5)2implies y2(cid:2)x(cid:7)2,so y(cid:2)(cid:12)sx(cid:7)2. Thus,the upper and lower halves of the parabola are the graphs of the functions f(cid:2)x(cid:3)(cid:2)sx(cid:7)2 [from Example 6(a)] and t(cid:2)x(cid:3)(cid:2)(cid:5)sx(cid:7)2.[See Figures 18(b) and (c).] We observe that if we reverse the roles of x and y, then the equation x(cid:2)h(cid:2)y(cid:3)(cid:2)y2(cid:5)2 does define x as a function of y (with y as the independent variable and xas the dependent variable) and the parabola now appears as the graph of the function h. y y y _2 (_2, 0) 0 x _2 0 x 0 x FIGURE 18 (a) x=¥-2 (b) y=œ„x+„„2„ (c) y=_œ„x+„„2„ Piecewise Defined Functions The functions in the following four examples are defined by different formulas in different parts of their domains. EXAMPLE 7 A function f is defined by (cid:13) 1(cid:5)x if x(cid:8)1 f(cid:2)x(cid:3)(cid:2) x2 if x(cid:10)1 Evaluate f(cid:2)0(cid:3), f(cid:2)1(cid:3),and f(cid:2)2(cid:3)and sketch the graph. SOLUTION Remember that a function is a rule. For this particular function the rule is the following:First look at the value of the input x. If it happens that x (cid:8)1,then the value of f(cid:2)x(cid:3)is 1(cid:5)x. On the other hand,if x(cid:10)1,then the value of f(cid:2)x(cid:3)is x2. Since 0(cid:8)1, we have f(cid:2)0(cid:3)(cid:2)1(cid:5)0(cid:2)1. Since 1(cid:8)1, we have f(cid:2)1(cid:3)(cid:2)1(cid:5)1(cid:2)0. y Since 2(cid:10)1, we have f(cid:2)2(cid:3)(cid:2)22(cid:2)4. How do we draw the graph of f? We observe that if x(cid:8)1,then f(cid:2)x(cid:3)(cid:2)1(cid:5)x,so the part of the graph of f that lies to the left of the vertical line x(cid:2)1must coincide with 1 the line y(cid:2)1(cid:5)x,which has slope (cid:5)1and y-intercept 1. If x(cid:10)1,then f(cid:2)x(cid:3)(cid:2)x2,so the part of the graph of f that lies to the right of the line x(cid:2)1must coincide with the 1 x graph of y(cid:2)x2,which is a parabola. This enables us to sketch the graph in Figure l9. The solid dot indicates that the point (cid:2)1,0(cid:3)is included on the graph; the open dot indi- FIGURE 19 cates that the point (cid:2)1,1(cid:3)is excluded from the graph. ❙❙❙❙ SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION 19 The next example of a piecewise defined function is the absolute value function. Recall (cid:6) (cid:6) that the absolute valueof a number a,denoted by a ,is the distance from ato 0on the real number line. Distances are always positive or 0,so we have |||| For a more extensive review of absolute (cid:6)a(cid:6)(cid:3)0 for every number a values, see Appendix A. For example, (cid:6) (cid:6) (cid:6) (cid:6) (cid:6) (cid:6) (cid:6) (cid:6) (cid:6) (cid:6) 3 (cid:2)3 (cid:5)3 (cid:2)3 0 (cid:2)0 s2(cid:5)1 (cid:2)s2(cid:5)1 3(cid:5)(cid:2) (cid:2)(cid:2)(cid:5)3 In general,we have (cid:6) (cid:6) a (cid:2)a if a(cid:3)0 (cid:6) (cid:6) a (cid:2)(cid:5)a if a(cid:4)0 (Remember that if ais negative,then (cid:5)ais positive.) (cid:6) (cid:6) EXAMPLE 8 Sketch the graph of the absolute value function f(cid:2)x(cid:3)(cid:2) x . y SOLUTION From the preceding discussion we know that y=|x| (cid:13) (cid:6) (cid:6) x if x(cid:3)0 x (cid:2) (cid:5)x if x(cid:4)0 0 x Using the same method as in Example 7,we see that the graph of f coincides with the line y(cid:2)xto the right of the y-axis and coincides with the liney(cid:2)(cid:5)xto the left of the FIGURE 20 y-axis (see Figure 20). EXAMPLE 9 Find a formula for the function f graphed in Figure 21. y 1 0 1 x FIGURE 21 SOLUTION The line through (cid:2)0,0(cid:3)and (cid:2)1,1(cid:3)has slope m(cid:2)1and y-intercept b(cid:2)0,so its equation is y(cid:2)x. Thus,for the part of the graph of f that joins (cid:2)0,0(cid:3)to (cid:2)1,1(cid:3),we have f(cid:2)x(cid:3)(cid:2)x if 0(cid:8) x(cid:8)1 |||| Point-slope form of the equation of a line: The line through (cid:2)1,1(cid:3)and (cid:2)2,0(cid:3)has slope m(cid:2)(cid:5)1,so its point-slope form is y(cid:5)y (cid:2)m(cid:2)x(cid:5)x (cid:3) 1 1 y(cid:5)0(cid:2)(cid:2)(cid:5)1(cid:3)(cid:2)x(cid:5)2(cid:3) or y(cid:2)2(cid:5)x See Appendix B. So we have f(cid:2)x(cid:3)(cid:2)2(cid:5)x if 1(cid:4)x(cid:8)2