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# Calculus I MT 100

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This 73 page Class Notes was uploaded by Mr. Halie Wilkinson on Saturday October 3, 2015. The Class Notes belongs to MT 100 at Boston College taught by Staff in Fall. Since its upload, it has received 48 views. For similar materials see /class/218071/mt-100-boston-college in Mathematics (M) at Boston College.

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Date Created: 10/03/15

PRECALCULUS REVIEW 11 Real Numbers Functions and Graphs Preliminary Questions 1 Give an example of numbers a and b such that a lt b and iai gt ibi SOLUTION Takea 73 andb 1 Thena lt b but iai 3 gt1 ibi 2 Which numbers satisfy iai a Which satisfy iai 7a What about i 7 ai a SOLUTION The numbers a 3 O satisfy iai a and i 7 ai a The numbers a 5 O satisfy iai 7a 3 Give an example ofnumbersa andb such that id bi lt iai ibi SOLUTION Take a 73 and b 1 Then iabii731ii72i2 but iaiibii73ii1i314 Thus in bi lt iai ibi 4 What are the coordinates of the point lying at the intersection of the lines 1 9 and y 74 SOLUTION The point 9 74 lies at the intersection of the lines 1 9 and y 74 5 In which quadrant do the following points lie 8 14 393 32 C 4 3 d 4r1 SOLUTION 2 Because both the x and ycoordinates of the point 1 4 are positive the point 1 4 lies in the rst quadrant b Because the xcoordinate of the point 73 2 is negative but the ycoordinate is positive the point 73 2 lies in the second quadrant c Because the xcoordinate of the point 4 73 is positive but the ycoordinate is negative the point 4 73 lies in the fourth quadrant d Because both the x and ycoordinates of the point 74 71 are negative the point 74 71 lies in the third quadrant 6 What is the radius of the circle with equation x 7 92 y 7 92 9 SOLUTION The circle with equation x 7 92 y 7 92 9 has radius 3 7 The equation fx 5 has a solution ifchoose one a 5 belongs to the domain of f b 5 belongs to the range of f SOLUTION The correct response is b the equation fx 5 has a solution if 5 belongs to the range of f 8 What kind of symmetry does the graph have if f7x 7fx 7 SOLUTION lff7x 7fx then the graph off is symmetric with respect to the origin Exercises 1 Use a calculator to nd a rational number r such that ir 7 nzi lt 104 SOLUTION r must satisfy 72 7104 lt r lt n2 104 or 9869504 lt r lt 9869705 r 98696 301 would be one such number 2 Leta 73 and b 2 Which of the following inequalities are true a a lt b b iai lt ibi c ab gt 0 1 1 d M lt 3b 9 74a lt 74b t Z lt E SOLUTION a True b False iai 3 gt 2 ibi c False 73x2 76 lt o d True 2 CHAPTER 1 I PRECALCULUS REVIEW 9 False 493 12 gt 78 74x2 0 True In Exercises 3 express the interval in terms of an inequality involving absolute value 3 722 SOLUTION 1x152 4 744 SOLUTION 1x1 lt 4 5 04 SOLUTION The midpoint of the interval is c 0 42 2 and the radius is r 4 7 02 2 therefore 04 can be expressedas 1x 7 21 lt 2 6 74 0 SOLUTION The midpoint of the interval is c 74 02 72 and the radius is r O 7 742 2 therefore the interval 74 O can be expressed as 1x 21 5 2 7 15 SOLUTION The midpoint of the interval is c 1 52 3 and the radius is r 5 7 12 2 therefore the interval 1 5 can be expressed as x 7 31 5 2 3 728 SOLUTION The midpoint ofthe interval is c 8 7 22 3 and the radius is r 8 7 722 5 therefore the interval 72 8 can be expressed as x 7 31 lt 5 In Exercises 9712 write the inequality in theform a lt x lt bfor some numbers a b 9 1x1 lt 8 SOLUTION 78 lt x lt 8 101x712lt8 SOLUTION 78 ltx712lt8so4ltxlt20 1112x1llt5 SOLUTION 75 lt2x1lt5so76lt2xlt4and73ltxlt2 1213x74llt2 SOLUTION 72 lt3x74lt2s02lt3x lt6and ltx lt2 In Exercises 13718 express the set of numbers x satisfying the given condition as an interval 13 1x1 lt4 SOLUTION 74 4 14 1x1 59 SOLUTION 79 9 15 1x 7 41 lt 2 SOLUTION The expression 1x 7 41 lt 2is equivalent to 72 lt x 7 4 lt 2 Therefore 2 lt x lt 6 which represents the interval 2 6 16 1x 71 lt 2 SOLUTION The expression 1x 71 lt 2 is equivalent to 72 lt x 7 lt 2 Therefore 79 lt x lt 75 which represents the interval 79 75 17 4x 711 5 8 SOLUTION The expression 14x 7 11 5 8 is equivalent to 78 5 4x 7 1 5 8 or 77 5 4x 5 9 Therefore 7 5 x 5 which represents the interval 73 18 3x 51 lt 1 SOLUTION The expression 3x 51 lt 1 is equivalent to 71 lt 3x 5 lt 1 or 76 lt 3x lt 74 Therefore 72 lt x lt 7 which represents the interval 72 7 g In Exercises 19722 describe the set as a union of nite or in nite intervals 19 lex74lgt2 S E C T10 N 11 I Real Numbers Functions and Graphs 3 SOLUTION x74gt20rx74lt72gtxgt6orxlt2gt7oo2U6oo 20 x12x41gt31 SOLUTION 2X4gt3or2x4lt73gt2xgt7lor2xlt77gt7oo7ZU7oo 21 xz1x2711gt21 SOLUTION x271gt20rx271lt725x2gt3orx2lt71thiswillneverhappengtxgtJ30rxlt7 3 700 7x5 U 3 00 221xz1x22x1gt2 SOLUTION x22xgt20rx22xlt72gtx22x72gt00rx22x2lt0Forthe rstcasethezeroesare x71i 3gt7oo717J3U71 3oo For the second case note there are no real zeros Because the parabola opens upward and its vertex is located above the xaxis there are no values of for which 12 2x 2 lt 0 Hence the solution set is 700 71 7 3 U 71 V3 00 23 Match the inequalities a7f with the corresponding statements i7vi aagt3 b1a751lt c a7 lt5 d1a1gt5 e1a741lt3 f1ltalt5 i a lies to the right of3 ii a lies between 1 and 7 iii The distance from a to 5 is less than iv The distance from a to 3 is at most 2 v a is less than 5 units from vi a lies either to the left of 75 or to the right of 5 SOLUTION a On the number line numbers greater than 3 appear to the right hence a gt 3 is equivalent to the numbers to the right of 3 i b 1a 7 51 measures the distance from a to 5 hence 1a 7 51 lt is satis ed by those numbers less than of a unit from 5 iii c 1a 7 1 measures the distance froma to hence 1a 7 1 lt 5 is satis ed by those numbers less than 5 units from v d The inequality 1a1 gt Sis equivalent to a gt 5 or a lt 75 that is either a lies to the right of 5 or to the left of 75 vi 9 The interval described by the inequality 1a 7 41 lt 3 has a center at 4 and a radius of 3 that is the interval consists of those numbers between 1 and 7 ii f The interval described by the inequality 1 lt x lt 5 has a center at 3 and a radius of 2 that is the interval consists of those numbers less than 2 units from 3 iv 1 1 24 Describe the set x lt O as an interval 1 SOLUTION Case 1 x lt Oandx 1gt O This implies that lt O and gt 71 5 71 lt x lt 0 Case 2 gt O and lt 71 for which there is no such 1 Thus solution set is therefore 710 25 Show that ifa gt b then b 1 gt a 1provided thata andb have the same sign What happens ifa gt O and b lt O SOLUTION Case 1a lfa andb are both positive then a gt b 5 1 gt 5 gt Case 1b lfa and b are both negative then a gt b 5 1 lt since a is negative 5 gt again since b is negative Case 2 lfa gt O andb lt 0 then gt O and lt 0 so lt See Exercise 2ffor an example ofthis 26 Which satisfy 1 7 31 lt 2 and 1x 7 51lt17 SOLUTION 1x731lt2gt72 ltx73 lt2gt1ltx lt5Also1x 751 lt 1 54ltx lt6 Sincewewantanx that satis es both of these we need the intersection of the two solution sets that is 4 lt x lt 5 27 Show that if 1a 7 51 lt and 1b 7 81 lt then 1a b7131 lt 1 Hint Use the triangle inequality 4 CHAPTER 1 I PRECALCULUS REVIEW SOLUTION lab7l3l la75b78l 5 la 7 5l lb 7 8l by the triangle inequality 1 l lt7 71 2 2 28 Suppose that lal 5 2 and lbl 5 3 a What is the largest possible value of la bl l b What is the largest possible value of la bl ifa and b have opposite signs SOLUTION a With lal 5 2 and lbl 5 3 it follows from the triangle inequality that la bl 5 lal lbl 2 3 5 b lfa and b have opposite signs then la bl lt 3 with the maximum occurring for b i3 and a arbitrarily close to O 29 Suppose thath 7 4l 5 l a What is the maximum possible value of lx 4l b Show that lxz e1ol 5 9 SOLUTION a lx 74l 5 l guarantees3 5x 55Thus75x4 5 9so lx4l 59 b lx2716llx74llx4l 5109 30 Prove that lxl 7 lyl 5 lx 7 yl Hint Apply the triangle inequality to y and 7 y SOLUTION First note lxl lx7yyl lx7yllyl by the triangle inequality Subtracting l yl fromboth sides of this inequality yields le M S lxi ill 31 Express r1 of as a fraction Hint lOOr1 7 r1 is an integer Then express r2 02666 as a fraction SOLUTION Let r1 5 We observe that 100r1 27 Therefore 100r1 7 r1 27E 7 27 and 27 3 r1 99 11 Now let r2 2 Then 10r2 2 and 100r2 2o Therefore 100r2 e 10r2 26 7 2 24 and r 7 24 7 4 2 7 90 7 15 32 Represent 17 and427 as repeating decimals SOLUTION l 142857 1 1 8 7 27 33 The text states the following If the decimal expansions of two real numbers a and b agree to k places then the distance la 7 bl 5 10 Show that the converse is not true that is for any k we can nd real numbers a and b whose decimal expansions do not agree at all but la 7 bl 5 lO k SOLUTION Let a l and b 5 see the discussion before Example 1 The decimal expansions ofa and b do not agreebut l1 7 9l lt 10k for all k 34 Plot each pair of points and compute the distance between them a 14 and 32 03 21and 2 4 c 00 and723 d 73 3a d23 SOLUTION a The points 14 and 32 are plotted in the gure below The distance between the points is d37l22742 22722 2 2 S E C T t O N 11 I Real Numbers Functions and Graphs 5 b The points 2 1 and 2 4 are plotted in the gure below The distance between the points is d 27224712 53 c The points 0 0 and 72 3 are plotted in the gure below The distance between the points is d 727023702 4T E d The points 73 73 and 72 3 are plotted in the gure below The distance between the points is d 73 7 722 73 732 W m 35 Determine the equation of the circle with center 2 4 and radius 3 SOLUTION The equation of the indicated circle is x 7 22 y 7 42 32 9 36 Determine the equation of the circle with center 2 4 passing through 1 71 SOLUTION First determine the radius as the distance from the center to the indicated point on the circle r 2 7 12 4 7 712 JR Thus the equation of the circle is x 7 22 y 7 42 26 37 Find all points with integer coordinates located at a distance 5 from the origin Then nd all points with integer coordinates located at a distance 5 from 2 3 SOLUTION To be located a distance 5 from the origin the points must lie on the circle 12 y2 25 This leads to 12 points with integer coordinates 5 0 05 075 34 734 3 74 7374 4 3 4 73 7473 6 CHAPTER 1 I PRECALCULUS REVIEW To be located a distance 5 from the point 2 3 the points must lie on the circle x 7 22 y 7 32 25 which implies that we must shift the points listed above two units to the right and three units up This gives the 12 points 77 3 3 3 2 8 27 2 5 7 47 54 L l 6 6 2 6 6 0 2 0 38 Determine the domain and range of the function fttrstu3 ltABCDiE de nedby fr A fs Bft B fu E SOLUTION The domain is the set D r s t u the range is the set R A B E 39 Give an example of a function whose domain D has three elements and range R has two elements Does a function exist whose domain D has two elements and range has three elements SOLUTION De ne fby f a b c gt 12 where fa l fb l fc 2 There is no function whose domain has two elements and range has three elements If that happened one of the domain elements would get assigned to more than one element of the range which would contradict the de nition of a function In Exercises 4048 nd the domain and range of the function 40 fx 7x SOLUTION D all reals R all reals 41 gt t4 SOLUTION D all reals R y y 3 O 42 fx x3 SOLUTION D all reals R all reals 43 gt m SOLUTION Dt t 2Ryy30 14 fX lxl SOLUTION D all reals R y y 3 O l 45 hs s SOLUTION Dss7 0Ryy0 l 46 fx 2 x SOLUTION Dxx0 RyygtO 47 gt 2 t2 SOLUTION D all reals R y y 3 J2 l 48 gt cos 7 SOLUTION Dtt7 0Ryzil yfl In Exercises 49752 nd the interval on which the function is increasing 49 fx ix ll SOLUTION A graph of the function y ix ll is shown below From the graph we see that the function is increasing on the interval 71 00 S E C T t O N 11 I Real Numbers Functions and Graphs 7 50 fx x3 SOLUTION A graph of the function y x3 is shown below From the graph we see that the function is increasing for all real numbers 51 fx x4 SOLUTION A graph of the function y x4 is shown below From the graph we see that the function is increasing on the interval 0 00 1 52 for m SOLUTION A graph of the function y 2 1 is shown below From the graph we see that the function is increasing 1 on the interval 700 0 In Exercises 53758 nd the zeros of the function and sketch its graph by plotting points Use symmetry and in creasedecrease information where appropriate 53 fx x2 7 4 54 fx 2x2 7 4 Decreasing x lt O X ymmetry f7 fx even function So yaxis symmetry 8 CHAPTER 1 I PRECALCULUS REVIEW 55 fx x3 7 4x SOLUTION Zeros 0 i2 Symmetry f7x 7fx odd function So origin symmetry 56 fx x3 SOLUTION Zeros 0 Increasing for all 1 Symmetry f7x 7fx odd function So origin symmetry s7 fx 2 7 x3 SOLUTION This is an xaxis re ection of3 translated up 2 units There is one zero at x 1 53 fx m 1 1 translated to the right 1 unit The function has no zeros SOLUTION This is the graph of 2 x 59 Which of the curves in Figure 26 is the graph of a function y y MX QX A B y y C D FIGURE 26 SOLUTION B is the graph of a function A C and D all fail the vertical line test 60 State Whether the function is increasing decreasing or neither S E C T l O N 11 I Real Numbers Functions and Graphs 9 a Surface area of a sphere as a function of its radius b Temperature at apoint on the equator as a function of time c Price of an airline ticket as a function of the price of oil d Pressure of the gas in a piston as a function of volume SOLUTION 2 Increasing b Neither c Increasing d Decreasing In Exercises 61456 let fx be the metion whose graph is shown in Figure 27 FIGURE 27 61 What are the domain and range of fx 7 SOLUTION D 04 R 04 62 Sketch the graphs of fx 2 and fx 2 SOLUTION The graph of y fx 2 is obtained by shifting the graph of y fx two units to the left see the graph below on the left The graph of y fx 2 is obtained by shifting the graph of y fx two units up see the graph below on the right fx2 fx2 63 Sketch the graphs of f2x fex and2fx SOLUTION The graph of y f 2 is obtained by compressing the graph of y f x horizontally by a factor of 2 see the graph below on the left The graph of y f x is obtained by stretching the graph of y f x horizontally by a factor of 2 see the graph below in the middle The graph of y 2 f x is obtained by stretching the graph of y f x vertically by a factor of 2 see the graph below on the right fax fX2 2 fx 64 Sketch the graphs of f7x and 7f7x SOLUTION The graph of y f7x is obtained by re ecting the graph of y fx across the yaxis see the graph below on the left The graph of y 7f7x is obtained by re ecting the graph of y fx across both the x and yaxes or equivalently about the origin see the graph below on the right f X if X 10 CHAPTER 1 I PRECALCULUS REVIEW 65 Extend the graph of fx to 74 4 so that it is an even function SOLUTION To continue the graph of f x to the interval 74 4 as an even function re ect the graph of f 1 across the yaxis see the graph below 66 Extend the graph of fx to 74 4 so that it is an odd function SOLUTION To continue the graph of fx to the interval 74 4 as an odd function re ect the graph of fx through the origin see the graph below 67 Suppose that fx has domain 4 8 and range 2 6 What are the domain and range of a fX3 b fX3 C f3X d 3fX SOLUTION a f x 3 is obtained by shifting f x upward three units Therefore the domain remains 48 while the range becomes 5 9 h f x 3 is obtained by shifting f 1 left three units Therefore the domain becomes 1 5 while the range remains c f 3 is obtained by compressing f x horizontally by a factor of three Therefore the domain becomes while the range remains 2 6 d 3 f x is obtained by stretching f x vertically by a factor of three Therefore the domain remains 4 8 while the range becomes 6 8 68 Let fx x2 Sketch the graphs of the following functions over 72 2 a fX1 b fX1 C f5X d 5fX SOLUTION a The graph of y fx l is obtained by shifting the graph of y fx one unit to the left 10 s 6 4 2 43444 72 71 1 2 f6D b The graph ofy fx l is obtained by shifting the graph of y fx one unit up 10 s 6 4 2 72 71 fx 1 8 EC T t O N 11 I Real Numbers Functions and Graphs 11 c The graph of y f 5 is obtained by compressing the graph of y f x horizontally by a factor of 5 f5X d The graph of y 5 f x is obtained by stretching the graph of y f x vertically by a factor of 5 100 so 60 40 20 4 X 72 71 1 2 5f X 69 Suppose that the graph of f x sin is compressed horizontally by a factor of 2 and then shifted 5 units to the right a What is the equation for the new graph b What is the equation if you rst shift by 5 and then compress by 2 c SOLUTION Verify your answers by plotting your equations a Let f x sinx After compressing the graph of f horizontally by a factor of 2 we obtain the function gx f2x sin2x Shifting the graph 5 units to the right then yields hx gx 7 5 sin2x 7 5 sin2x 710 b Let fx sinx After shifting the graph 5 units to the right we obtain the function gx fx 7 5 sinx 7 5 Compressing the graph horizontally by a factor of 2 then yields 1115 82 sin2x 7 5 c The gure below at the top left shows the graphs of y sinx the dashed curve the sine graph compressed horizontally by a factor of 2 the dash double dot curve and then shifted right 5 units the solid curve Compare this last graph with the graph of y sin2x 7 10 shown at the bottom left The gure below at the top right shows the graphs of y sinx the dashed curve the sine graph shifted to the right 5 units the dash double dot curve and then compressed horizontally by a factor of 2 the solid curve Compare this last graph with the graph of y sin2x 7 5 shown at the bottom right l 0 Figure 28 shows the graph of f x lxl 1 Match the functions a7e with their graphs i7V a for 1 b fOC C fX2 d f1 12 efX1 12 CHAPTER 1 I PRECALCULUS REVIEW 3 3 3 2 2 l l 1 73727171 12 3 73727171 12 3 73727171 2 3 yfX IXI 1 i ii 3 3 3 2 2 2 l l l 737 r1 1 3 73727171 12 3 73727171 12 3 2 2 72 73 73 73 iii iv V FIGURE 28 SOLUTION a Shift graph to the right one unit V b Re ect graph across xaxis iv c Re ect graph across xaxis and then shift up two units iii d Shift graph to the right one unit and down two units ii 9 Shift graph to the left one unit i 71 Sketch the graph of f2x and fx where fx lxl 1 Figure 28 SOLUTION The graph of y f 2 is obtained by compressing the graph of y f x horizontally by a factor of 2 see the graph below on the left The graph of y f x is obtainedby stretching the graph of y f x horizontally by a factor of 2 see the graph below on the right y y 6 6 4 4 2 4724123 4724123 fax fX2 72 Find the function f 1 whose graph is obtained by shifting the parabola y 12 three units to the right and four units down as in Figure 9 FIGURE 29 SOLUTION The new function is fx x 7 32 7 4 73 De ne fx to be the larger of and2 7 1 Sketch the graph of fx What are its domain and range Express fx in terms of the absolute value function SOLUTION 8 EC T l O N 11 I Real Numbers Functions and Graphs 13 The graph of y f x is shown above Clearly the domain of f is the set of all real numbers while the range is y l y 3 1 Notice the graph has the standard Vshape associated with the absolute value function but the base of the V has been translated to the point 1 1 Thus fx ix 7 1 l 74 For each curve in Figure 30 state whether it is symmetrical with respect to the yaxis the origin both or neither y y X lt lt X A B y y C D FIGURE 30 SOLUTION A Both B Neither C yaxis D Origin 75 Show that the sum of two even functions is even and the sum of two odd functions is odd SOLUTION Even f ggtlt7xgt fox gm for gm f W dd Odd f gx fX SEX a fOC 806 f gx 76 Suppose that f x and gx are both odd Which of the following functions are even Which are odd a room b N fx c NO 80 d gov SOLUTION a fXgX fXgX fXgX EV H b fx3 Hm fx3 3 Odd C fX SEX fOC 806 fOC 806 5 Odd fox 7m m 5 Even 87 ego go 77 Give an example of a curve that is symmetrical with respect to both the yaxis and the origin Can the graph of a function have both symmetries Hint Prove algebraically that f x O is the only such function SOLUTION A circle of radius 1 with its center at the origin is symmetrical both with respect to the yaxis and the origin The only function having both symmetries is fx O For if f is symmetric with respect to the yaxis then f7x fx If f is also symmetric with respect to the origin then f7x 7fx Thus fx 7fx or 2fx 0 Finally fx O Further Insights and Challenges 78 Prove the triangle inequality by adding the two inequalities flat 5 a 5 lat itbt 5 b tbl SOLUTION Adding the indicated inequalities gives W W 5 a b 5 W W and this is equivalent to ta bl 5 lat lbl 79 Show that ifr ab is a fraction in lowest terms then r has a nite decimal expansion if and only ifb 2 for some n m 3 O Hint Observe that r has a nite decimal expansion when 101V r is an integer for some N 3 O and hence b divides 10 14 CHAPTER 1 I PRECALCULUS REVIEW SOLUTION Suppose r has a nite decimal expansion Then there exists an integer N 3 0 such that 101v r is an integer call it k Thus r k1ON Because the only prime factors of 10 are 2 and 5 it follows that when r is written in lowest terms its denominator must be of the form 2 5m for some integers n m 3 0 Conversely suppose r ab in lowest with b 2quot for some integers n m 3 0 Then r g or azmi 2quot5mr a If m 3 n then 2quot 5mr aZm or r 10m and thus r has a nite decimal expansion less than or 5 equal to m terms to be precise On the other hand ifn gt m then 2 5 r a5 m or r a and once again r has a nite decimal expansion 80 Let p p1 ps be an integer with digits p1 ps Show that p 7 107170p1ps v 2 Use this to nd the decimal expan51on of r Note that 2 18 r 11 10271 SOLUTION Let p p1ps be an integer with digits p1 ps andlet 171 ps Then 10S5 5P1H PSApl uPsi luAPsPl upsl7 Thus P 7 10371 ipi pl H ps Consider the rational number r 211 Because 18 18 r 11 99 10271 it follows that the decimal expansion of r is OK 81 A function fx is symmetrical with respect to the vertical line 1 a if fa 7 x fa x 2 Draw the graph of a function that is symmetrical with respect to x 2 b Show that if fx is symmetrical with respect to x a then gx fx a is even SOLUTION 2 There are many possibilities two of which may be b Letgx fx a Then SEX fx 11 fd ix fa x symmetry with respect to x a gov Thus gx is even 82 E Formulate a condition for fx to be symmetrical with respect to the point a O on the xaxis SOLUTION In order for fx to be symmetrical with respect to the point a 0 the value of f at a distance x units to the right ofa must be opposite the value of f at a distance x units to the left ofa In other words fx is symmetrical with respect to a 0 if fa x 7fa 7 x S E C T10 N 12 I Linear and Quadratic Functions 15 12 Linear and Quadratic Functions Preliminary Questions 1 What is the slope of the line y 74x 7 9 SOLUTION The slope of the line y 74x 7 9 is 74 given by the coef cient ofx 2 Are the lines y 2x 1 and y 72x 7 4 perpendicular SOLUTION The slopes of perpendicular lines are negative reciprocals of one another Because the slope of y 2x 1 is 2 and the slope of y 72x 7 4 is 72 these two lines are not perpendicular 3 When is the line ax by c parallel to the yaxis To the xaxis SOLUTION The line ax by c will be parallel to the yaxis when b O and parallel to the xaxis when a O 4 Suppose y 3x 2 What is Ay ifx increases by 3 SOLUTION Because y 3x 2 is a linear function with slope 3 increasing x by 3 will lead to Ay 33 9 5 What is the minimum of fx x 32 7 4 SOLUTION Because x 32 3 0 it follows that x 32 7 4 3 74 Thus the minimum value of x 32 7 4 is 74 6 What is the result of completing the square for fx x2 1 SOLUTION Because there is no x term in x2 1 completing the square on this expression leads to x 7 02 1 Exercises In Exercises 14 nd the slope the yintercept and the xintercept of the line with the given equation 1 y 3x 12 SOLUTION Because the equation of the line is given in slopeintercept form the slope is the coef cient of x and the yintercept is the constant term that is m 3 and the yintercept is 12 To determine the xintercept substitute y O and then solve for x O 3x 12 or x 74 2 y 4 7 x SOLUTION Because the equation of the line is given in slopeintercept form the slope is the coef cient of x and the yintercept is the constant term that is m 71 and the yintercept is 4 To determine the xintercept substitute y O andthen solve forx O 4 7 x orx 4 3 4x 9y 3 SOLUTION To determine the slope and yintercept we rst solve the equation for y to obtain the slopeintercept form This yields y 73x From here we see that the slope is m 7g and the yintercept is To determine the xintercept substitute y O and solve for x 4x 3 orx Z 4 y73x76 SOLUTION The equation is in pointslope form so we see that m Substituting x 0 yields y 7 3 73 or y 0 Thus the x and yintercepts are both 0 In Exercises 578 nd the slope ofthe line 5 y 3x 2 SOLUTION m 3 6 y3x792 SOLUTION m 3 7 3x 4y 12 SOLUTION First solve the equation for y to obtain the slopeintercept form This yields y 73x 3 The slope of the line is therefore m 7K 8 3x 4y 78 SOLUTION First solve the equation for y to obtain the slopeintercept form This yields y 72x 7 2 The slope of the line is therefore m 7K In Exercises 9720 nd the equation ofthe line with the given description 16 CHAPTER 1 I PRECALCULUS REVIEW 9 Slope 3 yintercept 8 SOLUTION Using the slopeintercept form for the equation of a line we have y 3x 8 10 Slope 72 yintercept 3 SOLUTION Using the slopeintercept form for the equation of a line we have y 72x 3 11 Slope 3 passes through 7 9 SOLUTION Using the pointslope form for the equation ofaline we have y 7 9 3x 7 7 or y 3x 712 12 Slope 75 passes through 0 0 SOLUTION Using the pointslope form for the equation ofaline we have y 7 O 75x 7 O or y 75x 13 Horizontal passes through 0 72 SOLUTION A horizontal line has a slope of 0 Using the pointslope form for the equation ofa line we have y 7 72 0x 7 O or y 7 14 Passes through 71 4 and 2 7 SOLUTION The slope of the line that passes through 71 4 and 2 7 is 7 7 4 m m 1 Using the pointslope form for the equation of a line we have y 7 7 1x 7 2 or y X 5 15 Parallel to y 3x 7 4 passes through 1 1 SOLUTION Because the equation y 3x 7 4 is in slopeintercept form we can readily identify that it has a slope of 3 Parallel lines have the same slope so the slope of the requested line is also 3 Using the pointslope form for the equation ofalinewehavey71 3x 71ory 3 72 16 Passes through 14 and 12 73 SOLUTION The slope of the line that passes through 14 and 12 73 is 73 7 4 7 77 m 7 1271 11 Using the pointslope form for the equation ofaline we have y 7 4 7171x 7 1 or y 717 1 17 Perpendicular to 3x 5y 9 passes through 2 3 SOLUTION We start by solving the equation 3 5y 9 for y to obtain the slopeintercept form for the equation ofa line This yields 3 y g35 g from which we identify the slope as 7 Perpendicular lines have slopes that are negative reciprocals of one another so the slope of the desired line is my Using the pointslope form for the equation ofaline we have y 7 3 go 7 2 or y g 7 g 18 Vertical passes through 74 9 SOLUTION A vertical line has the equation x c for some constant 6 Because the line needs to pass through the point 74 9 we must have c 74 The equation of the desired line is then x 74 19 Horizontal passes through 8 4 SOLUTION A horizontal line has slope 0 Using the point slope form for the equation of a line we have y 7 4 0x 7 8 or y 4 20 Slope 3 xintercept 6 SOLUTION If the xintercept is 6 then the line passes through the point 6 0 Using the pointslope form for the equation ofaline wehave y 7 O x 7 6 or y x 718 21 Find the equation of the perpendicular bisector of the segment joining 12 and 54 Figure 11 Him The a c b d T T midpoint Q of the segmentjoining a b and c d is lt y Perpendicular bisector FIGURE 11 S E C T10 N 12 I Linear and Quadratic Functions 17 SOLUTION The slope of the segmentjoining 12 and 5 4 is m 7 4 7 2 7 1 T 5 71 T 2 and the midpoint of the segment Figure 11 is 1 5 2 4 midpoint 33 The perpendicular bisector has slope 71m 72 and passes through 33 so its equation is y 7 3 72x 7 3 or y 7 x 22 Interceptintercept Form Show that ifa b 0 then the line with xintercept x a and yintercept y b has equation Figure 12 y 71 b SIN lt 11 FIGURE 12 Line withequation 2 1 SOLUTION The line passes through the points a 0 and 0 b Thus m 7 Using the pointslope form for the equationofalineyieldsy707x7agty7xbgt xybgt 1 23 Find the equation of the line with xintercept x 4 and yintercept y 3 SOLUTION FromExercise 22 32 1 or 3 4y 12 24 A line of slope m 2 passes through 1 4 Find y such that 3 y lies on the line SOLUTION In order for the point 3 y to lie on the line through 14 of slope 2 the slope of the segment connecting 14 and 3 y must have slope 2 Therefore y 47y 4 m73717 2 7251 474 178 25 Determine whether there exists a constant c such that the line X cy 1 b Passes through 3 1 2 Has slope 4 d Is vertical C Is horizontal SOLUTION a Rewriting the equation of the line in slopeintercept form gives y 7 To have slope 4 requires 7 4 or c 7 b Substituting 3 and y 1 into the equation of the line gives 3 c 1 or c 72 c From a we know the slope of the line is 7 There is no value for c that will make this slope equal to 0 CD With 6 0 the equation becomes 1 1 This is the equation ofa vertical line 26 Assume that the number N of concert tickets which can be sold at a price of P dollars per ticket is a linear function NP for 10 5 P 5 40 Determine NP called the demand function if N10 500 and N40 0 What is the decrease AN in the number of tickets sold if the price is increasedby A P 5 dollars SOLUTION We rst determine the slope of the line mi 50070 7 500 7 50 710740 T 730 3 Knowing that N40 0 it follows that 50 50 2000 N P 7 P740 7 P lt gt 3 lt gt 3 3 Because the slope of the demand function is 7 53 0 a 5 dollar increase in price will lead to a decrease in the number of tickets sold of 5305 2 0 83 or about 83 tickets 18 CHAPTER 1 I PRECALCULUS REVIEW 27 Materials expand when heated Consider a metal rod of length L0 at temperature To If the temperature is changed by an amount AT then the rods length changes by AL aLOAT where a is the thermal expansion coef cient For steel 0 124 X1075 C 1 a A steel rod has length L0 40 cm at T0 40 C What is its length at T 90 C b Find its length at T 50 C ifits length at T0 100 C is 65 in c Express length L as a function ofT if L0 65 in at T0 100 C SOLUTION a With T 90 C and T0 40 C AT 50 C Therefore AL aLoAT 124 x 10 54050 0248 and L L0 AL 400248 cm b With T 50 C and T0 100 C AT 750 C Therefore AL aLoAT 124 x107565750 70403 and L L0 AL 649597 in c L L0 AL L0 aLoAT L01 aAT 651 aT 7100 28 Do the points 05 1 1 12 2 2 lie on aline SOLUTION Examine the slope between consecutive data points The rst pair of data points yields a slope of 12 7 1 2 4 1 7 5 5 while the secondpair of data points yields a slope of 2 7 12 8 8 2 7 1 1 Because the slopes are not equal the three points do not lie on a line 29 Findb such that 2 71 3 2 and b 5 lie on aline SOLUTION The slope of the line determined by the points 2 71 and 3 2 is 2 A A A 3 3 7 2 To lie on the same line the slope between 3 2 and b 5 must also be 3 Thus we require 5 7 2 3 3 b 7 3 b 7 3 or b 4 30 Find an expression for the velocity v as a linear function oft that matches the following data 392 586 78 974 SOLUTION Examine the slope between consecutive data points The rst pair of data points yields a slope of 586 7 392 7 9 7 2 7 0 7 while the second pair of data points yields a slope of 78 7 586 97 4 7 2 and the last pair of data points yields a slope of 974 7 78 97 6 7 4 Thus the data suggests alinear function with slope 97 Finally 11 7 392 97g 7 0 7 v 97x 392 31 The period T of a pendulum is measured for pendulums of several different lengths L Based on the following data does T appear to be a linear function of L S E C T t O N 12 I Linear and Quadratic Functions 19 157 192 222 248 SOLUTION Examine the slope between consecutive data points The rst pair of data points yields a slope of 192 7157 7 35 3 7 2 7 while the second pair of data points yields a slope of 222 7 192 3 4 7 3 and the last pair of data points yields a slope of 248 7 222 26 5 7 4 Because the three slopes are not equal T does not appear to be a linear function of L 32 Show that fx is linear of slope m if and only if fx h 7 fx mh forallx andh SOLUTION First suppose fx is linear Then the slope between 1 fx and X 1 fx 11 is 7fXhfX 7 h mhfxh7fx Conversely suppose fxh7fx mh for allxandfor allhThen m fXhfX fXhfX h xh7x which is the slope between 1 fx and X 1 fx h Since this is true for all x and h f mustbe linear it has constant slope 33 Find the roots of the quadratic polynomials 24x273x71 bx272x71 SOLUTION a xji W mi bx2i W2i2 1i In Exercises 3441 complete the square ana na the minimum or maximum value of the qmdratie netion 34 yx22x5 SOLUTION y X2 2x l 7l 5 x l2 4 therefore the minimum value of the quadratic polynomial is 4 and this occurs at x 7 3s yx276x9 SOLUTION y x 7 32 therefore the minimum value of the quadratic polynomial is O and this occurs at x 3 36 y 79x2 x SOLUTION y 79Jc2 7 x9 79Jc2 7 79x 7 2 therefore the maximumvalue of the quadratic polynomial is 31 6 and this occurs at x 18 37 yx26x2 SOLUTION y x2 6x 9 7 9 2 x 32 7 7 therefore the minimum value of the quadratic polynomial is 77 and this occurs at 73 38 y2x274x77 SOLUTION y 2o2 7 2 1717 7 2o2 7 2x 1 7 7 7 2 2x 712 7 9 therefore the minimum value of the quadratic polynomial is 79 and this occurs at x l 39 y74x23x8 20 CHAPTER 1 I PRECALCULUS REVIEW SOLUTION y 74x2 3x 8 74Jc2 7 ix 8 1 96 74x 7 2 therefore the maximum value of the quadraticpolynomial is and this occurs at x g 40 y3x212x75 SOLUTION y 3162 4 4 7 5 712 3x 22 7 17 therefore the minimum value ofthe quadratic polynomial is 717 and this occurs at 72 41 y 4x 712x2 SOLUTION y 712Jc2 7 g 712162 7 31 6 712x 7 2 therefore the maximumvalue ofthe quadratic polynomial is and this occurs at x 42 Sketch the graph of y x2 7 6x 8 by plotting the roots and the minimum point SOLUTION y x2 7 6X 9 7 9 8 x 7 32 7 1 so the vertex is located at 3 71 and the roots are x 2 and x 4 This is the graph of2 moved right 3 units and down 1 unit 43 Sketch the graph of y X2 4x 6 by plotting the minimumpoint the yintercept and one other point SOLUTION y x2 4X 4 7 4 6 x 22 2 so the minimum occurs at 72 2 If 0 then y 6 and if x 74 y 6 This is the graph of2 moved left 2 units and up 2 units 44 If the alleles A and B of the cystic brosis gene occur in a population with frequencies p and 1 7 p Where p is a fraction between 0 and 1 then the frequency of heterozygous carriers carriers with both alleles is 2p1 7 p Which value of p gives the largest frequency of heterozygous carriers SOLUTION Let 1 2 f2p72p272ltp27p4 gt 72ltp7 gt Then p yields a maximum 45 For which values ofc does fx X2 C 1 have a double root No real roots SOLUTION A double root occurs when 82 7 411 O or 82 4 Thus 6 i2 There are no real roots when 82 7 411 lt O or 82 lt 4 Thus 72 lt c lt 2 46 Let f x be a quadratic function and c a constant Which is correct Explain graphically 2 There is a unique value of c such that y fx 7 c has a double root b There is a unique value of c such that y f x 7 c has a double root SOLUTION First note that because f x is a quadratic function its graph is a parabola a This is true Because fx 7 c is a vertical translation of the graph of fx there is one and only one value ofc that Will move the vertex of the parabola to the xaxis b This is false Observe that f x 7 c is a horizontal translation of the graph of f x If f x has a double root then f x 7 c Will have a double root for any value of c on the other hand if f 1 does not have a double root then there is no value of c for which f x 7 c Will have a double root S E C T l O N 12 I Linear and Quadratic Functions 21 l 47 Prove that E 3 2 for all gt 0 Hint Consider XI2 7 x 122 SOLUTION Letx gt 0 Then 2 l XI2 7x 12 1 72 1 Because XI2 7 x 122 3 0 it follows that l x72 0 or x732 x x d b 48 Leta b gt O Show that the geometric mean Mb is not larger than the arithmetic mean Hint Use a variation of the hint given in Exercise 47 SOLUTION Leta b gt O and note 2 05 a72xa b4rb Therefore M 5 2 49 If objects of weights 1 and wl are suspended from the balance in Figure 13A the crossbeam is horizontal if bx awl If the lengths a and b are known we may use this equation to determine an unknown weight 1 by selecting wl so that the crossbeam is horizontal If a and b are not known precisely we might proceed as follows First balance 1 by wl on the left as in A Then switch places and balance 1 by wz on the right as in B The average I 7w1 wz gives an estimate for x Show that J is greater than or equal to the true weight 1 A 13 FIGURE 13 SOLUTION First note bx awl andax bwz Thus 7 l X 7 2wI M 7 l bx ax 2 a b 7 ba 2 a b Ea by Exercise 47 i V x 50 Find numbers 1 and y with sum 10 and product 24 Hint Find a quadratic polynomial satis ed by x SOLUTION Let x and y be numbers whose sum is 10 and product is 24 Then X y 10 and xy 24 From the second equation y 274 Substituting this expression for y in the rst equation gives 1 10 or x2 7 10 24 x 74x e 6 Owhencex 4 or 61fx 4then y 6 Onthe otherhandifx 6 theny 4 Thus the two numbers are 4 and 6 51 Find a pair of numbers whose sum and product are both equal to 8 SOLUTION Let and y be numbers whose sum and product are both equal to 8 Then y 8 and xy 8 From the second equation y Substituting this expression for y in the rst equation gives 1 8 or x2 7 8x 8 0 By the quadratic formula 8ix64732 x 2 4i2 22 CHAPTER 1 I PRECALCULUS REVIEW Ifx 4 z then 8 8 4722 y 472 42 42 44 Ontheotherhandifx472 men 7 8 7 8 42 y 472 472 42 Thus the two numbers are 4 Z and 4 7 Z 52 Show that the graph of the parabola y 12 consists of all points P such that d1 d2 Where d1 is the distance from P to O b and rig is the distance from P to the horizontal line y 7 Figure 14 42 ML ML FIGURE 14 SOLUTION Let P be apoint on the graph of the parabola y x2 Then P has coordinates 1 x2 for some real number xNowd2X2 and 2 212 2 4121 212 21 x70x7Z xx 75x E XZ x Zd2 Further Insights and Challenges 53 Show that if fx and gx are linear then so is fx gx Is the same true of fxgx SOLUTION If fx mx b and gx nx d then fxgx mxbnxd mnxbd which is linear fxgx is not generally linear Take for example fx gx x Then fxgx x2 54 Show that if fx and gx are linear functions such that f0 g0 and fl gl then fx gx SOLUTION Suppose fx mx b and gx nx d Then f0 b and g0 d which implies b d Thus fx mxbandgx nxbNowfl mbandgl nbsomb nbandm n Thus fX 806 55 Show that the ratio AyAx for the function fx 12 over the interval 11 12 is not a constant but depends on the interval Determine the exact dependence of AyAx on x1 and x2 2 2 A x 7 x SOLUTION For x2 y x2 x1 Ax x2 7 x1 56 Use Eq 2 to derive the quadrch formula for the roots oftDc2 bx c O SOLUTION Consider the equation axz bx c 0 First complete the square to obtain b 2 4ac7b2 a x 0 2a 4a Then b 2 b2 7 4ac b b2 74 b2 7 4ac x and X 2a 4amp2 2a 4amp2 2a Dropping the absolute values yields b b2 7 4ac 7b b2 7 4M 7b i b2 7 4M x i or x i 2a 2a 2a 2a 2a S E C T l O N 13 I The Basic Classes of Functions 23 57 Leta e O Show that the roots ofax2 bx e O and ex2 bx a O are reciprocals of each other SOLUTION Let r1 and r2 be the roots of ax2 bx e and r3 and r4 be the roots of ex2 bx a Withoutloss of generality let 77bb274aegti7 2a 7b7b774ae 2a r1 7bb274ae 7b7b274ae 2a7b7b274ae 7b7b274ae r 4 b2 7 b2 4ae 20 1 1 Similarly you can show r3 72 58 Complete the square to show that the parabolas y ax2 bx e and y ax2 have the same shape show that the rst parabola is congruent to the second by a vertical and horizontal translation SOLUTION 7a x2bx b2 e7b27a x b 24a87b2 yi a 4a2 4amp7 2a 4a 4ac7b2 Thus the rst parabola is just the second translatedhorizontally by 7 and vertically by 411 59 Prove Vi te s Formulas which state that the quadratic polynomial with given numbers a and 3 as roots is x2 bxewhereb 707 ande a SOLUTION Ifa quadratic polynomial has roots or and3 then the polynomial is x7ax73x27ax7Bxa x27a73xa3 Thus b 707 3 ande 043 13 The Basic Classes of Functions Preliminary Questions 1 Give an example of a rational function 3x2 7 2 7x3 x 7 1 h 2 Is lxl apolynomial function What about ix2 ll SOLUTION One example is SOLUTION lxl is not a polynomial however because x2 1 gt O for all x it follows that lxz 1l x2 1which is a polynomial 3 What is unusual about the domain off 0 g for fx x12 and gx 717le SOLUTION Recall that f o gx fgx Now for any real number x gx 71 7 lxl 5 71 lt 0 Because we cannot take the square root of a negative number it follows that f gx is not de ned for any real number In other words the domain of fgx is the empty set 4 Is fx increasing or decreasing SOLUTION The function fx C is an exponential function with base b lt 1 Therefore f is a decreasing function 5 Give an example of a transcendental function SOLUTION One possibility is fx eC 7 sinx Exercises In Exercises I712 determine the domain 0fthe meti0n 1 fx x14 SOLUTION x 3 O 2 gt t23 24 CHAPTER 1 I PRECALCULUS REVIEW SOLUTION All realS 3 fx x33x 74 SOLUTION All realS 4 hz z3 f3 SOLUTION z 0 1 5 gt Jr 2 SOLUTION t 72 1 m SOLUTION A11 realS 6 x 7 Ga 2 7 SOLUTION a i2 J x2 7 9 SOLUTION x 3 0 3 3 x 9 for f4 x 713 SOLUTION X 75 0 1 S 10 Fs Sinlt gt s 1 SOLUTION S 71 71 11 gy104yy71 SOLUTION y gt 0 x x 1239 f m SOLUTION x 75 0 3 74 In Exercises 13724 identify each of the following functions as polynomial rational algebraic or transcendental 13 fx 4x3 9x2 7 8 SOLUTION Polynomial 14 fxx 4 SOLUTION Rational 15 for 7 b SOLUTION Algebraic 16 fx 17x2 SOLUTION Algebraic 17 f x2 x x Sinx SOLUTION Transcendental 18 f x 2C SOLUTION Transcendental 2x3 3x 9 7 7x2 SOLUTION Rational 19 for 3x 7 9x 12 20 fx W S E C T t O N 13 I The Basic Classes of Functions 25 SOLUTION Algebraic 21 fx sinx2 SOLUTION Transcendental x 22 f x W SOLUTION Algebraic 23 fx x2 3f1 SOLUTION Rational 24 fx sin3x SOLUTION Transcendental 25 Is f x 2CZ a transcendental function SOLUTION Yes 26 Show that fx x2 31F1 and gx 3x3 7 9 x 2 are rational functions show that each is a quotient of polynomials 3 3 3 SOLUTION fx x23x 1 x2 35793 1 8X3X379xx 2 X 216 X In Exercises 27734 calculate the composite mctions f o g and g o f and determine their domains 27 fx gx 1 1 SOLUTION fgx W D x 3 71 gfx 1Dx 0 28 fx gx 4 SOLUTION fgx 14sz 0 gfx x4 D 1 7e 0 29 fx 2X gx x2 SOLUTION fgx 2 D R gfx 262 22X D R 30 fx pct g9 sin9 SOLUTION fg9 sinei D R gfx sin pct D R 31 f9 cos 9 gx x3 x2 SOLUTION fgx cosx3 x2 D R gf9 cos3 9 cos2 9 D R 1 32 100032 gltxgtx 2 1 7 1 7 1 72 2 2 SOLUTION fgx7M 2H7WDX7EO gfx7ltmgt 7x 1DR 1 72 33fl7 807 t SOLUTION fgt J12 D Not validfor anyt gft 7 7 D t gt 0 it 34 ft 7 glttgt1it3 SOLUTION fgt 17130151 gft 1 7132 D t 3 0 35 The population in millions of a country as a function of time t years is Pt 30 21 with k 01 Show that the population doubles every 10 years Show more generally that for any nonzero constants a and k the function gt a2 doubles after lk years SOLUTION Let Pt 30 20 Then Pt 10 30201 10gt 30 201 1 230 201 2Pt Hence the population doubles in size every 10 years In the more general case let gt a2 Then 1 g t E a2kt1k a2kt1 2 ier 28 Hence the function g doubles after 1 k years 26 CHAPTER 1 I PRECALCULUS REVIEW 36 Find all values of e such that the domain of x 1 x f x2 2e 4 is R SOLUTION The domain of f will consist of all real numbers provided the denominator has no real roots The roots of x22ex40are 72 i 42716 xieie274 There will be no real roots when e2 lt 4 or when 72 lt e lt 2 Further Insights and Challenges In Exercises 3743 we de ne the rst difference 5f ofa metion fx by 5fx fx 1 7 fx 37 Show thatiffx x2 then 5fx 2X 1 Calculate 5f for fx x and fx x3 SOLUTION fx x215fx fx 1 7 fx 1 12 7 x2 2 1 fxx5fxx1ix1 fx x3 5fx x13 7x3 3x23x1 38 Show that 510quot 9 10quot and more generally 5bquot e b C for some constant e SOLUTION 610X10x1710X 1010X 710 1Ox1071910X 5W bX17 bx bquotb 71 39 Show that for any two functions f and g 5f g 5f 5g and 5ef e5f where e is any constant SOLUTION 5f g fX 1 806 1 fX 806 fX 1 fX 806 1 i 806 5fX 5806 5Cf CfX 1 CfX CfOC 1 fX C5fXA 40 First differences can be used to derive formulas for the sum of the kth powers Suppose we can nd a function Px such that 5P x 1k and PO O Prove that P1 1k PZ 1quot 2k and more generally for every whole numbern Pn1k2knk I SOLUTION Suppose we have found a function Px such that 5Px X 1k and PO 0 Taking x O we have 5P0 P1 7 PO 01k 1quot Therefore P1 PO 1k 1quot Next take 1 1 Then 5P1 P2 7 P1 1 1k 2k and PZ P1 2k 1k 2k To prove the general result we will proceed by induction The basis step proving that P1 1k is given above so we move on to the induction step Assume that for some integer j Pj 1quot 2quot jk Then 5Pj Pj1 PJ39 11k and Pj1 Pjj1k 1k2kjkj1k Therefore by mathematical induction for every whole number n Pn 1quot 2quot nk 41 First show that pm satis es 5 P X 1 Then apply Exercise 40 to conclude that 123nw SOLUTION Let Px xx 12 Then 51 Px1 7 Px x1x2 xx1 X1X27X 2 7 2 2 x1 Also note that PO 0 Thus by Exercise 40 with k 1it follows that pm 7 123n S E C T l O N 13 I The Basic Classes of Functions 27 42 Calculate 503 502 and 50 Then nd a polynomial PX of degree 3 such that 5P x 12 and P0 0 Conclude that Pn1222n2 SOLUTION From Exercise 37 we know 5x1 5x22x1 and 5x33x23x11 Therefore 1 3 1 2 1 2 2 5516 5X 65XX 2XlXli Now using the properties of the rst difference from Exercise 39 it follows that 1 1 1 1 1 1 1 1 1 2x33x2x 53 52 55 3 5 2 5 5 3 2 5 1 3X2X6X 3 2x 6x 3x 2X6X 6 Finallylet 2X3 3X2 X PX T1 Then 5PX X l2 and PO 0 so by Exercise 40 withk 2it follows that 2 3 3 2 PM W 122232n2 43 This exercise combined with Exercise 40 shows that for all k there exists a polynomial PX satisfying Eq 1 The solution requires proof by induction and the Binomial Theorem see Appendix C a Show that sock k 1xk where the dots indicate terms involving smaller powers of X b Show by induction that for all whole numbers k there exists a polynomial of degree k l with leading coef cient 1 k l l k1 Pm xk1 such that 5P x 1k and P0 0 SOLUTION a By the Binomial Theorem 5Xnlx1nlixrtlltXrtlltquotT1gtXr1lt 1gtxnilu1gtixrtl quotT1gtxquotlt 1gtx 11 6ltxquot1gtltn1gtxquot Thus where the dots indicate terms involving smaller powers of X b Fork Onote that PX X satis es 5P X 10 1 and PO 0 Now suppose the polynomi 1 PX EXk pkllxk 1 p1X which clearly satis es PO 0 also satis es 5P X 1quot1 We try to prove the existence of l k1xk1wkqlx QX such that 5Q X lk Observe that QO O 28 CHAPTER 1 I PRECALCULUS REVIEW If 5Q x 1k and 5P x 1k1 then 5Q x 1k x 15P x5Px 5P By the linearity of 5 Exercise 39 we nd 5Q 7 SP XSP or 5Q 7 P XSP By de nition 1 1 QiPk1xk1ltq 7 gtxkqi7p1x so by the linearity of 5 l l L 5Q 7 P 7 m6ltxk1gtqk 7 sock r11 7 p1 7xltx1gtk 1 1 By pama 5xk1 k lxk Lk717k71xk 1 qule 1 sock kxk l Lk vk xk 2 Lk le 1 502 2 1 where the Livj are real numbers for each i j To construct Q we have to group like powers of x on both sides of 1 This yields the system of equations 611k1xkxk 1 k71 1 k71 k71 L L L 7 k k 7 1 1 k1k1x ltQk k X X 1 1 7 0 1 q k Qk71 Pk71 q1 P17 The rst equation is identically true and the second equation can be solved immediately for qk Substituting the value of qk into the third equation of the system we can then solve for quI We continue this process until we substitute the values of qk quI 112 into the last equation and then solve for ql 14 Trigonometric Functions Preliminary Questions 1 How is it possible for two different rotations to de ne the same angle SOLUTION Working from the same initial radius two rotations that differ by a whole number of full revolutions will have the same ending radius consequently the two rotations will de ne the same angle even though the measures of the rotations will be different 2 Give two different positive rotations that de ne the angle SOLUTION The angle 714 is de ned by any rotation of the form 271k where k is an integer Thus two different positive rotations that de ne the angle 714 are 71 971 71 4171 21 d 25 471 4 an 471 4 3 Give a negative rotation that de nes the angle SOLUTION The angle 713 is de ned by any rotation of the form 271k where k is an integer Thus a negative rotation that de nes the angle 713 is 271 1 5n 3 7 3 4 The de nition of cos 9 using right triangles applies when choose the correct answer S E C T t O N 14 I Trigonometric Functions 29 71 a0lt9lt b0lt9lt71 c0lt9lt271 SOLUTION The correct response is a O lt 9 lt 5 What is the unit circle de nition of sin 9 SOLUTION Let 0 denote the center of the unit circle and let P be a point on the unit circle such that the radius W makes an angle 9 with the positive xaxis Then sin 9 is the ycoordinate of the point 6 How does the periodicity of sin 9 and cos 9 follow from the unit circle de nition SOLUTION Let 0 denote the center of the unit circle and let P be a point on the unit circle such that the radius W makes an angle 9 with the positive xaxis Then cos 9 and sin 9 are the x and ycoordinates respectively of the point P The angle 9 271 is obtained from the angle 9 by making one full revolution around the circle The angle 9 271 will therefore have the radius W as its terminal side Thus cos9271 cos9 and sin9271 sin 9 In other words sin 9 and cos 9 are periodic functions Exercises 1 Find the angle between 0 and 271 that is equivalent to 13714 SOLUTION Because 13714 gt 271 we repeatedly subtract 271 until we arrive at a radian measure that is between 0 and 271 After one subtraction we have 13714 7 271 5714 Because 0 lt 5714 lt 271 5714 is the angle measure between 0 and 271 that is equivalent to 13714 2 Describe the angle 9 g by an angle ofnegative radian measure SOLUTION lfwe subtract 271 from 716 we obtain 9 711716 Thus the angle 9 716 is equivalent to the angle 9 711 71 6 3 Convert from radians to degrees 71 5 371 1 b d 7 3 3 C 12 4 SOLUTION a 1 E m 5710 b E 600 71 71 3 71 0 5 180 7 75 N23 870 d 371 180 7 1350 c 12 71 7 71 N 4 71 7 4 Convert from degrees to radians 2 1 b 30 c 25 d 120 SOLUTION 71 71 71 71 571 71 271 1 b30 25 d120 a 180 180 180 6 c 180 36 180 3 5 Find the lengths of the arcs subtended by the angles 9 and radians in Figure 20 FIGU RE 20 Circle of radius 4 SOLUTION s r9 49 36 s r 42 8 6 Calculate the values of the six standard trigonometric functions for the angle 9 in Figure 21 15 FIGURE 21 30 CHAPTER 1 I PRECALCULUS REVIEW SOLUTION Using the de nition of the six trigonometric functions in terms of the ratio of sides of a right triangle we nd sin9 817 cos9 1517 tan9 815 csc9 178 sec9 1715 cote 158 7 Fill in the remaining values of cos 9 sin 9 for the points in Figure 22 SOLUTION osmium M 71 TV 3 710 75 71 cos 9 sin 0 32g 32g 71325 0 71gt G 325 3 32g 4 8 Find the values of the six standard trigonometric functions at 9 11716 SOLUTION From Figure 22 we see that 117 1 117 J 51H 7 and cos Then tan 7 6 117 cosT 117 1171 COS cot 116 73 sinT r 117 1 csc 2 6 shill 7 117 1 2J5 sec 6 cosu 3 In Exercises 9714 use Figure 22 to nd all angles between 0 and 2a satisfying the given condition 7 1 9 cos9 7 7 7 It 57 SOLUTION 9 7 g 10 tan9 1 7 II 57 SOLUTION 9 7 17 11 tan 9 71 SOLUTION 9 3quot IT 12 csc9 2 SOLUTION 9 13 39 s1nx 2 S E C T l O N 14 I Trigonometric Functions 31 7 It 27 SOLUTION X 7 g 14 sect 2 7 It 57 SOLUTION t 7 g 15 Fill in the following table of values 9 n n 7 27 37 57 6 4 3 2 3 4 6 tan 9 sec 9 SOLUTION 9 n n n 7 27 37 57 6 4 3 2 3 4 6 tan 9 1 1 und 7 3 71 7 1 J5 ya sec 9 2 5 2 und 2 J2 2 J5 J5 16 Complete the following table of signs for the trigonometric functions 9 sin cos tan cot sec csc O 9 n lt lt2 n 9 n 2lt lt n 9 37 lt lt 2 3n lt9lt2n 2 SOLUTION 9 Sin cos tan cot sec csc n 0lt9lt n lt9ltn 7 i 2 3 nlt9lt 7 7 2 3n 7lt9lt2n 7 7 17 Show that if tan 9 c and O 5 9 lt 712 then cos 9 11 2 Hint Draw a right triangle whose opposite and adjacent sides have lengths c and 1 SOLUTION Because 0 5 9 lt 712 we can use the de nition of the trigonometric functions in terms ofright triangles tan 9 is the ratio of the length of the side opposite the angle 9 to the length of the adjacent side With 6 we label the length of the opposite side as c and the length of the adjacent side as 1 see the diagram below By the Pythagorean theorem the length of the hypotenuse is 1 62 Finally we use the fact that cos 9 is the ratio of the length of the adjacent side to the length of the hypotenuse to obtain 32 CHAPTER 1 I PRECALCULUS REVIEW V1c2 18 Suppose that cos 9 a Show that ifO 5 9 lt nz then sin9 z 3 and tan 9 z b Find sin 9 and tan 9 if3n2 5 9 lt 27 SOLUTION 2 Because 0 5 9 lt 712 we can use the de nition of the trigonometric functions in terms ofright triangles cos 9 is the ratio of the length of the side adjacent to the angle 9 to the length of the hypotenuse so we label the length of the adjacent side as 1 and the length of the hypotenuse as 3 see the diagram below By the Pythagorean theorem the length of the side opposite the angle 9 is 32 7 12 2J2 Finally we use the de nitions of sin 9 as the ratio of the length of the opposite side to the length of the hypotenuse and of tan 9 as the ratio of the length of the opposite side to the length of the adjacent side to obtain 2 2 2 2 f f 399 d t 9 2 sm 3 an an 1 b If 3712 5 9 lt 271 then 9 is in the fourth quadrant and sin 9 and tan 9 are negative but have the same magnitude as found in part a Thus sin 9 7 and tan9 72x2 Ni 3 In Exercises 19724 assume that O 5 9 lt 712 19 Find sine and tan9 ifcos9 SOLUTION Consider the triangle below The lengths of the side adjacent to the angle 9 and the hypotenuse have been labeled so that cos 9 The length of the side opposite the angle 9 has been calculated using the Pythagorean theorem V132 7 52 12 From the triangle we see that 12 3995 d 9712 Sin 713 an an 75 20 Find cos 9 and tan 9 if sin 9 g SOLUTION Consider the triangle below The lengths of the side opposite the angle 9 and the hypotenuse have been labeled so that sin 9 The length of the side adjacent to the angle 9 has been calculated using the Pythagorean theorem 52 7 32 4 From the triangle we see that 954 d t 953 cos 75 an an 74 S E C T10 N 14 I Trigonometric Functions 33 21 Find sin 9 sec 9 and cot 9 if tan 9 SOLUTION If tan9 then cote For the remaining trigonometric functions consider the triangle below The lengths of the sides opposite and adjacent to the angle 9 have been labeled so that tan 9 The length of the hypotenuse has been calculated using the Pythagorean theorem 22 72 453 From the triangle we see that 2 2 53 253 s1n9 and sec9 153 53 7 2 22 Find sin 9 cos 9 and sec 9 if cot 9 4 SOLUTION Consider the triangle below The lengths of the sides opposite and adjacent to the angle 9 have been labeled so that cote 4 The length of the hypotenuse has been calculated using the Pythagorean theorem 42 12 17 From the triangle we see that 9 1 17 9 4 4 17 d 9 17 s1n cos an sec 117 17 117 17 4 W 23 Find cos29ifsin9 g SOLUTION Using the double angle formula cos29 cos2 9 7 sin2 9 and the fundamental identity sin2 9 cos2 9 1we nd that cos29 17 2 sin2 9 Thus cos29 17 2125 2325 24 Find sin29 and cos29 if tan 9 SOLUTION By the double angle formulas sin29 2sin 9cos 9 and cos29 cos2 9 7 sin2 9 We can determine sin 9 and cos 9 using the triangle shown below The lengths of the sides opposite and adjacent to the angle 9 have been labeled so that tan 9 The hypotenuse was calculated using the Pythagorean theorem 12 52 Thus 9 5 xE d 9 1 J3 SlH 3H COS 5 3 73 3 Finally 6 3 22 smze2 i 3 3 3 2071 2 1 COS 73 3 F 43 25 Find cos 9 and tan 9 if sin 9 04 and 712 5 9 lt n 34 CHAPTER 1 I PRECALCULUS REVIEW SOLUTION We can determine the magnitude of cos 9 and tan 9 using the triangle shown below The lengths of the side opposite the angle 9 and the hypotenuse have been labeled so that sin 9 04 The length of the side adjacent to the angle 9 was calculated using the Pythagorean theorem 52 7 22 V21 From the triangle we see that 21 2 2 2l lcos 9t and ltanel 5 m 21 Because 712 5 9 lt it both cos 9 and tan 9 are negative consequently cos97 and tan97 5 21 5 2 J71 26 Find cos 9 and sin9 if tan9 4 and n 5 9 lt 3712 SOLUTION We can determine the magnitude of cos 9 and sin 9 using the triangle shown below The lengths of the sides opposite and adjacent to the angle 9 have been labeled so that tan 9 4 The length of the hypotenuse was calculated using the Pythagorean theorem x l2 42 17 From the triangle we see that l 17 4 4 17 lcosQl and ls1n9l 17 17 v14 17 Because it 5 9 lt 3712 both cos 9 and sin 9 are negative consequently l7 4 17 cos97 and sin97 l7 17 J 4 27 Find the values ofsin 9 cos 9 and tan 9 at the eight points in Figure 23 03965 0913 03965 0913 A 13 FIGURE 23 SOLUTION Let s start with the four points in Figure 23A The point in the rst quadrant has coordinates 03965 0918 Therefore 0918 23153 03965 sin9 0918 cos 9 03965 and tan 9 The coordinates of the point in the second quadrant are 70918 03965 Therefore 03965 704319 s1n0 03965 cos9 0918 and lame 70918 7 S E C T t O N 14 I Trigonometric Functions 35 Because the point in the third quadrant is symmetric to the point in the rst quadrant with respect to the origin its coordinates are 703965 70918 Therefore sin 9 70918 cos 9 703965 and tan 9 23153 70918 703965 Because the point in the fourth quadrant is symmetric to the point in the second quadrant with respect to the origin its coordinates are 0918 703965 Therefore 703965 sin 9 703965 cos 9 0918 and tan 9 704319 0 918 Now consider the four points in Figure 23B The point in the rst quadrant has coordinates 03965 0918 Therefore 0 918 s1n 9 0918 cos 9 03965 and tan9 23153 03965 The point in the second quadrant is a re ection through the yaxis of the point in the rst quadrant Its coordinates are therefore 703965 0918 and 723153 03965 sine 0918 cos 9 703965 and tan9 Because the point in the third quadrant is symmetric to the point in the rst quadrant with respect to the origin its coordinates are 703965 70918 Therefore 70918 23153 703965 sin 9 70918 cos 9 703965 and tan 9 Because the point in the fourth quadrant is symmetric to the point in the second quadrant with respect to the origin its coordinates are 03965 70918 Therefore 0 918 7 723153 sin 9 70918 cos 9 03965 and tan 9 7 03965 28 Refer to Figure 24A Express the functions sin 9 tan 9 and csc 9 in terms of c SOLUTION By the Pythagorean theorem the length of the side adjacent to the angle 9 in Figure 24 A is 1 7 62 Consequently c sin9Tc cos9 1762 and tan9 29 Refer to Figure 24B Compute cos 11 sin 11 cot 11 and csc 11 03 03 A 13 FIGURE 24 SOLUTION By the Pythagorean theorem the length of the side opposite the angle 11 in Figure 24 B is 1 7 032 091 Consequently 03 091 03 cos 03 sin v091 cot and csc W 1 W 1 W 091 W 091 36 CHAPTER 1 I PRECALCULUS REVIEW 30 Express cos 9 and sin 9 in terms of cos 9 and sin 9 Hint Find the relation between the coordinates a b and e d in Figure 25 FIGURE 25 SOLUTION Note the triangle in the second quadrant in Figure 25 is congruent to the triangle in the rst quadrant rotated 900 clockwise Thus e 7b andd a But a cos 9 b sin 9 e cos 9 andd sin 9 therefore cos 9 7 sin9 and sin 9 cos 9 In Exercises 31734 sketch the graph Over 0 27 31 2 sin49 SOLUTION 32 cos 2 9 7 SOLUTION 33 cos 29 7 SOLUTION 34 sin29ig 2 SOLUTION y 35 How many points lie on the intersection of the horizontal line y e and the graph of y sin for O 5 x lt 271 Him The answer depends on e S E C T10 N 14 I Trigonometric Functions 37 SOLUTION Recall that for anyx 71 5 sinx 5 1 Thus if ict gt 1 the horizontal line y c and the graph ofy sinx never intersect lf c 1 then y c and y sinx intersect at the peak of the sine curve that is they intersect at x 7 On the other hand ifc 71 then y c and y sinx intersect at the bottom of the sine curve that is they intersect atx Finally if 1c1 lt 1 the graphs ofy c and y sinx intersect twice 36 How many points lie on the intersection of the horizontal line y c and the graph of y tanx for O 5 x lt 27 SOLUTION Recall that the graph of y tanx consists of an in nite collection of branches each between two consecutive vertical asymptotes Because each branch is increasing and has a range of all real numbers the graph of the horizontal line y c will intersect each branch of the graph of y tanx once regardless of the value of c The interval 0 5 x lt 271 covers the equivalent of two branches of the tangent function so over this interval there are two points of intersection for each value of c In Exercises 3740 solvefor O 5 9 lt 27 see Example 6 37 sin29 sin39 O SOLUTION sinot 7 sinB when a 7B 271k or a n B 271k Substituting 0 29 and B 39 we have either29 739 271k or 29 n 39 271k Solving each ofthese equations for 9 yields 9 nk or 9 7n 7 271k The solutions on the interval 0 5 9 lt 271 are then 2 4 6 9o 557T5 wt 2 38 sin9 sin29 SOLUTION Using the double angle formula for the sine function we rewrite the equation as sin 9 2 sin 9 cos 9 or sin 91 7 2 cos 9 0 Thus either sin 9 O or cos 9 The solutions on the interval 0 5 9 lt 271 are then 5 9 0 E n 3 3 39 cos49 cos29 O SOLUTION cos a 7 cosB when a B 7H 271k or a B n27rk Substituting 0 49 andB 29 we have either 69 72 271k or 49 29 7H 271k Solving each of these equations for 9 yields 9 k or 9 nk The solutions on the interval 0 5 9 lt 271 are then 977 7 57 77 371117 76 2 6 6 2 6 40 sin9 cos29 SOLUTION Solving the double angle formula sin2 9 1 7 cos29 for cos29 yields cos29 1 7 2 sin2 9 We can therefore rewrite the original equation as sin 9 1 7 2 sin2 9 or 2 sin2 9 sin 9 7 1 O The lefthand side of this latter equation factors as 2 sin9 7 1sin 9 1 so we have either sin9 or sin9 71 The solutions on the interval 0 5 9 lt 27 are 5 975 6 6 9 2 In Exercises 41750 derive the identities using the identities listed in this section 41 cos29 2cos2 9 71 SOLUTION Starting from the double angle formula for cosine cos2 9 1 cos29 we solve for cos29 This gives 2 cos2 9 1 cos29 and then cos29 2 cos2 9 71 9 1 9 42 0052 i 2 2 SOLUTION Substitute x 92 into the double angle formula for cosine cos2x 1cos2x to obtain 2 9 1 cos 9 cos 2 2 43 Sing Nico 2 2 9 SOLUTION Substitute x 92 into the double angle formula for sine sin2x 1 7 cos2x to obtain sin2 lt gt 2 1 7 cos 9 9 1 7 cos 9 T Taking the square root of both s1des yields sm 5 38 CHAPTE R l I PRECALCULUS REVIEW 44 sin9 n 7 sin 9 SOLUTION From the addition formula for the sine function we have sin9 n sin9cos cos9sinn 7 sin9 45 cos9 n 7 cos 9 SOLUTION From the addition formula for the cosine function we have cos9 n cos9cosn7 sin9sinn cos 971 7 cos9 46 tanx cotg 7x SOLUTION Using the Complementary Angle Identity tanx Iit cos7r2 7 1 sin co 7x 7 2 sin7r2 7 x cosx 47 tan7r 7 9 7 tan 9 SOLUTION Using Exercises 44 and 45 sin7r7 9 7 sin7r 79 7 7 sin79 7 sin9 7 I 9 cos7r7 9 7 cos7r 79 7 7cos79 7 7cos9 7 an tan7r7 9 The second to last equality occurs because sin is an odd function and cosx is an even function 48 tan 2 l 7 tan2 1 SOLUTION Using the de nition of the tangent function and the double angle formulas for sine and cosine we nd sin 2 2 sin cos x l cos2 x 2 tan 1 tan2x 7 005215 coszx 7 Sinzx 1 coszx l 7 tanzx sin2x l cos 2 SOLUTION Using the addition formula for the sine function we nd 49 tanx sin2x sinx x sinx cosx cosx sinx 2 sin cosx By Exercise 41 we know that cos 2 2 coszx 7 1 Therefore sin 2 2sinx cosx 2sinxcosx sinx 2 2 tanx lcos2x l2cos x71 2cos x cosx l7 cos4x 50 sinzx coszx SOLUTION Using the double angle formulas for sine and cosine we nd 1 1 l sinzx coszx 517 cos2x 51 cos2x 117 cos22x 71111 4 711 4 74 2 2cos x 78 cos x 51 Use Exercises 44 and 45 to show that tan 9 and cot 9 are periodic with period 7 SOLUTION By Exercises 44 and 45 I 9 isin9n7 7sin97I 9 am icos9n77cos97 an and 9 n 7 9 cot9nw COS cot9 sin9 n 7sin9 Thus both tan 9 and cot 9 are periodic with period 7 S E C T 1 O N 14 I Trigonometric Functions 39 n n n n 52 Use the addition formula to compute cos noting that 7 12 12 3 4 SOLUTION eoslt9gt7eos97 gt7cosltzgt994999 1 3 M T 2 4 2 2 53 Use the Law of Cosines to nd the distance from P to Q in Figure 26 Q P FIGURE 26 SOLUTION By the Law of Cosines the distance from P to Q is 7 102 82 7 2108 cos 1619281 9 Further Insights and Challenges 54 Use the addition formula to prove cos39 4cos3 97 3cos9 SOLUTION cos 39 cos29 9 cos29 cos 9 7 sin29 sin 9 2 cos2 9 7 1 cos 9 7 2 sin 900s 9 sin 9 cos 92 cos2 9 717 2 sin2 9 cos 92 cos2 9 717 217 cos2 9 cos92cos2971722cos294cos3973cos9 55 Use the addition formulas for sine and cosine to prove I b tanatanb an a 17 tanatanb cotacotb1 cota 7 b cotb 7 cota SOLUTION mm b 7 sina b 7 sinacosb cosasinb 7 7 land laub 7 7 7 39 39 7 0051100517 sinusinb 7 7 cosa b cosacosb s1nas1nb my 7 1 tana tanb coma 7 b cosa 7 b cosacosbsinasinb W 22 I cotd001b1 sina 7 b sinacosb 7 cosasinb Si llCOSb 7 00511 Sl b cotb 7 cota asinb sinusinb 56 Let 9 be the angle between the line y mx b and the xaxis Figure 27A Prove that m tan 9 FIGURE 27 40 CHAPTER 1 I PRECALCULUS REVIEW SOLUTION Using the distances labeled in Figure 27A we see that the slope of the line is given by the ratio rs The tangent of the angle 9 is given by the same ratio Therefore m tan 9 57 Let L1 and L2 be the lines of slope m1 and m2 Figure 273 Show that the angle 9 between L1 and L2 satis es mzml l cote m2 7 ml SOLUTION Measured from the positive xaxis let a and B satisfy tana m1 and tanB m2 Without loss of generality let B Z 0 Then the angle between the two lines will be 9 B 7 0 Then from Exercise 55 cothotal 7mi121 71m1m2 cota7cotB mLimL m27m1 1 Z cote cotB 7 a 58 Perpendicular Lines Use Exercise 57 to prove that two lines with nonzero slopes m1 and m2 are perpendicular if and only ifmz 7lm1 SOLUTION If lines are perpendicular then the angle between them is 9 712 5 l com2 m ml 7 m2 0 l mlmz ml 7 m2 gt mlmz 7 gt m1 7 m2 59 Apply the doubleangle formula to prove l 2 cos 2 J2 l b 3 5 22 Guess the values of cos l and of cos 1 for all n 32 2 SOLUTION 7 714 1 cos 1 g 1 2 2 2 cos 8 cos 2 2 2 2 f lcos ll22 1 b 005 MTquot 7 2f 5 2y2 2 c Observe that 8 23 and cos involves two nested square roots of 2 further 16 24 and cos involves three nested square roots of 2 Since 32 25 it seems plausible that n 1 2 2 2 2 C0532 2V and that cos 2 involves n 7 l nested square roots of 2 Note that the general case can be proven by induction 15 Inverse Functions Preliminary Questions 1 Which of the following satisfy f71x fx 7 a N 7 x b fltxgt717 x 0 NC 71 d N 7 w e fx 7 m t for 7 fl SOLUTION The functions 2 fx x b fx l 7x and f fx x71 satisfy f71x fx 2 The graph of a function looks like the track of aroller coaster Is the function onetoone SOLUTION Because the graph looks like the track of a roller coaster there will be several locations at which the graph has the same height The graph will therefore fail the horizontal line test meaning that the function is not onetoone S E C T l O N 15 I Inverse Functions 41 3 Consider the function f that maps teenagers in the United States to their last names Explain why the inverse of f does not exist SOLUTION Many different teenagers will have the same last name so this function will not be onetoone Conse quently the function does not have an inverse 4 View the following fragment of a train schedule for the New Jersey Transit System as de ning a function f from towns to times ls f onetoone What is f 627 Trenton 6 2 l Hamilton Township 627 Princeton Junction 634 New Brunswick 6 38 SOLUTION This function is onetoone and f 1 627 Hamilton Township 5 A homework problem asks for a sketch of the graph of the inverse of f x x cosx Frank after trying but failing to nd a formula for f 1x says it s impossible to graph the inverse Bianca hands in an accurate sketch without solving for f 1 How did Bianca complete the problem SOLUTION The graph of the inverse function is the re ection of the graph of y fx through the line y x 6 Which of the following quantities is unde ned a sin 17 b cos 12 c csc l g d csc l 2 SOLUTION b and c are unde ned sin 1 7 and csc 12 g 7 Give an example of an angle 9 such that cos 1cos 9 9 Does this contradict the de nition of inverse function SOLUTION Any angle 9 lt O or 9 gt n will work No this does not contradict the de nition of inverse function Exercises 1 Show that f x 7x 7 4 is invertible by nding its inverse 4 4 SOLUTION Solving y 7x 7 4 for 1 yields 1 Thus f 1x 2 Is fx X2 2 onetoone lfnot describe a domain on which it is onetoone SOLUTION f is not onetoone because f7l fl 3 However if the domain is restricted to x 2 O or x 5 0 then f is onetoone 3 What is the largest interval containing zero on which f x sin is onetoone SOLUTION Looking at the graph of sinx the function is onetoone on the interval 7n2 712 x72 4 Show that f x is invertible by nding its inverse x 3 a What is the domain of fx The range of f 1x b What is the domain of f 1x The range of fx SOLUTION We solve y fx for as follows 7 x 7 2 7 x 3 yx 3y x 7 2 y yx7x73y72 3 7 x Y2 y7l Therefore 3x2 f 10 T a Domain offx xix 73 Range off 1x b Domain off 1x xix 73 Range offx 5 Verify that fx X3 3 and gx x 7 3 3 are inverses by showing that fgx x andgfx x 42 CHAPTER 1 I PRECALCULUS REVIEW SOLUTION 3 fltgltxgtgtltx73gt13 3x733x 80 x3 37 313 X313 X t 1 t 1 6 Repeat ExerciseSfor ft t1 and g0 SOLUTION t1 1 tlt7l t71 t t fg 371 t1iltt7h The calculations for gft are identical ZGM 7 The escape velocity from a planet of mass M andradius R is vR T Where G is the universal gravitational constant Find the inverse of vR as a function of R SOLUTION To nd the inverse we solve 7F for R This yields R 7 ZGM yz Therefore 20M 71 7 v R 7 In Exercises 8715 nd a domain on which f is onetoone and aformulafor the inverse off restricted to this domain Sketch the graphs of f and f 1 8 f x 3x 7 2 SOLUTION The linear function fx 3x 7 2 is onetoone for all real numbers Solving y 3x 7 2 for 1 gives 1 y 23 Thus 1 2 lm 3 9 fx 4 7 x SOLUTION The linear function fx 4 7 x is onetoone for all real numbers Solving y x 7 4 for 1 gives 1 4 7 yThusf 1x 47x fXfquotX4X SECT1ON 15 I Inverse Functions 43 1 10 for X SOLUTION The graph of f x 1X 1 given below shows that f passes the horizontal line test and is therefore 1 1 1 onetoone on its entire domain 1 x y 71 Solving y 1 for givesx 7 1 Thus f 1x 71 x y x y 1 1 yf391x i yfX 7 2 4 3 2 W 1 1 1 74 7 1 1 11 for 7X 7 3 SOLUTION The graph of f x 17x 7 3 given below shows that f passes the horizontal line test and is therefore onetoone on its entire domain 1 x g Solving y 17x 7 3 for gives 1 3 3 1 x thus f 1x i 7x 7y7 7 1 12 fs S Z SOLUTION To make fs s z onetoone we must restrict the domain to either 5 s gt O or s s lt 0 If we 1 1 1 choose the domain 5 s gt 0 then solving y for s yields s Hence f 1s Had we chosen the s2 5 1 domain 5 s lt 0 the inverse would have been f 1s 77 s 13 for m SOLUTION To make the function fx onetoone we must restrict the domain to either 1 x 3 0 or 1 x21 x x 5 0 If we choose the domain 1 x 3 0 then solving y for 1 yields 1 x2 1 44 CHAPTER 1 I PRECALCULUS REVIEW 14 fz z3 SOLUTION The function f z Z3 is onetoone over its entire domain see the graph below Solving y Z3 for Z yields yl3 z Thus f 1z Z13 15 fx x3 9 SOLUTION The graph of f x V163 9 given below shows that f passes the horizontal line test and therefore is onetoone on its entire domain 1 x 3 7913 Solving y X3 9 for 1 yields 1 y2 7 913 Thus f 1xx 7 91 E FIGURE 19 SOLUTION Here we apply the rule that the graph of f71 is obtainedby re ecting the graph of f across the line y x For C and D we must restrict the domain of f to make f onetoone a y b y c y SECTlON 15 I Inverse Functions 45 d V e V t y D E F 17 Which of the graphs in Figure 20 is the graph ofa function satisfying f 1 f 74 A 7 x W X C FIGURE 20 Y SOLUTION Figures B and C would not change when re ected around the line y 1 Therefore these two satisfy f 1 7 f 18 Let n be a nonzero integer Find a domain on which f x 1 7 x 1 coincides with its inverse Hint The answer depends on whether n is even or odd SOLUTION First note 5 n1nn1 5 nIn5 711715 ffX711X 711X 7X 716 so f x coincides with its inverse For the domain and range of f let s rst consider the case when n gt 0 If n is even then fx is de ned only when 1 7 x 3 0 Hence the domain is 71 5 x 5 1 The range is O 5 y 5 1 lfn is odd then fx is de ned for all real numbers and the range is also all real numbers Now suppose n lt 0 Then 7n gt O and 1 71771 X771 1771 to lfn is even then fx is de ned only whenx 7 1 gt 0 Hence the domain is 111 gt 1 The range is y gt 1 1er is odd then f x is de ned for all real numbers exceth 1 The range is all real numbers except y 1 19 Letfx x7x 1 a Show that f 1 exists but do not attempt to nd it Hint Show that f is strictly increasing b What is the domain of f l c Find f 13 SOLUTION a The graph of f x X7 X 1 is shown below From this graph we see that f x is a strictly increasing function by Example 3 it is therefore onetoone Because f is onetoone by Theorem 3 f exists b The domain off 1x is the range of fx 700 00 c Note that f1 17 11 3 therefore f 13 1 20 Show that fx X2 1 1 is onetoone on 700 O and nda formula for f 1 for this domain 46 CHAPTER 1 I PRECALCULUS REVIEW SOLUTION Notice that the graph of fx X2 1 1 over the interval 700 0 shown above passes the horizontal line test Thus fx is onetoone on 700 0 To nd a formula for f l we solve y x2 1 1 for x which yields 1 i 7 1 Because the domain of f was restricted to x 5 O we choose the negative sign in front of the radical Therefore f 1x 7 21 Let fx x2 7 2 Determine a domain on which f 1 exists and nd a formula for f 1 on this domain SOLUTION From the graph of y x2 7 2 shown below we see that if the domain of f is restricted to either 1 5 1 or x 3 1 then f is onetoone and f 1 exists To nd a formula for f 1 we solve y x2 7 2 for x as follows y1x272x1x712 x 7 1 i y 1 1 1 i y 1 If the domain of f is restricted to x 5 1 then we choose the negative sign in front of the radical and f 1x 1 7 X 1 If the domain of f is restricted to x 3 1 we choose the positive sign in front of the radical and f 1x 1 X1 22 Show that fx x x 1 is onetoone on 1 00 and nd a formula for f 1 on this domain What is the domain of f l SOLUTION The graph of fx X x 1 on 1 00 is shown below From this graph we see that for x gt 1 the function is increasing which implies that the function is onetoone Also note that since f is increasing for x gt 1 fx 3 f1 2 for gt1 123456739 To nda formula for f 11et y X x 1 Then xy X2 1or1c2 7 xy 1 0 Using the quadratic formula we nd x 7 y i y2 7 4 7 2 To have x 3 1 for y 3 2 we must choose the positive sign in front of the radical Thus 2 7 f4 JHZ X for 3 2 S E C T T O N 15 I Inverse Functions 47 In Exercises 23728 evaluate without using a calculator 23 cos 1 1 SOLUTION cos ll 0 v 71 1 24 Sin 7 SOLUTION sin 1 25 cot ll SOLUTION cor1 1 g 26 sec 1 1 SOLUTION sec 1 g 27 tan 1 J5 SOLUTION tanil J5 tan 1 28 sin 171 SOLUTION Sin 171 7 In Exercises 29738 compute without using a calculator n 29 sin 1 sin 3 SOLUTION sin 1sin 4a 30 sin 1 sin SOLUTION sin 1sin 43 sin 17 7 The answer is not 43 because 43 is not in the range of the inverse sine function 31 cos 1 cos 2 SOLUTION cos 1 cos 37 cos 10 The answer is not 37 because 37 is not in the range of the inverse cosine function 32 sin 1 sin 6 SOLUTION sin 1sin75 sin 17 7 The answer is not 756 because is is not in the range of the inverse sine function 37 33 tan 1 tan SOLUTION tan 1tan 34 tan 171 7 The answer is not 37quot because 37 is not in the range of the inverse tangent function 34 tan 1tan a SOLUTION tan 1tan a tan 10 O The answer is not it because a is not in the range of the inverse tangent function 35 sec 1sec37r SOLUTION sec 1 sec 37 sec 1 71 tr The answer is not 37 because 37 is not in the range of the inverse secant function 37 36 sec 1 ltsec gt 2 SOLUTION No inverse since sec3T r 13 H 00 005 T 37 csc 1 csc7 71 SOLUTION No inverse since csc7n a oo 38 cor1 cot7 48 CHAPTER 1 I PRECALCULUS REVIEW SOLUTION cot 1 cot cot1 l The answer is not because is not in the range of the inverse cotangent function In Exercises 39 42 simplify by referring to the appropriate triangle or trigonometric identity 39 tancos1 x SOLUTION Let 9 cos 1 x Then cos 9 x and we generate the triangle shown below From the triangle 1 Vl X2 xtan9 x l all x2 6 x SOLUTION Let 9 tan 1 x Then tan 9 x and we generate the triangle shown below From the triangle tancos 40 costan1 x l costan1x cos 9 x2 1 lx2 41 cotsec1 x SOLUTION Let 9 2 sec 1 x Then sec 9 x and we generate the triangle shown below From the triangle cotsec1x cot 9 x l x lxz l 6 l SOLUTION Let 9 sin 1 x Then sin 9 x and we generate the triangle shown below From the triangle xl x2 x 42 cotsin1 x cotsin1 x cot 9 In Exercises 43 50 refer to the appropriate triangle or trigonometric identity to compute the given value 1 2 43 cos sm 3 SOLUTION Let 9 sin 1 Then sin 9 and we generate the triangle shown below From the triangle 12 J3 cos s1n cos9 3 3 SECTION 15 I Inverse Functions 49 44 tancos 1 SOLUTION Let 9 cos 1 Then cos 9 and we generate the triangle shown below From the triangle tan cos 1g tan9 3 239 45 tansin 1 08 SOLUTION Let 9 sin 1 08 Then sin 9 08 g and we generate the triangle shown below From the triangle 1 4 tans1n 08 tan 9 46 cos cot 1 l SOLUTION cot 1 l Hence coscot1 l cos 2 Na 47 cotcsc1 2 SOLUTION csc 1 2 Hence cotcsc1 2 cotg oxla 48 tansec1 2 SOLUTION sec1 2 Hence tansec1 2 tan 23 2 49 cottan1 20 SOLUTION Let 9 tan 1 20 Then tan 9 20 so cottan1 20 cot 9 2 tall 9 50 sincsc120 SOLUTION Let 9 csc 1 20 Then csc 9 20 so sincsc1 20 sin 9 2 CS 9 Further Insights and Challenges 51 Show that if f x is odd and f1x exists then f1x is odd Show on the other hand that an even function does not have an inverse SOLUTION Suppose fx is odd and f1x exists Because fx is odd f x fx Let y f1x then fy x Since fx is odd f y fy x Thus f 1 x y f 1x Hence f 1 is odd On the other hand if f x is even then f x f x Hence f is not onetoone and f 1 does not exist 52 A cylindrical tank of radius R and length L lying horizontally as in Figure 21 is lled with oil to height it Show that the volume Vh of oil in the tank as a function of height h is Wk 2 L R2 cos 1 1 R h2hR I12 S E C T10 N 16 I Exponential and Logarithmic Functions 51 5 Explain the phrase the logarithm converts multiplication into addition SOLUTION This phrase is a verbal description of the general property of logarithms that states logab loga logb 6 What are the domain and range of lnx SOLUTION The domain of lnx is x gt O and the range is all real numbers 7 Which hyperbolic functions take on only positive values SOLUTION coshx and sechx take on only positive values 8 Which hyperbolic functions are increasing on their domains SOLUTION sinhx and tanhx are increasing on their domains 9 Describe three properties of hyperbolic functions that have trigonometric analogs SOLUTION Hyperbolic functions have the following analogs with trigonometric functions parity identities and deriva tive formulas Exercises 1 Rewrite as a whole number without using a calculator a 70 b 10222 5 2 W C 453 9 8 13 853 t 34147122 32 d 2743 SOLUTION a 70 1 b 1022 2 5 2 10014 125 25 4 29 c mos4513 415415 1 d 27W3 27134 34 81 9 8 13 853 8135813 252 24 16 t 34147122 32 321273222 32 0 In Exercises 2710 solve for the unknown variable 2 92x 98 SOLUTION 119 93 then 2 8 and 4 3 82x ex1 SOLUTION Ifezquot ex1then2x x 1 and 1 4 er2 e4t73 SOLUTION Iret2 e4t 3 thentz 417 3 ort2 7 41 3 17 3171 0 Thus 1 1 or 1 3 5 3X 7 er SOLUTION Rewrite X1 as 3 1X1 3 X 1 Then 3C 3 X 1which requiresx 7 71Thusx 712 6 y 125 SOLUTION Rewrite n as SI2C Sic2 and 125 as 53 Then Sic2 53 so xZ 3 and 6 7 47x 2x1 SOLUTION Rewrite 4 as 22 X 2 Then 2 2C 2X which requires 72 X 1 Solving for 1 gives 1 71 3 s b4 1012 SOLUTION b4 1012 is equivalent to b4 1034 so b 103 Alternately raise both sides of the equation to the onefourth power This gives b 1012 4 103 9 k32 27 SOLUTION Raise both sides of the equation to the twothirds power This gives k 2723 27132 32 9 10 b2xl b76 52 CHAPTER 1 I PRECALCULUS REVIEW SOLUTION Rewrite b2x1 as hm Then 2x 1 76 and 74 In Exercises 11722 calculate directly without using a calculator 11 log 27 SOLUTION log 27 log 33 3log3 3 3 12 logs 1 SOLUTION log5 E log 5 2 7210g5 5 72 13 log2253 5 5 SOLUTION log2 253 glogzz 5 14 log2853 5 SOLUTION log2853 glogz 23 Slog22 5 15 log644 1 1 SOLUTION log644 log64 6413 5 log64 64 5 16 log7492 SOLUTION log7 492 210g7 72 2 2 log7 7 4 17 log 2 log4 2 5 1 1 SOLUTION lOggZ log42 lOgg 813 log4 412 5 E g 18 log25 30 log25 g 5 5 SOLUTION log25 30 log25 g log25 lt30 log25 25 1 19 log 48 7log412 48 SOLUTION log4 48 7log412 log4 E log44 1 20 1n e75 19 SOLUTION 1nEe75 1ne12 e75 Ine1275 1ne1910 E 21 1ne3 1ne4 SOLUTION 1ne3 1ne4 3 4 7 22 log2 g log2 24 4 4 SOLUTION log2 g log2 24 log2 lt 24 log2 32 log2 25 Slog22 5 3 23 Write as the natural log of a single expression a 21n531n4 b 51nX121n9x SOLUTION a 21n5 31n4 Ins2 1n43 ln25 1n64 1n25 64 1n1600 b 51nx12 1n 9x Inx52 1n 9x 1HX52 9x 1n9x72 24 Solve forx 1nc2 1 7 31nx 1n2 SOLUTION Combining terms on the lefthand side gives 12 1 X3 1nx2 1 e 3lnx 1nx2 17lnx3 1n X2 1 2 or 213 7 x2 7 1 O x 1 is the only real root to this equation Substituting 1 1 into the original equation we nd Therefore 1n2731n11n2701n2 as needed Hence 1 1 is the only solution S E C T10 N 16 I Exponential and Logarithmic Functions 53 In Exercises 25730 solve for the unknown 25 7e5t 100 SOLUTION Divide the equation by 7 and then take the natural logarithm of both sides This gives 5 71 100 till 100 in T Of 73H T l 26 6e 4 2 SOLUTION Divide the equation by 6 and then take the natural logarithm of both sides This gives 74tlnl or IE 3 4 27 t 8 SOLUTION Since 8 23we have2 7 2x 7 3 O or x 7 3X 1 O Thusx 71 or 3 28 82tl 9817t SOLUTION Taking the natural logarithm of both sides of the equation gives 211 1n9e1 1n91ne1 t ln91 7t Thus 3t 1n9 ort ln9 29 1nx4 7lnx2 2 4 SOLUTION lnx4 7lnx2 ln lnx2 21nx Thus21nx 2 or lnx 1 Hence 1 e x 30 log y 3log3y214 SOLUTION 14 log y 3log3y2 log y log y6 log3 y7 Thus y7 314 or y 32 9 31 Use a calculator to compute sinhx and coshx for x 73 O 5 SOLUTION x 73 O 5 L 7x 73 7 3 0 7 0 5 7 75 sinhx e e l 7100179 8 e 0 L 74203 2 2 2 2 x 7x 7 3 5 7 coshx i 100677 8 1 8 M 74210 2 2 32 Compute sinhln5 and tanh3ln 5 without using a calculator SOLUTION 1quot57 1quot5 5715 245 Sinhans L 125 2 2 2 31 5 731 5 sinh3 1n5 M 53 7153 56 71 tauh3ln5 cosh3ln5 quot5 quot5 53 153 56 1 33 For which values of x are y sinhx and y coshx increasing and decreasing SOLUTION The graph of y sinhx is shown below on the left From this graph we see that sinhx is increasing for all 1 On the other hand from the graph of y coshx shown below on the right we see that coshx is decreasing for x lt O and is increasing for gt O 25 ycoshx 54 CHAPTER 1 I PRECALCULUS REVIEW 34 Show that y tanhx is an odd function 7x 7 77x 7x 7 x x 7 7x SOLUTION tanh7x L L 7 7tanhx e x e PX e x ex ex e 35 The population ofa city in millions at time t years is Pt 24800 where t 0 is the year 2000 When will the population double from its size att 0 l 2 SOLUTION Population doubles when 48 2483906t Thus 006 ln2 or t N 1155 years 36 According to the Gutenberg7Richter law the number N of earthquakes worldwide of Richter magnitude M ap proximately satis es a relation ln N 1617 7 bM for some constant b Find b assuming that there are 800 earthquakes of magnitude M 5 per year How many earthquakes of magnitude M 7 occur per year SOLUTION Substituting N 800 and M 5 into the Gutenberg7Richter law and solving for b yields 1617 7 In 800 b N 1897 5 The number N of earthquakes of Richter magnitude M 7 then satis es lnN 1617 718977 2891 Finally N 8239891 N 18 earthquakes 37 The energy E in joules radiated as seismic waves from an earthquake of Richter magnitude M is given by the formula loglo E 48 15M 21 Express E as a function of M b Show that when M increases by 1 the energy increases by a factor of approximately 31 SOLUTION 2 Solving loglo E 48 15M for E yields E 104815M b Using the formula from part a we nd EltM 1 104815M1 1063l5M 15N EM 104815M 7104815M 10 31 6228 38 Refer to the graphs to explain why the equation sinhx t has a unique solution for every t and cosh x t has two solutions for every t gt 1 SOLUTION From its graph we see that sinhx is a onetoone function with lim sinhx 700 and lim sinhx x7700 X700 00 Thus for every real number I the equation sinhx t has a unique solution On the other hand from its graph we see that coshx is not onetoone Rather it is an even function with a minimum value of cosh0 1 Thus for every t gt 1 the equation coshx t has two solutions one positive the other negative 39 Compute coshx and tanhx assuming that sinhx 08 SOLUTION Using the identity coshzx 7 sinhzx 1it follows that coshzx 7 g2 1 so that coshzx and v41 coshx 5 Then by de nition 4 1nhx g 4 tanhx hx J47 A41 T 40 Prove the addition formula for coshx SOLUTION Hy 7xy 2 My 2 70H coshx 7y L L 2 4 ex 87X eX y e O ex 7 e x 7 eX y e O 4 4 lt gtlt7gtlt gt coshx cosh y sinhx sinh y S E C T10 N 16 I Exponential and Logarithmic Functions 55 41 Use the addition formulas to prove sinh2x 2 coshx sinhx cosh2x coshzx sinhzx SOLUTION sinh2x sinhx x sinhx coshx coshx sinhx 2 coshx sinhx and cosh2x coshx x coshx coshx sinhx sinhx coshzx sinhzx 42 An imaginary train moves along a track at velocity v and a woman walks down the aisle of the train with velocity u in the direction of the train s motion Compute the velocity 11 of the woman relative to the ground using the laws of both Galileo and Einstein in the following cases a v 100 mph and u 5 mph Is your calculator accurate enough to detect the difference between the two laws b v 140000 mph and u 100 mph SOLUTION a By Galileo s law 1000 50 1050 mph The speed of light is c 186000 milessec 669600000 mph Using Einstein s law and a calculator 1 000 50 tanhil 3 tanhil tanhil 1568100358 x 10quot c 669600000 669600000 so 11 1050 mph No the calculator was not accurate enough to detect the difference between the two laws b By Galileo s law 100000 50 100050 milessec By Einstein s law 100 000 tanhil 3 tanhil 4 tanhil 0601091074 6 86000 186000 so 11 N 100 0355 milessec Further Insights and Challenges 43 Show that log b logba 1 SOLUTION 1 bilnb dl 71M Th 1 b 1 lm In 7 0g 71nd an 0gba7 lnb us 0g Ogba7 lna lnb 7 log 1 44 Verify the formula log 1 for a b gt 0 oga b log 1 SOLUTION Let y logbx Then x by andloga x log by yloga b Thus y 1 b 0 45 2 Use the addition formulas for sinhx and coshx to prove tanh u tanh v 1 tanh u tanh u b Use a to show that Einstein s Law of Velocity Addition Eq 3 is equivalent to tanhu v uv 7 no 1 62 SOLUTION a 3mm v sinhu v sinhu coshv cosh u sinhv coshu v cosh u cosh v s1nh u s1nh v 7 sinhu cosh v coshu sinh v 1cosh u coshv 7 tanhu tanhv 7 coshu cosh v sinhu sinh v 1cosh u coshv 7 1 tanh u tanh v b Einstein s law states tanh 1wc tanh 1uc tanh 1vc Thus 71 71 3 tanhtam1uCtanh1vC tanhtanh vc tanhtanh uc c 1 tanhtanh 1vc tanhtanh 1uc 1cuv 1 H BI Hence u 1 le 9 56 CHAPTER 1 I PRECALCULUS REVIEW 46 Prove that every function f x is the sum of an even function fx and an odd function fx Hint39fix fx i f7x Express fx 5eC 8e in terms of coshx and sinhx SOLUTION Let fx XHZ x and fioc X 7XThen f f 2 fx Moreover mix flt7xgt 2flt7lt7xgtgt flt7xgt2 fltxgt mm so fx is an even function while 117 fX 2fx flt7xgt27 for 7fx 72flt7xgtgt if so fx is an odd function For fx 5eC 8e we have fx w 8coshx 5coshx 13coshx and fx w 5sinhx 7 8sinhx 73mm Therefore fx fx fx 13 coshx 7 3sinhx 17 Technology Calculators and Computers Preliminary Questions 1 Is there a de nite way of choosing the optimal viewing rectangle or is it best to experiment until you nd a viewing rectangle appropriate to the problem at hand SOLUTION It is best to experiment with the window size until one is found that is appropriate for the problem at hand 2 Describe the calculator screen produced when the function y 3 x2 is plotted with viewing rectangle a 111X1021 b 011X104l SOLUTION 2 Using the viewing rectangle 71 1 by 0 2 the screen will display nothing as the minimum value of y 3 x2 b Using the viewing rectangle 0 1 by 0 4 the screen will display the portion of the parabola between the points 0 3 and 14 3 According to the evidence in Example 4 it appears that fn 1 1n never takes on a value greater than 3 for n 3 0 Does this evidence prove that fn 5 3 forn 3 0 SOLUTION No this evidence does not constitute a proof that fn 5 3 for n 3 0 4 How can a graphing calculator be used to nd the minimum value of a function SOLUTION Experiment with the viewing window to zoom in on the lowest point on the graph of the function The ycoordinate of the lowest point on the graph is the minimum value of the function Exercises The exercises in this section should be done using a graphing calculator or computer algebra system 1 Plot fx 2x4 3x3 7 142 7 9x 18 in the appropriate viewing rectangles and determine its roots SOLUTION Using a viewing rectangle of 74 3 by 720 20 we obtain the plot below S ECT l O N 17 I Technology Calculators and Computers 57 Now the roots of f x are the xintercepts of the graph of y f x From the plot we can identify the xintercepts as 73 715 1 and2 The roots offx are therefore 73 x 715 1 and 2 2 How many solutions does 13 7 4X 8 0 have SOLUTION Solutions to the equation x3 7 4X 8 O are the xintercepts of the graph of y x3 7 4X 8 From the gure below we see that the graph has one xintercept between x 74 and x 72 so the equation has one solution Y 3 How many positive solutions does 13 7 12 8 0 have SOLUTION The graph of y x3 7 12 8 shown below has two xintercepts to the right of the origin therefore the equation x3 7 12 8 O has two positive solutions 4 Does cosx x 0 have a solution Does it have a positive solution SOLUTION The graph of y cosx 1 shown below has one xintercept therefore the equation cosx x O has one solution The lone xintercept is to the left of the origin so the equation has no positive solutions 5 Find all the solutions of sin forx gt O SOLUTION Solutions to the equation sinx correspond to points of intersection between the graphs of y sin and y J The two graphs are shown below the only point of intersection is at x 0 Therefore there are no solutions ofsinx forx gt 6 How many solutions does cosx 12 have SOLUTION Solutions to the equation cosx 12 correspond to points of intersection between the graphs of y cosx and y 12 The two graphs are shown below there are two points of intersection Thus the equation cosx x2 has two solutions CHAPTE R l I PRECALCULUS REVIEW 7 Let fx x 7 1002 1000 What will the display show if you graph fx in the viewing rectangle 710 10 by 710 710 What would be an appropriate viewing rectangle SOLUTION Because 1 7 1002 3 0 for allx it follows that fx x 7 1002 1000 3 1000 for allx Thus using a viewing rectangle of 710 10 by 710 10 will display nothing The minimum value of the function occurs when by 1000 2000 58 x 100 so an appropriate viewing rectangle wouldbe 50 150 8 1 8 in an appropriate v1ew1ng rectangle What are the vertical and horizontal asymp x 8 Plot the graph of fx totes see Example 3 1 shown below we see that the vertical asymptote is x and the horizontal 8x SOLUTION From the graph ofy 8X 4 asymptote is y 1 x 9 Plot the graph of f x 4 X in a viewing rectangle that clearly displays the vertical and horizontal asymptotes SOLUTION From the graph of y 4 X shown below we see that the vertical asymptote is x 4 and the horizontal 7 x asymptote is y 71 10 Illustrate local linearity for fx x2 by zooming in on the graph atx 05 see Example 6 SOLUTION The following three graphs display fx 12 over the intervals 71 3 03 07 and 045 055 The nal graph looks like a straight line y y y 3 05 03 04 39 6 03 025 4 02 02 2 01 015 0 035 045 055 065 046 043 05 052 054 11 Plot fx cosx2 sin for 0 5 x 5 271 Then illustrate local linearity at x 38 by choosing appropriate viewing rectangles SOLUTION The following three graphs display fx cosx2 sinx over the intervals 0271 35 41 and 375 385 The nal graph looks like a straight line Y Y 04 02 x 36 33 39 4 76 373 33 332 334 X 12 A bank pays r 5 interest compounded monthly If you deposit P0 dollars at time t 0 then the value of your gt Find to the nearest integer N the number of months after which the account account after N months is P0 1 value doubles S ECT l O N 17 I Technology Calculators and Computers 59 SOLUTION PN P01 Oi gSW This doubles when PN 2P0 or when 2 1 N The graphs of y 2 and y 1 N are shown below they appear to intersect at N 167 Thus it will take approximately 167 months for money earning r 5 interest compounded monthly to double in value gt t t t ti 1 0 155 160 165 170 175 X In Exercises 13718 investigate the behavior of the function as n or x grows large by making a table of function values and plotting a graph see Example 4 Describe the behavior in words 13 fn 711quot SOLUTION The table and graphs below suggest that as n gets large nu approaches 1 n nln 10 1258925412 102 1047128548 103 1006931669 104 1000921458 105 1000115136 106 1000013816 y y 1 1 0 2 46 310 0 2004006008001000 4n 1 14 n f 6n 7 5 4n 1 2 SOLUTION The table and graphs below suggest that as n gets large 6 5 approaches n L n 4n 1 6n 7 5 10 7454545455 106 6666673889 N1 N1 0 2 4 6 3 10 0 20 40 60 30 100 n2 1s fn 1 SOLUTION The table and graphs below suggest that as n gets large fn tends toward 00 60 CHAPTER 1 I PRECALCULUS REVIEW n2 n 1 n 1378061234 1635828711 x 1043 1195306603 x 10434 5341783312 x 104342 1702333054 x 1043429 106 1839738749 x 10434294 HHHHH O 10000 1 gtlt 1043 0 246310 0 20406030100 6 X 16 fx Ct SOLUTION The table and graphs below suggest that as 1 gets large f x roughly tends toward 22026 x 6 X x x 7 4 1818391210 102 2011236934 H O 103 2180933633 104 2200443568 105 2202426311 106 2202536451 107 2202635566 y V 21000 22000 20000 21000 19000 190000000 13000 13000 17000 17000 0 50 100 150 200 0 200 400 600 300 1000 17 fx 1 tan 16 x SOLUTION The table and graphs below suggest that as 1 gets large f x approaches 1 1 x X tau gt X X 10 1033975759 102 1003338973 103 1000333389 104 1000033334 105 1000003333 106 1000000333 Y Y 15 15 14 14 13 13 12 12 11 11 1 1 r t 5 10 15 20 20 40 60 30 100 S ECT 1 O N 17 I Technology Calculators and Computers 61 1 X2 18 fx 1 tan gt x SOLUTION The table and graphs below suggest that as 1 gets large fx approaches 139561 1 2 X 13H gt X 1396701388 H HHHHHH OOOOOO ourpr HHHHH quotLn39mmmm xoxoxoxoxo LIILIILIILIILII 00000 HHHHN mmmmm JgtJgtJgtan mmmmoo Lamaho 152 152 143 143 144 144 14 14 0 20 40 60 30 100 19 The graph of f9 A cos 9 B sin 9 is a sinusoidal wave for any constants A and B Con rm this for A B 11 12and34byplom g f9 SOLUTION The graphs of f9 cos 9 sin 9 f9 cos 9 2 sin9 and f9 3 cos 9 4sin 9 are shown e ow y y y 2 4 1 A I 2 7 245 3 R2 2V 3 jil 246 3 1 2 4 A3 L 1 A B L 2 AB 34 20 Find the maximum value of f 9 for the graphs produced in Exercise 19 Can you guess the formula for the maxi mum value in terms of A and B SOLUTION ForA 1 andB 1max14 w J ForA1andB 2max225 ForA3andB 4max53242 Max A2 32 21 Find the intervals on which fx XX 2x 7 3 is positive by plotting a graph SOLUTION The function fx xx 2x 7 3 is positive when the graph of y xx 2x 7 3 lies above the xaxis The graph of y xx 2x 7 3 is shown below Clearly the graph lies above the xaxis and the function is positive for E 72 O U 3 00 22 Find the set of solutions to the inequality 12 7 4c2 7 1 lt O by plotting a graph SOLUTION To solve the inequality 12 7 4c2 7 1 lt O we can plot the graph ofy x2 7 4 x2 7 1 and identify when the graph lies below the xaxis The graph of y x2 7 4c2 7 1 is shown below The solution set for the inequality 12 7 4x2 71 lt Ois clearlyx e 72 71 u 1 2 62 CHAPTER 1 I PRECALCULUS REVIEW Further Insights and Challenges 23 FIE Let f1x x and de ne a sequence of functions by fn1x fx xfnx For example f2x x 1 Use a computer algebra system to compute fnx for n 3 4 5 and plot fnx together with J for x 3 0 What do you notice SOLUTION With f1x x and f2x x l we calculate 2 f3xltxl x 7 6xl xl 7 4X1 1 x26xl x x428x370x228x1 x 4 2 4x1 T1 8lxl6xx2 and 1 120 1820x2 8008x3 12870x4 8008x5 18206 120x7 x8 161 xl 6xx2128x70x228x3 x4 f5X A plot of f1x f2x f3x f4x f5 1 and is shown below with the graph of J shown as a dashed curve It seems as if the f are asymptotic to 20406030100 24 Set P0x l and P1x x The Chebyshev polynomials useful in approximation theory are de ned succes sively by the formula Pn1x ZXPnoC 7 Pn1x a Show that P2x 2x2 7 1 b Compute Pnx for 3 5 n 5 6 using a computer algebra system or by hand and plot each P 1 over the interval 71 l c Check that your plots con rm two interesting properties of Chebyshev polynomials a Pnx has n real roots in 71 l and b for e 71 l Pnx lies between 71 and l SOLUTION a With P0x l and P1x x we calculate P2x 2xP1x 7 P0x 2xx 71 2x2 711 b Using the formula Pn1x ZXPnOc 7 Pn1x with n 2 3 4 and 5 we nd P3x 2x2x2 7 1 7 x 4x3 7 3x P4x 2x4x3 7 310722 71 8x4 7 812 1 P5X16X5 7 2013 5x P6x 32x6 7 4814 18x2 71 The graphs of the functions Pnx for O 5 rt 5 6 are shown below Chapter Review Exercises 63 P20 P30 FAX c From the graphs shown above it is clear that for each n the polynomial PnOC has precisely n roots on the interval 71 1 andthat 71 5 PnX 1fOIX e 711 CHAPTER REVIEW EXERCISES 1 Express 4 10 as a set X 1X 7 at lt c for suitablea and c SOLUTION The center of the interval 4 10 is 41 210 7 and the radius is 3 Therefore the interval 4 10 is equivalent to the set X 1X 7 71 lt 3 2 Express as an interval aX1X751lt4 bX15X3152 SOLUTION 2 Upon dropping the absolute value the inequality 1X 7 51 lt 4 becomes 74 lt X 7 5 lt 4 or 1 lt X lt 9 The set X 1X 7 51 lt 4 can therefore be expressed as the interval 1 9 b Upon dropping the absolute value the inequality 15X 31 5 2 becomes 72 5 5X 3 5 2 or 71 5 X 5 7 The set X 15X 31 5 2 can therefore be expressed as the interval 71 7 3 Express X 2 5 1X 7 11 5 6 as a union of two intervals SOLUTION The set X 2 5 1X 7 11 5 6 consists of those numbers that are at least 2 but at most 6 units from 1 The numbers larger than 1 that satisfy these conditions are 3 5 X 5 7 While the numbers smaller than 1 that satisfy these conditions are 75 5 X 5 71 Therefore X 2 51X 7115 6 75 71 U 37 4 Give an example ofnumbers X y such that 1X1 lyl X 7 y SOLUTION LetX 3 andy 71Then1X1lyl 31 4 andX 7 y 3 7 71 4 5 Describe the pairs of numbers X y such that 1X yl X 7 y SOLUTION First consider the case when X y 3 0 Then 1X yl X y and we obtain the equation X y X 7 y The solution of this equation is y 0 Thus the pairs X 0 With X 3 O satisfy 1X yl X 7 y Next consider the case when X y lt 0 Then 1X yl 7X y 7 and we obtain the equation 7X 7 y X 7 y The solution of this equation is X 0 Thus the pairs 0 y With y lt 0 also satisfy 1X yl X 7 y 6 Sketch the graph ofy fX 271WherefXX2 for 72 5 X 5 2 64 CHAPTER 1 I PRECALCULUS REVIEW SOLUTION The graph of y fx 2 7 l is obtained by shifting the graph of y fx two units to the left and one unit down In the gure below the graph of y f x is shown as the dashed curve and the graph of y f x 2 7 l is shown as the solid curve In Exercises 7710 let f x be the metion whose graph is shown in Figure I FIGURE 1 7 Sketch the graphs ofy fx 2 and y fx 2 SOLUTION The graph of y fx 2 is obtained by shifting the graph of y fx up 2 units see the graph below at the left The graph of y fx 2 is obtained by shifting the graph of y fx to the left 2 units see the graph below at the right fX 2 fx 2 8 Sketch the graphs ofy fx and y fx SOLUTION The graph of y fx is obtained by compressing the graph of y fx vertically by a factor of2 see the graph below at the left The graph of y f x is obtained by stretching the graph of y f x horizontally be a factor of 2 see the graph below at the right fX2 f XZ 9 Continue the graph of fx to the interval 74 4 as an even function SOLUTION To continue the graph of f x to the interval 74 4 as an even function re ect the graph of f 1 across the yaxis see the graph below 74737271 1234 Chapter Review Exercises 65 10 Continue the graph of f x to the interval 74 4 as an odd function SOLUTION To continue the graph of fx to the interval 74 4 as an odd function re ect the graph of fx through the origin see the graph below In Exercises 11714 nd the domain and range of the function 11 fx Ax 1 SOLUTION The domain of the function fx NC 1 is x x 3 71 and the range is y y 3 0 4 12 x 14 1 4 SOLUTION The domain of the function fx 4 1 is the set of all real numbers and the range is y O lt y 5 4 x 13 f 2 x 3 7 x SOLUTION The domain of the function fx 14 fx xx27x5 SOLUTION Because 2 37X ispczx 3andtherangeisyy7 0 1 1 12 19 x27x5 X2X 57 xi 4 4 x27x53uforallxltfollows thatthe domain ofthe function fxV1627X5isallrealnumbersandthe range is y y 3 192 15 Determine whether the function is increasing decreasing or neither 1 a N 3 b N m c gm t2 t d gm P H SOLUTION a The function fx 3 can be rewritten as fx a This is an exponential function with a base less than 1 therefore this is a decreasing function b From the graph of y 1 Jc2 1 shown below we see that this function is neither increasing nor decreasing for all 1 though it is increasing for x lt O and decreasing for x gt O c The graph of y t2 t is an upward opening parabola therefore this function is neither increasing nor decreasing for all I By completing the square we nd y t 2 7 i The vertex of this parabola is then at t 7 so the function is decreasing fort lt 7 and increasing for t gt 7 d From the graph of y t3 t shown below we see that this is an increasing function 66 CHAPTER 1 I PRECALCULUS REVIEW 16 Determine whether the function is even odd or neither 2 fx x4 7 312 b gx sinx 1 c for 7 2 SOLUTION a f7x 74 7 372 x4 7 3x2 fx so this function is even b g7x sin7x 1 which is neither equal to gx nor to 7gx so this function is neither even nor odd c f7x 277 2 27CZ fx so this function is even In Exercises 17722 nd the equation of the line 17 Line passing through 71 4 and 2 6 SOLUTION The slope of the line passing through 71 4 and 2 6 is 6 7 4 2 m m g The equation of the line passing through 714 and 2 6 is therefore y 7 4 x 1 or 2 7 3y 714 18 Line passing through 71 4 and 71 6 SOLUTION The line passing through 71 4 and 71 6 is vertical with an xcoordinate of 71 Therefore the equa tion of the line is x 7 19 Line of slope 6 through 9 1 SOLUTION Using the pointslope form for the equation of a line the equation of the line of slope 6 and passing through 91isy716x79or6x7y53 3 20 Line ofslope 75 through 4 12 SOLUTION Using the pointslope form for the equation of a line the equation of the line of slope 7 and passing through 4 712 is y 12 730 7 4 or 3 2y 712 21 Line through 2 3 parallel to y 4 7 x SOLUTION The equation y 4 7 x is in slopeintercept form it follows that the slope of this line is 71 Any line parallel to y 4 7 x will have the same slope so we are looking for the equation of the line of slope 71 and passing through 2 3 The equation of this line is y 7 3 7x 7 2 orx y 5 22 Horizontal line through 73 5 SOLUTION A horizontal line has a slope of O the equation of the speci ed line is therefore y 7 5 0x 3 or y 5 23 Does the following table of market data suggest a linear relationship between price and number of homes sold during a oneyear period Exp ain Price thousands of l 180 l 195 l 220 l 240 l l Noofhomessold l 127 l 118 l 103 l 91 l SOLUTION Examine the slope between consecutive data points The rst pair of data points yields a slope of 1187127 7 9 19571807 E7 5 while the second pair of data points yields a slope of 1037118 7 15 7 3 2207195 7 25 7 5 and the last pair of data points yields a slope of 917103 7712 773 2407220 7 20 7 5 Because all three slopes are equal the data does suggest a linear relationship between price and the number of homes 24 Does the following table of yearly revenue data for a computer manufacturer suggest a linear relation between revenue and time Explain Chapter Review Exercises 67 Year 1995 l 1999 l 2001 l 2004 l l Revenue billionsof l 13 l 18 l 15 l 11 l SOLUTION Examine the slope between consecutive data points The rst pair of data points yields a slope of 18 7 13 7 5 1999 71995 T 4 while the second pair of data points yields a slope of 15 7 18 3 m 5 and the last pair of data points yields a slope of 11 7 15 7 i 2004 7 2001 3 Because the three slopes are not equal the data does not suggest a linear relationship between revenue and time 25 Find the roots of f x x4 7 4x2 and sketch its graph On which intervals is f x decreasing SOLUTION The roots of fx x4 7 412 are obtained by solving the equation x4 7 4x2 x20 7 2x 2 0 which yields 1 72 x 0 and 2 The graph of y fx is shown below From this graph we see that fx is decreasing for 1 less than approximately 714 and for 1 between 0 and approximately 14 26 Let hz 2Z2 12z 3 Complete the square and nd the minimum value ofhz SOLUTION Let hz 2Z2 12z 3 Completing the square yields hz 2z26z32z26z937182z32715 Because Z 32 3 0 for all z it follows thathz 2z 32 7 15 3 715 for all z Thus the minimum value ofhz is 5 27 Let fx be the square of the distance from the point 2 1 to apoint x 3 2 on the line y 3X 2 Show that f x is a quadratic function and nd its minimum value by completing the square SOLUTION Let fx denote the square of the distance from the point 2 1 to a point x 3 2 on the line y 3x 2 Then fx x7223x2712 x2 74x49x26x110x22x5 which is a quadratic function Completing the square we nd 710 211 5 1710 1249 fg X 51 100 10 x 10 10 Because X 2 3 0 for allx it follows that fx 3 for all 1 Hence the minimum value of fx is 28 Prove that2 3x 3 3 0 for all x SOLUTION Observe that x23x37 x23x9 37 9 7 x3 2 3 T 4 4 T 2 4 Thusx23x3 3 g gt 0forallx In Exercises 29734 sketch the graph by hand 29 yt4 68 CHAPTER 1 I PRECALCULUS REVIEW SOLUTION 30 y t5 SOLUTION 31 739 9 yism2 SOLUTION 32 y 10 SOLUTION 33 y x13 SOLUTION 35 Show that the graph of y fx 7 b is obtained by shifting the graph of y fx to the right 3b units Use this observation to sketch the graph of y x 7 4 SOLUTION Letgx f x Then gx73bfltxi3bgt fltxibgt Chapter Review Exercises 69 Thus the graph of y fx 7 b is obtained by shifting the graph of y fx to the right 3b units The graph of y g 7 4 is the graph of y xl shiftedright 12 units see the graph below was 36 Let hx cosx and gx x71 Compute the composite functions hgx and ghX and nd their domains SOLUTION Lethx cosx and gx x71 Then hgx 1104 cosxil The domain of this function is x 73 0 On the other hand 1 cos x ghx gcosx secx The domain of this function is X 2n31m for any integer n 37 Find functions f and g such that the function fgt m 94 SOLUTION One possible choice is ft t4 and gt 12 9 Then fgl f12t 9 12t 94 as desired 38 Sketch the points on the unit circle corresponding to the following three angles and nd the values of the six standard trigonometric functions at each ang e 7 b 77 77 a 7 C SOLUTION a 1 IS H 0 8 IS H l NIH a 27c 27 J5 tan77ix5 cotgiiT 2n 2J5 sec 72 sc 3 77 J2 77 J2 Sin 7 cos 2 4 b 39 tan 71 cot 71 77 77 secT 2 cscT7 2 7n 1 77 J5 SID i cos 7 6 2 6 2 7 3 7 c tan cot 7r xE 6 6 77 NE 77 sec 7 csc 72 6 2 6 70 CHAPTER 1 I PRECALCULUS REVIEW 9 39 What is the period of the function g9 sin29 sin 5 SOLUTION The function sin29 has a period of 71 and the function sin92 has a period of 471 Because 471 is a multiple of 71 the period of the function g9 sin29 sin 92 is 471 40 Assume that sin 9 2 where g lt 9 lt 71 Find 9 a tan9 b sin29 c csci SOLUTION lf sin 9 45 then by the fundamental trigonometric identity 4 2 9 2 v 2 9 17 9 17 cos Sin 5 25 Because 712 lt 9 lt 71 it follows that cos 9 mustbe negative Hence cos 9 735 sin9 7 45 19 7 r a an cos9 735 3 b39292399 924 3 24 s1n s1n cos 7 7 5 5 25 c We rstnote that 9 7 licos97 1773572 smlt gt T T Thus 41 Give an example of values a b such that a cosa b 73 cosa cos b b cos g 7E SOLUTION 2 Take a b 712 Then cosa b cos 71 71 but 71 71 cosacosbcos cos 000 b Take a 71 Then cos cos 0 1 1 2 2 2 42 Letfx cosx Sketch the graph ofy 2f 7 forO 5 x 5 671 SOLUTION 43 Solve sin2x cosx O for O 5 x lt 271 SOLUTION Using the double angle formula for the sine function we rewrite the equation as 2 sin cosx cosx cosx2 sinx l 0 Thus either cosx O or sinx 712 From here we see that the solutions are x 712 1 7716x 3712 andx 11716 2 44 How does hn 21 behave for large whole number values ofn Does hn tend to in nity n 2 SOLUTION The table below suggests that for large whole number values ofn hn 2 tends toward 0 Chapter Review Exercises 71 n hn n22quot 10 009765625000 102 7888609052 gtlt10 27 103 9332636185 X 10296 104 5012372749 x 10 3003 105 1000998904 gtlt10 30093 106 1010034059 x 10301018 45 Use a graphing calculator to determine whether the equation cosx 5x2 7 814 has any solutions SOLUTION The graphs of y cosx and y 5x2 7 814 are shown below Because the graphs do not intersect there are no solutions to the equation cosx 5x2 7 8x 46 Using a graphing calculator nd the number of real roots and estimate the largest root to two decimal places 2 fx 1814 7x5 7 x b gx 1714 7 x5 7 x SOLUTION a The graph of y 1814 7 x5 7 x is shown below at the left Because the graph has three xintercepts the function f x 1814 7 x5 7 x has three real roots From the graph shown below at the right we see that the largest root of fx 1814 7 x5 7 x is approximately 1 151 Y 2 1 1515 71 1 2 1505 m b The graph of y 1714 7 x5 7 x is shown below Because the graph has only one xintercept the function f x 1714 7 x5 7 x has only one real root From the graph we see that the largest root of fx 1714 7 x5 7 x is approximately 1 O 47 Match each quantity a7d with i ii or iii if possible or state that no match exists a 2 b 3 c 2 d 2H3 1 i 2111 ii 611 iii SOLUTION a No match b No match c i 2 17 21117 d iii392 b3b 2 b l H 3 H 3 3 48 Match each quantity a7d with i ii or iii if possible or state that no match exists 72 CHAPTER 1 I PRECALCULUS REVIEW 1 a 1n b 12 c em l d In a 1n b i lna 1n b ii 1nd 7 lnb iii SOLUTION V f 7 7 a 11 ln b 71nd lnb b No match c iii elna7lnb 81ml d No match 1 7a einb E39 49 Find the inverse of f x X3 7 8 and determine its domain and range SOLUTION To nd the inverse of fx x3 7 8 we solve y x3 7 8 for as follows y2 X3 3 8 x3y28 x 3 y2 8 Therefore f 1x 3x2 8 The domain of f 1 is the range of f namely 1 x 3 0 the range of f 1 is the domain off namely y y 3 2 x 7 2 50 Find the inverse of f x and determine its domain and range 1 7 l SOLUTION To nd the inverse of fx we solve y g for x as follows x72yx71yx7y x7yx27y 2 x y 17y Therefore 27 x72 f 10 7 l 7 x x 7 l The domain off 1 is the range off namely 1 1 1 the range off 1 is the domain off namely y y 1 51 Find a domain on which ht t 7 32 is onetoone and determine the inverse on this domain SOLUTION From the graph ofht t 7 32 shown below we see that h is onetoone on each of the intervals I 3 3 andt 5 3 ht 732 We nd the inverse ofht t 7 32 on the domain t t 5 3 by solving y t 7 32 fort First we nd I732lt73l Having restricted the domain to t t 5 3 it 7 3 7t 7 3 3 7 t Thus 3 7t t 3 7 The inverse function is Irla 3 7 J7 Fort 3 31140 3 J7 Chapter Review Exercises 73 X 52 Show thatgx is equal to its inverse on the domain 1 x 71 x1 SOLUTION To show that gx 71 is equal to its inverse we need to show that for 1 7E 1 g x First we notice that for 7E 1 gx y 1 Therefore gltgltxgtgtgxf1 53 Suppose that gx is the inverse of f 1 Match the functions a7d with their inverses i7iv a f X 1 393 f X 1 c 4f X d f 416 i 8004 ii 8064 iii go 7 1 W 80 1 SOLUTION 2 iii f x 1 and go 7 1 are inverse functions 7 x 7X7 7X7Q7D717 fgx711x711x gfX 1 i 1 8006 X b iv fx 1 and gx 7 1 are inverse functions fgX11 fgX X gfx171 X171x c ii 4 f x and go 4 are inverse functions 4fgX4 4064 X 84fX4 gfX Xv d i f 4 and gx 4 are inverse functions f4gX4 fgX X gf4x 7 lt4xgt 7 x 54 Plot fx Mix and use the zoom feature to nd two solutions of fx 05 SOLUTION The graph of fx xe is shown below Based on this graph we should zoom in near 1 05 and near 175 to nd solutions of fx 03 From the gure below at the left we see that one solution of f x 03 is approximately 1 049 from the gure below at the right we see that a second solution of fx 03 is approximately 1 178 74 CHAPTER 1 I PRECALCULUS REVIEW 032 031 029 028 027

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