SECTION 3-4 Rational Functions

Transcription

1 20 3 Polnomial and Rational Functions 0. Shipping. A shipping bo is reinforced with steel bands in all three directions (see the figure). A total of 20. feet of steel tape is to be used, with 6 inches of waste because of a 2-inch overlap in each direction. If the bo has a square base and a volume of 2 cubic feet, find its dimensions. SECTION 3-4 Rational Functions Rational Functions Vertical and Horizontal Asmptotes Graphing Rational Functions Rational Functions Just as rational numbers are defined in terms of quotients of integers, rational functions are defined in terms of quotients of polnomials. The following equations define rational functions: f() 2 6 p() g() q() 3 h() 3 r() 0 In general, a function f is a rational function if f() n() d() d() 0 where n() and d() are polnomials. The domain of f is the set of all real numbers such that d() 0. If a and d(a) 0, then f is not defined at a and there can be no point on the graph of f with abscissa a. Remember, division b 0 is never allowed. It can be shown that: If f() n()/d() and d(a) 0, then f is discontinuous at a and the graph of f has a hole or break at a. If a is in the domain of f() and n(a) 0, then the graph of f crosses the ais at a. Thus: If f() n()/d(), n(a) 0, and d(a) 0, then a is an intercept for the graph of f. What happens if both n(a) 0 and d(a) 0? In this case, we know that a is a factor of both n() and d(), and thus, f() is not in lowest terms (see Section A-4). Unless specificall stated to the contrar, we assume that all the rational functions we consider are reduced to lowest terms.

2 3-4 Rational Functions 2 EXAMPLE Finding the Domain and Intercepts for a Rational Function Find the domain and intercepts for f() Solution f() n() d() ( 2)( ) 2 9 ( 3)( 3) Since d(3) 0 and d(3) 0, the domain of f is 3 or (, 3) (3, 3) (3, ) Since n(2) 0 and n() 0, the graph of f crosses the ais at 2 and. Matched Problem Find the domain and intercepts for: f() Vertical and Horizontal Asmptotes Even though a rational function f ma be discontinuous at a (no graph for a), it is still useful to know what happens to the graph of f when is close to a. For eample, consider the ver simple rational function f defined b f() It is clear that the function f is discontinuous at 0. But what happens to f() when approaches 0 from either side of 0? A numerical approach will give us an idea of what happens to f() when gets close to 0. From Table, we see that as approaches 0 from the right, / gets larger and larger that is, / increases without bound. We write this smbolicall* as as 0 TABLE Behavior of / as approaches 0 from the right ( 0 + ) / 0,000,000 0,000,000, / increases without bound (/ ) *Remember, the smbol does not represent a real number. In this contet, is used to indicate that the values of / increase without bound. That is, / eceeds an given number N no matter how large N is chosen.

3 22 3 Polnomial and Rational Functions If approaches 0 from the left, then and / are both negative and the values of / decrease without bound (see Table 2). This is denoted as as 0 TABLE 2 Behavior of / as approaches 0 from the left ( 0 ) / 0,000,000 0,000,000, / decreases without bound (/ ) The graph of f() / for, 0, is shown in Figure. The behavior of f as approaches 0 from the right is illustrated on the graph b drawing a curve that becomes almost vertical and placing an arrow on the curve to indicate that the values of / continue to increase without bound as approaches 0 from the right. The behavior as approaches 0 from the left is illustrated in a similar manner. FIGURE f() near 0. EXPLORE-DISCUSS Construct tables similar to Tables and 2 for g() / 2, and discuss the behavior of the graph of g() near 0. The preceding discussion suggests that vertical asmptotes are associated with the zeros of the denominator of a rational function. Using the same kind of reasoning, we state the following general method of locating vertical asmptotes for rational functions. Theorem Vertical Asmptotes and Rational Functions Let f be a rational function defined b f() n() d()

4 3-4 Rational Functions 23 where n() and d() are polnomials. If a is a real number such that d(a) 0 and n(a) 0, then the line a is a vertical asmptote of the graph of f(). Now we look at the behavior of f() / as gets ver large that is, as and as. Consider Tables 3 and 4. As increases without bound, / is positive and approaches 0 from above. As decreases without bound, / is negative and approaches 0 from below. For our purposes, it is not necessar to distinguish between / approaching 0 from above and from below. Thus, we will describe this behavior b writing 0 as and as TABLE 3 Behavior of / as 0,000,000 0,000,000, increases without bound ( ) / / approaches 0 (/ 0) TABLE 4 Behavior of / as 0,000,000 0,000,000, decreases without bound ( ) / / approaches 0 (/ 0) The completed graph of f() / is shown in Figure 2. Notice that the behavior as and as is illustrated b drawing a curve that is almost horizontal and adding arrows at the ends. The curve in Figure 2 is an eample of a plane curve called a hperbola, and the coordinate aes are called asmptotes for this curve. The ais is a vertical asmptote for /, and the ais is a horizontal asmptote for /. FIGURE 2 f(), 0. Vertical f() asmptote Horizontal asmptote f()

5 24 3 Polnomial and Rational Functions DEFINITION Horizontal and Vertical Asmptotes The line a is a vertical asmptote for the graph of f() if f() either increases or decreases without bound as approaches a from the right or from the left. Smbolicall, f() or f() as a or a The line b is a horizontal asmptote for the graph of f() if f() approaches b as increases without bound or as decreases without bound. Smbolicall, f() b as or EXPLORE-DISCUSS 2 Construct tables similar to Tables 3 and 4 for each of the following functions and discuss the behavior of each as and as : (A) f() (B) g() (C) 2 2 h() In Section 3-, we saw that the behavior of a polnomial P() a n n... a a 0 as is determined b its leading term, a n n. In a similar manner, the behavior of a rational function is determined b the ratio of the leading terms of its numerator and denominator; that is, the graphs of f() a m m a a 0 b n n b b 0 and g() a m m b n n ehibit the same behavior as and as (see Fig. 3). FIGURE 3 Graphs of rational functions as. g g f f g f 2 7 f () g() f () g() f () 2 3 g() (a) (b) (c)

6 3-4 Rational Functions 2 In Figure 3(a), the degree of the numerator is less than the degree of the denominator and the ais is a horizontal asmptote. In Figure 3(b), the degree of the numerator equals the degree of the denominator and the line 2 is a horizontal asmptote. In Figure 3(c), the degree of the numerator is greater than the degree of the denominator and there are no horizontal asmptotes. These ideas are generalized in Theorem 2 to provide a simple wa to locate the horizontal asmptotes of an rational function. Theorem 2 Horizontal Asmptotes and Rational Functions Let f be a rational function defined b the quotient of two polnomials as follows: f() a m m a a 0 b n n b b 0. For m n, the line 0 (the ais) is a horizontal asmptote. 2. For m n, the line a m /b n is a horizontal asmptote. 3. For m n, the graph will increase or decrease without bound, depending on m, n, a m, and b n, and there are no horizontal asmptotes. EXAMPLE 2 Finding Vertical and Horizontal Asmptotes for a Rational Function Find all vertical and horizontal asmptotes for f() n() d() Solution Since d() 2 9 ( 3)( 3), the graph of f() has vertical asmptotes at 3 and 3 (Theorem ). Since n() and d() have the same degree, the line a 2 b a 2 2, b 2 is a horizontal asmptote (Theorem 2, part 2). Matched Problem 2 Find all vertical and horizontal asmptotes for f()

7 26 3 Polnomial and Rational Functions Graphing Rational Functions We now use the techniques for locating asmptotes, along with other graphing aids discussed in the tet, to graph several rational functions. First, we outline a sstematic approach to the problem of graphing rational functions: Graphing a Rational Function: f() n()/d() Step. Intercepts. Find the real solutions of the equation n() 0 and use these solutions to plot an intercepts of the graph of f. Evaluate f(0), if it eists, and plot the intercept. Step 2. Vertical Asmptotes. Find the real solutions of the equation d() 0 and use these solutions to determine the domain of f, the points of discontinuit, and the vertical asmptotes. Sketch an vertical asmptotes as dashed lines. Step 3. Sign Chart. Construct a sign chart for f and use it to determine the behavior of the graph near each vertical asmptote. Step 4. Horizontal Asmptotes. Determine whether there is a horizontal asmptote and if so, sketch it as a dashed line. Step. Complete the Sketch. Complete the sketch of the graph b plotting additional points and joining these points with a smooth continuous curve over each interval in the domain of f. Do not cross an points of discontinuit. EXAMPLE 3 Graphing a Rational Function Graph: f() 2 3 Solution f() 2 n() 3 d() Step. Intercepts. Find real zeros of n() 2 and find f(0): f(0) 0 intercept intercept The graph crosses the coordinate aes onl at the origin. Plot this intercept, as shown in Figure 4.

8 3-4 Rational Functions 27 FIGURE 4 Horizontal asmptote Vertical asmptote and intercepts Intercepts and asmptotes Step 2. Vertical Asmptotes. Find real zeros of d() 3: The domain of f is (, 3) (3, ), f is discontinuous at 3, and the graph has a vertical asmptote at 3. Sketch this asmptote, as shown in Figure 4. Test number Value of f Sign of f Step 3. Sign Chart. Construct a sign chart for f() (review Section 2-8), as shown in the margin. Since 3 is a vertical asmptote and f() 0 for 0 3, f() as 3 Since 3 is a vertical asmptote and f() 0 for 3, 0 3 FIGURE f() as 3 Notice how much information is contained in the sign chart for f. The solid dot determines the intercept; the open dot determines the domain, the point of discontinuit, and the vertical asmptote; and the signs of f() determine the behavior of the graph at the vertical asmptote. Step 4. Horizontal Asmptote. Since n() and d() have the same degree, the line 2 is a horizontal asmptote. Sketch this asmptote, as shown in Figure 4. Step. Complete the Sketch. B plotting a few additional points, we obtain the graph in Figure. Notice that the graph is a smooth continuous curve over the interval (, 3) and over the interval (3, ). As epected, there is a break in the graph at 3. f() 2 3

9 28 3 Polnomial and Rational Functions As ou gain eperience in graphing, man of the steps in Eample 3 can be done mentall (or on scratch paper) and the process can be speeded up considerabl. Matched Problem 3 3 Proceed as in Eample 3 and graph: f() 2 Remark: Refer to Eample 3. When we graph f() 2/( 3) on a graphing utilit [Fig. 6(a)], it appears that the graphing utilit has also drawn the vertical asmptote, but this is not the case. Most graphing utilities, when set in connected mode, calculate points on a graph and connect these points with line segments. The last point plotted to the left of the asmptote and the first plotted to the right of the asmptote will usuall have ver large coordinates. If these coordinates have opposite sign, then the graphing utilit ma connect the two points with a nearl vertical line segment, which gives the appearance of an asmptote. If ou wish, ou can set the calculator in dot mode to plot the points without the connecting line segments (Fig. 6(b)]. FIGURE 6 Graphing utilit graphs 2 of f(). 3 (a) Connected mode (b) Dot mode In the remaining eamples we will just list the results of each step in the graphing strateg and omit the computational details. EXAMPLE 4 Graphing a Rational Function Graph: f() Test number Value of f Sign of f Solution f() ( 3) 2 ( 2)( ) intercept: 3 9 intercept: f(0) 2 4. Domain: (, 2) (2, ) (, ) Points of discontinuit: 2 and Vertical asmptotes: 2 and

10 3-4 Rational Functions 29 f() as 2 f() as 2 f() as f() as Horizontal asmptote: Sketch in the intercepts and asmptotes (Fig. 7), then sketch the graph of f (Fig. 8). Intercepts and asmptotes f() FIGURE 7 FIGURE 8 Matched Problem 4 Graph: f() CAUTION The graph of a function cannot cross a vertical asmptote, but the same statement is not true for horizontal asmptotes. The graph in Eample 4 clearl shows that the graph of a function can cross a horizontal asmptote. The definition of a horizontal asmptote requires f() to approach b as increases or decreases without bound, but it does not preclude the possibilit that f() b for one or more values of. In fact, using the cosine function from trigonometr, it is possible to construct a function whose graph crosses a horizontal asmptote an infinite number of times (see Fig. 9). FIGURE 9 Multiple intersections of a graph and a horizontal asmptote. 2 f() cos f() as is a horizontal asmptote

11 260 3 Polnomial and Rational Functions EXAMPLE Graphing a Rational Function Graph: f() Solution Test number Value of f Sign of f 2 4 f() intercepts: and 4 intercept: f(0) 2 Domain: (, 2) (2, ) Points of discontinuit: 2 Vertical asmptote: 2 f() as 2 f() as 2 No horizontal asmptote ( )( 4) 2 Even though the graph of f does not have a horizontal asmptote, we can still gain some useful information about the behavior of the graph as and as if we first perform a long division: Quotient Remainder Thus, f() As or, 6/( 2) 0 and the graph of f approaches the line. This line is called an oblique asmptote for the graph of f. The asmptotes and intercepts are sketched in Figure, and the graph of f is sketched in Figure.

12 3-4 Rational Functions 26 Oblique asmptote Intercepts and asmptotes f() FIGURE FIGURE Generalizing the results of Eample, we have Theorem 3. Theorem 3 Oblique Asmptotes and Rational Functions If f() n()/d(), where n() and d() are polnomials and the degree of n() is more than the degree of d(), then f() can be epressed in the form f() m b r() d() where the degree of r() is less than the degree of d(). The line m b is an oblique asmptote for the graph of f. That is, [ f() (m b)] 0 as or Matched Problem Graph, including an oblique asmptotes: f() 2

13 262 3 Polnomial and Rational Functions Answers to Matched Problems. Domain: (, 3) (3, ) (, ); intercepts: 2, 2 2. Vertical asmptotes: 3, ; horizontal asmptote: f() 3 2 f() f() 2 EXERCISE 3-4 A In Problems 4, match each graph with one of the following functions: f() g() h() k() 2. 3.

14 3-4 Rational Functions g() f() f() f() g() 2 6 p() g() f() 2 In Problems 2, find the domain and intercepts. Do not graph f() h() s() 2 2 r() F() 2. G() In Problems 3 20, find all vertical and horizontal asmptotes. Do not graph f() 4. h() 4. s() p() 8. q() t() 20. g() B In Problems 2 40, use the graphing strateg outlined in the tet to sketch the graph of each function. Check Problems 2 40 on a graphing utilit. 2. f() 22. g() f() 24. f() h() 26. p() f() 28. f() g() k() r() f() 38. f() (3 ) 2 (2 3) f() f() If f() n()/d(), where n() and d() are quadratic functions, what is the maimum number of intercepts f() can have? What is the minimum number? Illustrate both cases with eamples. 42. If f() n()/d(), where n() and d() are quadratic functions, what is the maimum number of vertical asmptotes f() can have? What is the minimum number? Illustrate both cases with eamples. In Problems 43 48, find all vertical, horizontal, and oblique asmptotes. Do not graph f() 44. g() 2 4. p() 46. q() s() 32 9 r() 22 3 In Problems 49 2, use a graphing utilit to investigate the behavior of each function as and as, and to find an horizontal asmptotes f() 0. f() 2 2. f() C 3 f() 32 In Problems 3 8, use the graphing strateg outlined in the tet to sketch the graph of each function. Include an oblique asmptotes. Check Problems 3 8 with a graphing utilit g() 2 f() 2

15 264 3 Polnomial and Rational Functions. k() F() If f() n()/d(), where the degree of n() is greater than the degree of d(), then long division can be used to write f() p() q()/d(), where p() and q() are polnomials with the degree of q() less than the degree of d(). In Problems 9 62, perform the long division and discuss the relationship between the graphs of f() and p() as and as f() 60. f() f() 62. f() 2 3 In calculus, it is often necessar to consider rational functions that are not in lowest terms, such as the functions given in Problems For each function, state the domain, reduce the function to lowest terms, and sketch its graph. Remember to eclude from the graph an points with values that are not in the domain g() 2 f() r() APPLICATIONS 67. Emploee Training. A compan producing electronic components used in television sets has established that on the average, a new emploee can assemble N(t) components per da after t das of on-the-job training, as given b N(t) 0t t 4 t 0 Sketch the graph of N, including an vertical or horizontal asmptotes. What does N approach as t? 68. Phsiolog. In a stud on the speed of muscle contraction in frogs under various loads, researchers W. O. Fems and J. Marsh found that the speed of contraction decreases with increasing loads. More precisel, the found that the relationship between speed of contraction S (in centimeters per second) and load w (in grams) is given approimatel b w S(w) w h() G() 4 3 s() 2 w Sketch the graph of S, including an vertical or horizontal asmptotes. What does S approach as w? 69. Retention. An eperiment on retention is conducted in a pscholog class. Each student in the class is given da to memorize the same list of 40 special characters. The lists are turned in at the end of the da, and for each succeeding da for 20 das each student is asked to turn in a list of as man of the smbols as can be recalled. Averages are taken, and it is found that a good approimation of the average number of smbols, N(t), retained after t das is given b t Sketch the graph of N, including an vertical or horizontal asmptotes. What does N approach as t? 70. Learning Theor. In 97, L. L. Thurstone, a pioneer in quantitative learning theor, proposed the function to describe the number of successful acts per unit time that a person could accomplish after practice sessions. Suppose that for a particular person enrolling in a tping class, 0 where f() is the number of words per minute the person is able to tpe after weeks of lessons. Sketch the graph of f, including an vertical or horizontal asmptotes. What does f approach as? Using calculus techniques, it can be shown that the minimum value of a function of the form g() a b c t 30 N(t) t a( c) f() ( c) b 0( ) f() a 0, c 0, 0 is min g() g( c/a). Use this fact in Problems Replacement Time. A desktop office copier has an initial price of $2,00. A maintenance/service contract costs $200 for the first ear and increases $0 per ear thereafter. It can be shown that the total cost of the copier after n ears is given b C(n) 2,00 7n 2n 2 The average cost per ear for n ears is C(n) C(n)/n. (A) Find the rational function C. (B) When is the average cost per ear minimum? (This is frequentl referred to as the replacement time for this piece of equipment.) (C) Sketch the graph of C, including an asmptotes.

16 3- Partial Fractions Average Cost. The total cost of producing units of a certain product is given b C() 2 2 2,000 The average cost per unit for producing units is C() C()/. (A) Find the rational function C. (B) At what production level will the average cost per unit be minimal? (C) Sketch the graph of C, including an asmptotes. 73. Construction. A rectangular dog pen is to be made to enclose an area of 22 square feet. (A) If represents the width of the pen, epress the total length L() of the fencing material required for the pen in terms of. (B) Considering the phsical limitations, what is the domain of the function L? (C) Find the dimensions of the pen that will require the least amount of fencing material. (D) Graph the function L, including an asmptotes. 74. Construction. Rework Problem 73 with the added assumption that the pen is to be divided into two sections, as shown in the figure. SECTION 3- Partial Fractions Basic Theorems Partial Fraction Decomposition You have now had considerable eperience combining two or more rational epressions into a single rational epression. For eample, problems such as 2 3 2( 4) 3( ) 7 4 ( )( 4) ( )( 4) should seem routine. Frequentl in more advanced courses, particularl in calculus, it is advantageous to be able to reverse this process that is, to be able to epress a rational epression as the sum of two or more simpler rational epressions called partial fractions. As is often the case with reverse processes, the process of decomposing a rational epression into partial fractions is more difficult than combining rational epressions. Basic to the process is the factoring of polnomials, so the topics discussed earlier in this chapter can be put to effective use. We confine our attention to rational epressions of the form P()/D(), where P() and D() are polnomials with real coefficients and no common factors. In addition, we assume that the degree of P() is less than the degree of D(). If the degree of P() is greater than or equal to that of D(), we have onl to divide P() b D() to obtain where the degree of R() is less than that of D(). For eample, P() R() Q() D() D()