Chapter 1 FUNCTIONS AND GRAPHS 6—a? 6-12 5 . : 13. = sy = 0 1.1 introduction to Functions 9) = ge) 20) "2 _6-(=22_ 2 1 1. (a) Ar) = mr? HA) == ="3 AN 8 (b) A(d) == 4) =r 14. Hg) = RY 2. From geometry, c= 2u7 (b) c=nd Hy =Sravi=2420) =244=0 8 3. From geometry, H(0.16) = ist 2v/0.16 = 50 + 2(0.4) 4 4 (d\® =50+ 0.8 = 50.8 V= Fi =37 3) 48 1B 1 15. 9(t) = at® — ®t “FETT = ( ) ( 0) o ) gl-zl=al-%5) ~a?{~= 4 I 2 3 2 4. From geometry, A = 6e?; e? = “=; e = Vx beni (2) i 2 5. Al) = lw =5l oH -= a2 6. From geometry, V = Tart = FE) = Sart 1°73 g(a) = a(a?) —a?(a) =a® ~a®* = 0 7. From geometry, A = s* VA = VsE 16. sy) =64HF1-3 s=VA 3(8) =6yB+1-3=6V0-3 =6(8)-3=18—3=15 8. From geometry, p=4s s(a?) =6vaT + 1-38 45 p T51 17. K(s) =3s2 ~s+6; _r K(—8) =3(—8)? ~ (~8) + 6=38s +5 +6 5=4 K(25) = 3(28)2 — 25 +6 =125% ~ 25 +6 9. fx)=2+Lf(1)=2-1+1=3 18. TH) =st+7 f= =2(-)+1=-1 T(—2t) = 5(~2t) + 7=—10t +7 TE+1)=5t+1)+7=5t+5+7="5t+12 10. f(z) =5z-9 (2) =5(2)-9=10-9=1 19. J RAM = fA isa) =5(=2)—0 = — J Ye - = - x f(=2) =5(-2)—9=-10-9=-19 PAP 11. fl@)=5-3z =_8 f(-2)=5-3(=2)=5+6=11 . f(0.4) =5-3(0.4) =5-1.2= 3.8 20. f(z) = 22? +1 fle+2)— fm) —2=2z+2)% +1222 ~1-2 12. f(T) =72-25T = 2x? + dz +4) ~ 2% -2 (2.6) =7.2-25(2.6)=0.7 =2¢2 482+ 8-222 2 f=) =712-25(-2) = 172 =82+6 1 Downloaded from StudyXY.com ® + StudyXY Sd Ye. o> \ | iF ’ pr E \ 3 S Stu dy Anything This ContentHas been Posted On StudyXY.com as supplementary learning material. StudyXY does not endrose any university, college or publisher. Allmaterials posted are under the liability of the contributors. wv 8) www.studyxy.com 2 Chapter 1 FUNCTIONS AND GRAPHS 21. 3 is an integer (whole number); 3 is rational {may (b) Let 2 be a positive or negative integer, then be written as a ratio of integers, 3/1); 3 is real the reciprocal is £ which is the ratio of two in- (not the square root of a negative number). —7 is tegers, 1 and 2, and is thus a rational number. irrational; it is not a ration of integers; —7 is real. Yes. —V/=6 is imaginary; V7/3 is irrational (not the » ] . » ratio of integers) and real. 30. (a) Yes, |positive or negative rational] = positive rational 5 . : o 22. 7: real, rational (b) Yes, ET oe reciprocal YEE pick is : integer integer <4: imaginary rational ZT. real, rational 31. For z <0, |] > 0 which is to the right of zero on 3 number line. ZT. real, irrational [3 ’ 32. (a) |=} <1 describes 7) _7 a wm eS 23. B|=3. |5|=5 — Bl H 2 tooo s-l<zcl PY (b) lz] > 2 describes 2-2 -val= vs 24. |—4| 1) =4 4 bres - A= (0) = 2 0 2 szr<—2orz>2 V2 =v2 i ~~ n 33. 5 = f(t) = 17.5 — 4.9%; f(1.2) = 17.5 — 4.91.2)? —l=e{-=)== =104 | 3) ( 7) 2 " 18 (18) _19 34. C=0014(T —40) 41" 4/74 S(T) =0014(T — 40) F(15) = 0.014(15 — 40) = ~0.35 in, the change 25. (a) 4 <0 in length at T = 55°F. (b) 1 > —m since —7 =~ ~3.1d and 1 > —3.14 35. d=v +0057 (© IE fv) =v +0050 37 2 £(30) = 30+ 0.05(30) = 75 ft 26. (a) 3 <2 f(2v) = 204 0.05(20)% = 2v + 0.20° 60) = 60 + 0.05(60)% = 240 ft b) v2 > 1.42 fl ®) 712(30)] = 2(30) + 0.2(30)% = 240 ft. {c) ~|-3|=-3=>-4<-3=—{-3 36 p= 20R_ 27. (a) b-a;b> a, positive integer - ~ (10+ RP (b) a—bd> a, negative integer. _ __200R b . 760 = ioo + mp (c}) ———, positive rational number less than 1 10 bra’ f(R+10) = OREO) [100+ (R + 10)]2 28. (a) a+ b, positive integer _ 200(R + 10) (b) 2 positive rational integer. © (110+ R)? (ec) a xb, positive integer - 1.2 Algebraic Functions 29. (a) Let x be a positive integer, then [z] = z which 2 nct is positive integer and thus an integer. Yes. Let 1 FG) =Flz-1=/E-1)2+4 z be a negative integer, then jz] = —z which is a =a? _2z+1+%4 positive integer and thus an integer. Yes. =vzZ2 2x +5 + Studyxy Section 1.2 Algebraic Functions 3 2. [CEP =(z-12=(E-D)=z-Dz-1) nm E+ D2 —ae? +1)" (@2 41? = (2% -2z+1)(z—1) (= +1) (x2 + 1)1/2 =x —z 22+ 2 +z 1 2241-22 =2%—322+3z—1 “@ pr 1 x 12. & —22%)V/4(22) + 422(1 — 22)" ¥/4 (1 — 22%)%/4 4. {G[F(@)]}? = {GIVaT+ 4}? (1 — 222)/2 (1 —222)3/4 = _1)2 CY. Sg _ (1-290) +42 CARY x= 2a) _ 2-42 +42? = —972)5/4 5. VIA 1 = (2° + DVB)? = (28 + 11/8 (1-27 rei x (I= 227)5/4 6 (z-1)(VZ+z+1)=z-1){F +z +1) 13. The domain and range of f(z) = x + 5 are each =Valial tz -22—z—1 all real numbers. VAT 14. g(u) = 3 — u?; since 3 — ? is defined for all real (4 _ 5) 72 numbers, the domain is the set of all real numbers. 7. (dz — 5) =(1z-3) However, the range is all real numbers g(u) < 3, since u? is never negative. 8 AEHD aya apn m3 3:2 Cra (De? + )!% = 16. G(R) = % is not defined for R = 0. 2 “1/2 Domain: all real numbers except 0. 9. (0+ 1)" +(z+3)(2 +1) Range: ali real numbers except 0. z+3 =2z+1+ Worrsi 16. F(r) = \/r +4 is not defined for real numbers less VEFIVEI+2+3 than =4. = —_— Domain: all real numbers » > —4 and the range Vez +1 : ie l4z43 cannot be negative due to the principal square _mHliz+s root of +4. Vaz +1 Range: all real numbers F(r) > 0. 3z +4 viz +1 17. The domain of f(s) = 2 is all real numbers ex- cept zero since it gives a division by zero. The ~2/3(1 _ 2) — (3x — 1)/3 10. (3z~1)"*3(1-2)- (32-1) range is all positive real numbers because § is ~ t 1- So Bz 1) always positive. 3x —1 1-2 (3 — 1)V/3(3z —~ 13 18. T(t) = 2t* +¢* — 1; the domain is all real numbers = mon and the range is not defined for any real numbers . G21) less than —1. So the range is all real numbers —z— (3x = E07 T(t) > -1. _l-z-3z+1 19. JI{h) = 2h + VR + 1 where V/A is not defined for CBr) h <0. 24x Domain: ali real numbers i > 0. = 1\%/3 Range: all rea! numbers H(h) > 1. (3x — 1) — StudyXY 4 Chapter 1 FUNCTIONS AND GRAPHS 20. f(z) = = is not defined for real numbers 31. w= f(t) = 5500 — 2 greater than or equal to 2. 32. p = f(c) = 100c — 300 Domain: all real numbers z < 2, and the range 33. m{h) = 110+ 0.5(k — 1000) for & > 1000 cannot be negative due to the principal square root of 2 — x, thus the range will be greater than 34. n= f(2) = 0.52 + 0.7(100} = 0.52 + 70 0. 85. C = f(I) = 500+ 5(1 ~ 50) 1 - 21. The domain of Y(y) = yl y > 2 because 514250 W=2 8 the square root requires y—2 2 O ory > 2 and to 36. M = f(h)= = avoid a division by zero, ¥ > 2 is required. h n 37. (a) 0.12 +0.4y = 1200 = 120-0ls 22. f(n) = 55 since division by zero is undefined, - (2) 012 +0.4y = 1200 = y(z) = —— the domain must be restricted to exclude any val- 1200 — 0.1(400) ues when 6 — 2 = 0. In this case, n = 3 must (b) 9400) = ——=—=— = 20001 be excluded. So the Soman is the set of all real 38. 182 = (18 — d)? +12 pLn=s r= 36d & __D 4 D-38 A c=2nr=2n/36d—-& 23. f(D) = 53 Bra pr since division by zero is undefined, the domain must be restricted 39. For the square, p = x, side = z. to exclude any value for which D —2, D +4, or 4 D — 6 are equal to zero. In this case, D # 2, —4, 4 EE 6. So the domain is the set of all real numbers saunre 4 16 16 except 2, —4,6. Tor the circle, ¢ = 60 — xz = 271; _60—z _60-p 24. g(x) = 2 since division by zero is unde- TTT TT fined, the domain must be restricted to exclude Ar mm? (60 — p)? any value for which z — 3 = 0. In this case 7 = 3 circle = PEER must be excluded. And square roots aren't de- Tus, the total fined for negative values, so the domain is all real 2? (60—p) bers z > 2, except z = 3. A=Tg+—p numbers x > 2, Pp! 16 4m 25. F(t) =8i—t2 fort <2 F(2) =3.2-22=12 40. A= f(d) F(3) does not exist. This is a circle and a square. x —8) = 2(—8) = —16 (since — 8 < — 2 26. h{-8) =2(-8) (since <-1) a= (3) a 2(-3) Ll =! (since Zl >-1) 2 2 2 2 41. A = f(z) = 7(6~2)® with domain 0 < z < 6 since 27. J) = ITT = vVA=2 (since 1 2 1) = represents the radius and it must be greater than or equal to zero and less than or equal to six. f (3) Pp. 1 +1= 3 (since _ 1 <1) Using the end point values for the radius gives a 4 4 4 4 range 0 < A < 367. 1 1 ice 1 42. d= f{h) 2.4(3)-1-5 since = #0) - 5) 3700520 120% 4-12 = @? (0) = 0 {since 0 = 0) d= 14400 + 12 29. d(t) = 40(2) + 55¢ = 80 + 55¢ The domain ish> 0 since the distance above the ground is nonnegative. 30. C= f(r) =3(2nrh + 277%) = 6wr(2) + 6m? The range is d > 120 m since 120 m is the value = 1277 + 677? of d when h =0. + StudyXxy Section 1.3 Rectangular Coordinates 5 43. s= f(t) 4. d= st = 300 y 5 = 300/¢ (Note: cannot have negative time or R speed) B A Domain: can't divide by zero, so ali real numbers x -4 0 4 t>0 Range: all real numbers 8 > 0 (upper limits de- ce? pend on truck) 44. I= f(w) 8. Joining the points in the order A=lw=8 ABCA form an isosceles triangle. i= s (Note: cannot have negative width or length) Domain: can’t divide by zero, so all real numbers | 7] m= BG Range: all real numbers 0 <1 <8 f i, 45. The domain of f = rel is C > 0 because C' 6. Isosceles right triangle must be > 0 to avoid taking the square root of a y negative and > 0 to prevent division by zero. s 46. y= f(z) =550 — x A 4] Domain: all real numbers greater than zero and ra less than 550 (since distance cannot be negative) BL © 47. From Ex. 33, 7. Rectangle m= f(h) = 110 + 0.5(h — 1000) = 0.5h — 390 y m= 0.52 —390 for h> 1000 B “lio for 0<h<1000 D c 48. From Ex 33, C = f(I) = 5 + 250 i] Co [814250 for 1>350 ry vo me wt] 500 for 0<1<50 A B RP Se — 8, Paralleogram 1.3 Rectangular Coordinates & ¥ 1. A(2,1) B(-1,2); O(-2,-3) p ° c 2. D=(3,-2; B= (~35,05) = (-1, 1); was 23 Sy F=(0,-4) J A 3 B 3. y 9. The coordinates of the fourth 71 oA vertex, V, are (5, 4). : EEE H Be + StudyXxy 6 Chapter 1 FUNCTIONS AND GRAPHS 10. The abscissa is the z-coordinate, since this is an 25. The ratio 2 is positive in QI and QIIL equilateral triangle and we know the base is (2,1), (7,1) z then the third vertex must be equi-distant be- 26. The ratio y/r is negative in quadrants two and tween 2 and 7. So 37 = 3 four. 11. In order for the z-axis to be the perpendicular 27. (a) d=3—(~5)=8 bisector of the line segment join P and Q, Q must be (3,2). (b)d=4-(-2)=6 12. The line segment joining P and Q are bisected by 28. From exercise 27, the distance between (—5,—2) the origin gives point Q(4,—1). and (3,4) is 10 because it is the hypotenuse of a right triangle with legs of 6 and 8. 138. All points with abscissas of 1 are on a vertical line through (1, 0). The equation of this vertical line _— sz=1. 1.4 The Graph of a Function 14. Ordinates are y-coordinates; the points whose or- Lye dinates are —3 are all on a fine parallel to the sy=cr- z-axis, 3 units below. =| y 15. All points (z,3), where is any real number, are EA n points on a line parallel to the z-axis, 3 units above 1]-2 it. 2 0 —4, 16. All points (~2,), where y is any real number, are poiuts on a line parallel to the y-axis, 2 units to 2 y=6-1z the left. 3 17. All points whose abscissas equal their ordinates z | y H are on a 45° line through the origin. The equation —1]63 13 of this line is y = = and it bisects the first and 0.16 third quadrants. 156 6 35 0 [a 18. When the abscissa equals the negative of their or- 80 dinates, then £ = —y. These pairs lie on a line formed of points made by varying z. These points 3 y=3-22 form a line bisecting quadrants two and four. 3 19. Abscissas are T-coordinates; thus the abscissa of Ly all points on the y-axis is zero. = = py 2 20. Ordinates are y-coordinates; thus the ordinate of 0] 3 all points on the z-axis is zero. 1 2 311 21. All points for which z > 0 are to the right of tiie y-axis. 4, y=222 +1 22. All points below the z-axis. y zy 10 23. All points which lie to the left of a line that is —2 9 parallel to the y-axis, one unit to the left have ~1!3 z<-1. 01 24. All points which lie above a line parallel to the : : z-axis, 4 units above have y > 4. 40 4 + StudyXxy Section 14 The Graph of a Function sn 10. p= 2 H 7? +05 th on ) ii i —2]022 2 2m —1|067 ) 220 ilo AN do ! 1| 067 or ’ ’ 2 | 022 1. y=ViTz 6. y=2+3z+ z2 ny - y xT y . , i —-3 2 n EAL —2] 0 is “Ts ¢ 31 4 TTS AR 1] 6 HL * 12. y= VaZ=16 ra x y y —6 | 45 4 { x e Vv 8 E g 1 [ — 18 = [SE . 2h 2 6145 Ts 13 2 8 d UINO0w a oe Ne y= 3x — 2° Yhin=:g Wan oo " , yaerey NE : z Xres= NE -3| 18 ] is E75 : Hd Ti —2 2 N 7 1s A 7 i 2 2 Hee 14. DoW. = > wo we mE | MEE nen eh Mie .Y +2 Ymax=8: i] as pee vex xr y . y 2 Te z NN i] 0.66 - t- StudyXY 8 Chapter 1 FUNCTIONS AND GRAPHS 16. 19. TTROOT ps SEE Ti FREE 5 pe | isn Ge mi Hell Re gag NE Sess? Ne paid REE if EES = = rh ar i aE Wie Jeet ie | EEio REM =: xegtal Wie . 16. 20. — - TTT o 5 Mois Rehan Yin s Rh g g a = [ae Ee Ne Gass NIH = : mine-5° Sus HERG Shanes NH . Thats We isda 7 fg - Vacl=l Nem Arise! ays Xres=] [Yrs 17. 21. WI Fu ~ PUZE PRAT 71 = $l GS) ST a ghinse ERAT Unints Reale en Wis > Fes Wal feelet | Nie = fErish Wow hie mye Thins Wao Verte viz 4 Vocixt : Win Sze = xVrs Rreax] Yr 18. 22, WINDY FIots FWts PREY WINDOW PIE Plz Tt pins-o NE auins East He ie $2eitt Lois Es ie Ininrs Wim oe Ee BH Sie wigssl SES = Xe&ssi : Rh a : | y 1 StudyXY Section 1.4 The Graph of a Function 9 23. 27. V5R+2=3=V5R+2-3=0 i TER J X es his Graph y = v/5z + 2 — 3 and use zero feature to Shinai Ris solve. Inaxes’ Ps Sets - Row Apgas a NE = WINDOW TE fr pas fpeoz i pay Ea Eh — oe — i = if a R=14 28. z—-2= i Collect all terms on the left; z—~1-2=0. Let y =z— 1 —2. Using the trace, 24. we see that the solutions are approximately —0.4 , hod Lo arco and 2.4. With the zoom feature, the results may 7 Yi be read more accurately. iy Ee = - - LL 4 29. From the graph, y = pay has range y<-lory>0 25. To solve z? — 41 = 0 using a graphing calculator, we let y = 22 —41. Using the trace, we see that the UINOOY Jeane solutions for y = 0 sre approximately —6.4 and fey when 6.4. With the zoom feature these results may be yines hs read more accurately. se Ri 26. w(w—4) = 9; w?—4w = 9. Collect all terms on the 30. Set, the range at Zon = —2, Tmax = 2 © uP d—Q = 22 A - - min = —2, Tmax = 2, left; w?—4w—9=0. Lety = z?—4z—9. Using the Ymin = —2,¥max = 2. From the graph, using the trace we sce that the solutions are approximately trace, we see that the minimum y-value is approxi- —1.6 and 5.6. With the zoom feature these results mately —0.25. Therefore, the range is all numbers may be read more accurately. greater than —0.25. a Bali | Vv 1 StudyXY 10 Chapter 1 FUNCTIONS AND GRAPHS 31. Set the range at Tumin = —4, Tmax = 4, 35. Graphy = = PI 2-3 using a graphing Ymio = —6, Umax = 4. From the graph, using the z—2 +4 z-6 min 7 Jmax ? calculator, the range appears to be the set of all trace, we see that the range appears to be y > 0 real numbers. ory < —4. WI0u TRC EE me | [EE 13 feelza Ri inir=ria Wim i Eert az Ee EE Loss Sex La 32. Set the range at Tin = —3,Zmax = 3, bf _ Ymin = —3,Ymax = 3. From the graph the 0a appears to be the set of all real numbers. im 36. Using a graphing calculator with iE Zain = 0 Tmax = 5, Ymin = —4; Ymax = 4, the range appears to be the set of all real numbers. 33. Graph y = EL on graphing calculator and + Graph y = ——=—; on grap TU] use the minimum feature, then from the graph, y+1 Y(y} = ——=¢ has range Y(y) > 3.464. w= © port Hh Ry Auk nest Wie al a NE Xeesel Wie 87. ¢=0011r+4.0 c NN r_|c wv 500 ! 9.5 1000 | 15 Sine 0s 2000 | 26 95 , 500 3000 34. Set the range at Tmin = ~2, Tmax = 5, Ymin = —4 Ymax = 4. The range appears to be the set of all real numbers excluding —3. 38. H =2401" I H RE Tot refi 02] 96 0.4 38.4 06 | 86.4 0.8 | 153.6 rd + Studyxy Section 14 The Graph of a Function 1 39. p= 0.05(1 +m) 43. h=15+86t — 4.9¢% when h =0,¢ = 18s m Ie t h m P 0 50 01] 055 10] 1250 02] 03 20 | =750 03 | 0.216 ois 0.40175 - Te TE 7 TE To Ee J Ep NE 40. N=+/n?—169 Kress : Wes . N n N 13]0 By aN 15 | 0.748 1.7 | 1.085 $5 ay \ 2.0 | 1.520 N 13 20 41. €® + (e + 5.00) = 40,000 44. Graph y = 92° — 240022 + 240,000 — 8,000,000 e=244cm and use the zero feature to solve. +500 = 29.4 cm : oy TTR n oi RT Ef Te | [a ad | Ailosspe | RE Eee | [EE SEE | me Xpege a Bees % Casa hes | ed Ea 42, A=520 =lw = (w+ 12}w =w? + 12w => w? + 12w—520=0 45. P=2142w=200=1=100-w he SEA A= lw = (100 - ww = 100w — wv? ed Wi for 30 <w < 70 Bie i Ea ; Ni w|30 40 50 60 70 A [2100 2400 2500 2400 2100 [7 \ Li 2500 er | aol 7 The approximate dimensions, in cm, to 2 signifi- w cant digits are w = 18 cm, { =~ 30 cm. EY — StudyXY 12 Chapter 1 FUNCTIONS AND GRAPHS 46. y =x(10 — 22)(12 — 22) 48. To determine the maximum capacity graph y =z(120 — 442 + 42?) y = 120z — 4422 + 423 from Exercise 32 and use y = 120z — 4472 + 42° the maximum feature to solve. zy [AY Vive Te Vin 180 Tr ae 29% acd aw Salve 1a eo 3.72 Maen Wik 432 Xresa1 Ves = From the table y = 90 for « between 1 and © = for between 2 and 3. Graph NC v1 =120z — 442% + 42° — 90 Ee Bare and use the zero feature to solve. The maximum capacity is 96.8 in®. INDO: “Piets TIE plots poe 1 3y1Rj Rox dxteay F tei Nh 5 : iy = ys : Peels yes | 49. z= |y y = z is the same as 22 y=|z| forz>0. -1|1 y = |] is the same as bee I] [EE] 0|6 y=-zfrz<o. Nal if ; H 2|2 {Snr bees iS Ne Ag =H From the graphs, ¥ = 90 when z = 1.3 in and \ z=24in NR 47. s = VE—4t%. Graph y = vz — 427 and use the r= maximum feature to find the maximum cutting speed. Tor negative values of z, y = |z| becomes y = —z. pan SELES med Spex sclalPs i hae lot wan [BE 50. The graphs differ because the absolute value does not allow the graph to go below the z-axis. 5 y=2-z Bo z | y Ek —11 3 2 0] 2 1] 1 * From the graph the maximum cutting speed is 2] 0 y=3-x 0.25 ft/min. 3 1 + StudyXY Chapter 1 Review Exercises 13 y=12-2| 4. C= 200+ 20(10) + 100 s |y yo 2 xi = 301 + 20w = \ y = 30(w + 20) + 20w TTT 2 NL = 30w + 600 + 20w = 50w + 600 >To x + 3]1 5. flx)=Tz-5 3) =7(3)~5=21-5=16 _f3-z =x<1 f(~6) =7(=6) 5 = —42~5 = —47 51. @={%7 21 6. g()=8-3I Ty 1 ny _1 51 (2) =s-s(2) 2 —-1| 4 2 IIE ) 9-4) =8-3(-49=20 1 2 T= : 7. Hn) = VIZ 5 H(—4) = 1-2-4) =3 H(2h) = T= 22h) = /I— 4h 1 —_— z<0 3v— 52. flz)=4q v—1 8. $v) = bot vz+l z20 o-y=2CA2 z y —241 —2 [03 rm Bw+l)-2 _ 3u+l ¥ == — 1 [05 oD ==3T4T Tere -0.1| -09 1 2 5 I . 9. flx)=3%~2x+4 1 14 -1 . fla +n) — f(z) 3 2 =3(@ +h)? -2z+h)+4— (322 — 2x +4) : Co . =3(?+ 2h +h?) —2 —2h +4322 +224 53. the gph passes the vertical line test and is, =32%+6zh+3h% ~ 2x — 2h + 4-322 + 2x —4 erefore, a runction. = 6zh + 30% —- 2h 54. S rtical Ii ill int t £h h at mul- tiple points. Graph is that of a relation, 10 Fla) =a +22 3 Pie points. Lirap - F3+h) ~ F(3) 55. No. Some vertical lines will intercept the graph at = (3+) +2(3+h)2-3(3+h) multiple points. ~ (3° +2(3)* - 3(3)) = 33+ 3(3)%h + 3(8)h* + 13 56. Any vertical line will intercept ithe graph at only +2(9 + 6h + h2) — § — 3h — 36 one point. Graph is that of a function. — 274+ 27h +9h% + AS + 18 + 12h +2h%—9—3n—36 PT ES T= A —— — hd 2 Chapter 1 Review Exercises RE 1165+ 36k 11. f(g) =3-2 1. A=mr? = n(n)? A= drt? f(2x) —2f(z) = 3 — 2(2z) - 2(3 — 27) =3-dr—-6+4r=-3 2. A=nrs=arvhE +72 =31vh? +9 2 12. f@)=1~z 3. 2000(z) + 1800(y) = 50,000 f@)® - fe?) = (1 -a%)2 — (1 - (a%)?) 0 250 =1-2% +2 —1+2t =-gTt = 22% — 22% + StudyXxy 14 Chapter 1 FUNCTIONS AND GRAPHS 13. fz) =8.07-2¢ 22. y=5z—10 £(5.87) = 8.07 — 2(5.87) = —3.67 £(~4.29) = 8.07 — 2(—4.29) = 16.65 ~ 16.7 z| vy 0-10 14. g(x) = Tx ~ 2? al o 9(45.81) = 7(45.81) — (45.81)? =-1778 2 9(~21.85) = 7(—21.85) — (—21.85)? -10 = —6304 S — 0.087629 15. G9) = —3 5188 23. y = 4z — 2? 0.17427 — 0.087629 G(0.17427) = =e 0.17427) = =3 ose 172) zly = 0.16503 S : 4 0.053206 — 0.087629 G(0.053206) = — 2 214 (0.053206) 3.0125(0.053206) 3l3 y = —0.21476 4 | 0 2-4 16. WO) =e 24. y=a?-8c-5 8.912 — 4(8.91) MEI) = go sea =v =0.0344 ol _s (~4.91)% — 4(—4.91) 1-12 10 h(-4.91) = ——L A) { ) (—4.91)% + 564 2-17 h(—4.91) = 0.0082 3-20 had 4-21 17. The domain of f(z) = z* +1 is ~c0 < = < co. 51 —20 The range is f(z} > 1. 18. The domain of G(2) = 3 is all real numbers ex- 25. y=3-z-3? cept 0. The range is y # 0. = v | T 3 19. The domain of g(t) = 7 is all real numbers wo 2 t>—4. 1 0 3 1 The range is all real numbers g(t} > 0. 21 7 } 20. The domain of F(y) = 1-2/7 is all real numbers y 20. The range is all real numbers F(y) < 1. 26. y = 6+4z +22 21. y=dc+2 s |v xz |y -3| 3 02 —2| 2 6 “ilo -1| 3 2 : ol 6 2. 1]11 218 -2 3|27 + Studyxy Chapter 1 Review Exercises 15 27. y= 6x 32. Z=2B-2R° z|y o - 6 z |v B of ¥2 —3[ 265 z nd I —2| 412 R ol © ~1|480 Ly-® 6) 0500 X 2] —4 11480 =335 1 35 319 2412 3265 28. V =3 058° 3 \2 27 v 9 : . 3 33.72 -3=0 2] ~1 is’ Graph y = 7z—3 and use the zero feature to solve. = gt Tre] 20. y=2-2 Us aes Th foe RAE 2-14 2 set Vas Si La “1 1 A a 0 2 =1ERY {i Fi ol ua so he 30. y=a*— dz i fis z=04 z |y —2| un 1] 5 o| 0 1-3 34. 32 + 11 = 0. Graph y = 32 + 11 and use zero 2 8 3 feature to solve. x 31. y= —— v z+1 “8 BLL $0 7 ot WE 3] 3 ! ress aE. 2| 2 { Hi [or] 1 3 : I~ StudyXY