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Unformatted text preview: i FUNCTIONS AND nooELs M Four Ways to Represent a Function in exercises requiring estimations or anproximations, your answers may vary slightly from the answers given here.
1. (21} E‘he point (—1,—2) is on the graph off, so f{~1) 2 ~2.
(b) When 2: m ‘2, y is about 2.8, so ﬂ?) % 2.8.
{c} f(:r) : 2 is equivalent to y : 2. When 3; : 2, we have a: : «~3 and a: : 1.
(d) Reasonable estimates for m when 3; 2 (i are a: : $2.5 and a: : 0.3. {e} The domain off consists ofall :evaiues on the graph off. For this function, the domain is *3 3 1? g 3, or $3, The range off consists ofall yvvaiues on the graph of f. For this function, the range is m2 3 y g 3, or $2, 3]. (f) As $ increases from —1 to 3, y increases from —2 to 3. Thus, f is increasing on the interval P13]. r a) The point (—4, —Z) is on the graph of f: so f(—4) : ——Z. The point (3, 4} is on the graph ofg: so 9(3) : 4. (b) We are looking for the values of a? for which the y—vaiues are equal. The y—values for f and g are equal at the goints (W2F i) and {2, 2}, so the desired values ofz are WE and 2.
{c) ﬁx) 2 —E is equivaient to y a ——1. When y m —1, we have :5 z 73 and z z: 4.
(6) As :3 increases from U to 4, y decreases from 3 to v1. Thus, f is decreasing on the interval {Gt (e) The domain off consists ofall :c—values on the graph off. For this ﬁlnction, the domain is M4 3 :c S 4, or [M4, 4]. The range of f consists of all yvalues on the graph of f. For this function, the range is 72 g y g 3, or {72, 3].
(f) The domain ofg is {win 3] and the range is [0.5, 3. From Figure l in the text, the iowest point occurs at about (t, a) : (1‘2, WSS). The highest point occurs at about (17. 115). Thus, the range of the vertical ground acceieration is #85 S a g 115. Written in interval notation, we get [—85, 3.1.5]. 4. Example I . A car is driven at 60 tni/h for 2 hours. The distance of mm
traveled by the car is a fonction of the time t. The domain of the £20
function is {t  0 g t g 2}, where t is measured in hours. The range
ofthe function is {d D g d S 120}, where d is measured in miles. 0 2 Example 2: At a certain university, the number of students N on
campus at any time on a particular day is a function of the time t after of midnight. The domain ofthe function is {t  0 g t g 24}, where it is measured in hours. The range of the function is {N G g N _<_ )9},
. . . i o 6 12 is 24 time
where N 15 an integer and k is the lergest number 0t students on (midnight) CI} inpus at DBCﬁl. 14 [3 CHAPTER 1 FUNCTIONS AND MODELS 37. 9(5) 2 \/£ —_ 5 is deﬁned when x — 5 2 0 or :i: 2 5, so the domairz is {5, 00). Sinceyz 1m:%5 2; yz :LU—z'i :> :c :y2 +5,Weseethatgist§1e
top haif 9f 3 parabola. 2x+1 if2x+120
38, F{:c}z2;s~§~}§ I _ I ‘
—{2$+E) 1f 2$+1<1
2m+1 aura—é
_ ﬁh—l ﬁ$<w§ The domain is iii, er {—00, 00) 3a: +1113 , 35' if 3 2 G
39. Gm) : ‘. Smce ix! 2 _ , Wf: have
it“ my: if m < 0 3z+$ 42: if$>0 — if£8>0 4 if$>e
I”: if$<0 g ifzn<0 3 1”“)
CC Nate that G is not deﬁaed for m 2 0. The demaia is (“00, G) U {0, 00). I x imeO
5: 2 —;. Since {xi 2 _ s we have
.' IL” ~33 If a: < 0
as . E .
E; if x > 0 3; if :c > 0
gm 2 um i 1
P if z < O —— if 2: < 0 “E Noze {hat g is not deﬁneé for :3 2 0. The domain is (—00, {1) U (0, 00). 41 ﬂ :c—i—Z ifcc<0 42 ﬂ 3—ém K3332
. s‘ x .‘q: x
) lug: £3320 ) 2mw5 if$>2
"flue domain is R. "{he domain is R. 32+2 ifssg—l 43. fir) x { [\J $ ifzc>—l Note that for x x ml, both a: + 2 and $2 aye equal to 1. The domain is R. 3:3 1 ' CHANER‘! FUNCTEONSAN?) MODELS ct the Eength and width ofthe rectangie be L anti W. Then the area is L117 2 16. so that W 2: Iii/L. the perimeter is 1 2!. + 2W. so P(L) : 2L + EGG/L) : ’21. + 32/L. and the domain ofP is L > 0. since lengths must be positive quantities. ifwe further restrict L to be larger than l‘l". then L > 4 would he the domain. 53. Let the length ofa side of the equilateral triangle he :5. Then by the Pythagorean Theotem. the height 3; ofthe triangle satisﬁes ) '3, F < [T _ . .
y“ + (52:) z are. so that y“) : 1'2 m %;z; m: and y : Usmg the formula for the area A of a mangle, : ﬁbaseﬂheight}, we obtain Ali?) 2 z @552, with domain :2: > t}. 54. Let the volume ofthe eube be l—" and the length ol’an edge be L, Then V : L3 so L 2 and the surface azea is 9 50/) m 3 V) ’ x WW3, with domain V > 0. 55. Let each side of the ease ot'the box have length 2:. and let the height ofthe box be h. Since the volume is 2. we know that 2 : hzrg. so that it 2 2/3“). and the surface area is S m 1:2 + 43515. Thus, S(;t~) :2 x2 + 492(2/932) :: :12 + (8/3). with domain a; > 0. 27.1172 8 56. The area ot‘the window is A 2 3:51 4— m 3:}; + , where h is the height of the rectangular portion of the Window. The perimeter is P : 2h + a: 7‘» 2 30 ¢2> 2}: : 30 m :t' — 3157?; 42> it : :§(60 — 2x M 7m). E‘hus.
60729—73. 2—32 . W I ,  r is 4 new + g 2 153cm e {133:2 —E~ £12 2 153: e ﬁx? — g1“! 2152: —— :ez(ﬂr 8 Since the iehgths x and It must be positive quantities. we have :2: > 0 and it > 0. For it > O, we have 2h > I} 42> be . Hence, the domain ofA is 0 < a: < 2 ate 7r 2 + a 3f}e;t:_:i;7rzz>0 <26.» 60>21c+7r$ <52"; m< 57. The height of the box is .r and the length and width are L 2 20 A 2:10. W 2 12 7 2x. Then V :2 Lil's: and so : (20 w 2.1)(12 — 2 4(10 w :t}(5 ~ : 4.3:(60 — 163: + $2) I 41:3 w 643:2 + 2401'.
The sides L, ll", and 1: l‘llLZSI be positive. Thus. L > 0 4:; 20 — 21' > (3 ee .2: < 10; W > 0 4:9 12 — 2m > 0 ¢:> 1‘ < 6; and :L' > 0. Combining these restrictions gives us the domain 0 < :1: < 6. 58. $2.00 it 0.0 <1 g 1.0
2.20 if 1.0 < .5 31.1
2.40 if 1.1 <11". 31.2
2.60 if 1.2 (1' $1.3
2.80 if 1.3 < :26 S 1.4 00
0'.
0‘
O.
09
O"
0'8
0‘6
00 C
A
a
2 z 3.00 if 1.4 < :L’ 31.5 on
3.20 if 15 < :1". 3.1.6
3.40 if 1.6 < 331.?
3.60 it" 1.7 < x $18 3.80 if 1.8 < a: 31.0
4.00 if 1.9 < .I < 2.0 l
l
l Oi 2 .x wisxamewemsmwww $EC‘E‘IGN 1.1 FOin WAYS TO REPRESENT A FUNCTEOhE 1T 59. (a) RUE?” {h} On 2514.000. tax is assessed on 554600. and 1f)‘3"§~i($wt00tt) $400.
7i 0n 326.000. tax is assessed on 816.000. and
1‘s E o——————
E .4 ‘4 . s t u . 
,6 WM 10%{Stt}.t)(}0_} —: 1.5%rssnon; :2 81000 $900 : stone.
0} 50.000 when I (in deliam
(e) As in part (b). there is i 090 tax assessed on 320.300 of income. so Win doilztrst’I the graph of T is a tine segment from (10.000. 0} to (120.800. '1 G80}. “no”; The tax on $30,000 is $2500. so the graph of?" for .1’ > 20,000 is
, EGUU
the ray with initiai peint (20.008. E000) that passes through {30700012500} 0% 59.000 20,000 sneer) It‘m doliars) One example is the amount paid fer cabie or telephone system repair in the horne. usually measured to the nearest quaner hem: Another exampie is the arneent paici try a sturient in tuition fees ifthe fees vary according to the number ot‘eredits for which the student has registered. 61. f is an odd function because its graph is symmetric about the origin. g is an even function because its graph is symmetric with
respect t0 the y—axis. 62. f is not an even function since it is not symmetric with respect to the y—axis. f is not an 0st Function since it is not symmetric
about the origin. Hence. f is neither even 00? odd. 9 is an even function because its graph is symmetric with respect to the
yraxis. 63. (a) Becanse an eVen function is symmetric with respect to the yaxis. anti the point (5. 3) is or} the graph ofthis even function.
the point (“5, 3) must eiso be on its graph. (b) Because an odd function is symmetric with respect to the origin. and the point {5. 3) is on the graph of this odd function. the point {7.5, 73) must also he on its graph. 6%. (21) NJ" is even. we get the rest ofthe graph by reﬂecting about the y—axis. (b) If f is odd. we get the rest ofthe graph by rotating 180“ about the origin. " ' captain FUNCTIONSAND MODELS {lo [{3} l means that the point {2, l) is on the graph off. We eat] use the
point—slope form ofa line to obtain an equation for the famiiy of linear
limctions through the point (2‘1). 3/ m t : vols: — 2), which is equivaient to y : mx + {1 ~ 2771} in siopeiotereept form. (c) To betong to both families. an equation must have stope m t 2, so the equation in par: (b). y 2 m; + (I « 2m), becomes 3: : 2.1? — It is the oan function that belongs to both families. il members ofthe family oilinear ftinetions : 1 wt m(a: + 3) have graphs that are lines passing through the point (*3, l). 7. All members ot‘the family oi’linear functions : c: m :2", have graphs that are lines with siope m1. The yintereept is c. wmesmxmatosxwumwwwmmwmwmssmoa mmwmtmmwmwm 8. The vertex ofthe parabola on the left is (:10). so an equation is g; cm o(.L’ * 3V —: 0. Since the point (:1. 2) is on the parabola, we’il substitute =1 form ano‘ 2 for y to ﬁnd a. 2 = (1(4 w 3? 2;» o: m ‘2‘ so an equation is ﬁx} 2 2(37 —— 3}? The ywintercept ol‘the patabola on the right is (0‘ 1), so an equation is? r: (41:2 + by: 1. Since the points (“2.2) anti (1, w2.5} are on the parabola. we‘li substitute —2 for 3: and 2 for y as welt as l for and «n25 for y to obtain two equations with the unknowns 0. anti 1). [\3 }: 2mattzeZbE1 :r» @4252] (l)
(11* [\L‘ .5): 72.5:n+b+1 :3» oebz—QSI} (2) (2) —l— (1) gives us So a “6 ::> a. : —1_ From (2)? M} «t» b 2 73.5 :e» b 2 ~25, so an equation is gm : ax? — +1. SECTION 1.2 MATHEMATECAL MODELS: A CATALOG OF {ESSENTEAL FUNCTEONS {3 2i 9. Since f(~1) x {(0) x f(2) : 0, f has zeros ofml= 0, and 2, so an equation for f is ﬁzz) : aim — (wl}]($ we 0}($ A 2).
or ﬁx) 2 arm + 1X1” —— 2). Because f(}) x 6, we’il substitute 1 for mead 6 for ﬁe}.
6 :2 a(1)(2}(mi) m M20. 3 6 ﬁx» a z _3, so an equation forf is ﬁes) m —3:r:(:c Jr e 2). ) For T m 0.02% "l" 8.59, the slope is 0.02, which means that the average surface temperature of the worid is increasing at a rate of 0.02 0C per year. The Tintercept is 8.50, which represents the average surface temperature in °C in the year 1960.
(12} t z 2100 m 1900 m 200 2.» :F m 0.052(200) + 8.50 : 12.50 “C 11. (a) D : 200, so c :2 U.D417I)(a + 1) z 0.0417(208){a + 1) : 8.34s s 8.34. “the slope is 8.34, which represents the change in mg of the dosage for a child for each change of 1 year in age. (b) For a newborn, a 2 D, so c 2 8.34 mg.
12. (3} (ii) The slope of ——4 means that for each increase of 1 doilar for a
rental space, the number of spaces rented decreases by 4. The yiutercept of 200 is the number of spaces that would be occupied if there were no charge for each space. The sr~intercept of 50 is the smallest rental fee that results in no spaces renteti. 10 2C} 30 40 50 60 x (b) The slope of % means that F increases g degrees for each increase 13. {a} F
(100,212; . ‘
of 1°C. (Equwalcntly, F lucreases by 9 when G increases by 5
and F decreases by 9 when 0 decreases by 5.) The F—intercept of
32 is the Fahrenheit temperature corresponding to a Celsius temperature of 0. a) Let d 2 distance traveieé (in mites) and t n time elapsed (in hours). At (b)
t:0,dz€)andati= 50minutes : 50 516 m % h,d:40. Thuswe have two points: (6, 0) anti (%,40), so m 2 450 W3 3 48 and so if 2 48$. 6 (c) The slope is 48 and represents the car‘s speed in mi/h. T2 ~ T1 80 —— 7O 10 1
“2,1 ' T' " b———:...——mw_w_:—_ '
5 l 1) Usmg A in place ofm and T In piece ofy, we ﬁnd the slope to 6 N2 m N1 33 A 1:3 60 6 So a irriear equation is T — so : g(N ~ m) ¢> T w so a guy — A? «2:» T : gN+ 23—7 [£31 a 51.15]. (h) The slope of % means that the temperature in Fahrenheit éegrees increases one—sixth as rapidly as the number of cricket chirps per minute. Saié differentiy, each increase of 6 cricket chirps per minute corresponds to an increase of 10F. ' When N 2 150, the temperature is given approximateiy by T m %(E50} + 312:: :2 TSiEUF :2: 76°F. 22 E CHAPTER1 FUNCTIONS/3ND MOSELS 16. {a} Let .e denote the hamper of chairs produced in one day and y the associated cost. Using the points (108, 2200) and (300, 4800), we get the slope W z%‘§$ :13. Soy~2200m13($~108) <:> y 2 139: + 900. (b) The slope of the line in part (a) is 13 and it represents the cost {in dollars} of producing each additional chair. lGO zoo 300 I (c) The yintereept is 900 aad it represents the ﬁxed daiiy costs of operating
the factory. change in pressure 4.34 _ . .
z W— m . . 5 th t
10 feet change in depth 10 0 434 Using P for pressure and d or dep w1th the porn 17, (a) We are given (d, P) m (0, 15), we have the siope~intercept form of the line, P z (3.434d + 15. (b) When P 2 100, then 106 2 0.434d + 15 ¢=> 0.43401 m 85 ea d 2 0:24 W 195.85 feet. Thus, the pressure is
100 lb/in2 at a depth of approximately 196 feet. . . . (72 — Cl _ 460 m 380 _ so H 1
a) Ustng d in place ofe and C in place ofy, we find the slope to he ml— m m a 320 —. 4.
So a linear equation is C — 450 x i {d — 800) e: O —— 460 = it}. ~ 200 4:5 C 2 id —§— 260. (e) Letting d z 1500 we get 0 a g (1500} + 260 z 535.
The cost of driving 1500 miles is $635. The slope of the line represents the cost per mile, $0.25. (d) The y~intercept represents the fixed cost, $269. (e) A linear function gives a suitable model in this situation because you have ﬁxed monthly costs such as insurance and car payments, as well as costs that increase as you drive, such as gasoline, oil, and tires, and the cost of these for each additional mile driven is a constant. 19. (a) The data appear to be periodic and a sine or cosine ﬁanction would make the best model. A model of the form
ﬂat.) 2 a cos(bx} ~l~ (2 seems appropriate. (5} The data appear to be decreasing in a iinear fashion. A model of the form ﬁx) 2 me —l— b seems appropriate. 20. {a) The data appear to be increasing exponentially. A model of the form ﬁe) 2 a  61' or ﬁx) 2 a  bx + (2 seems appropriate. 03) The data appear to be decreasing similarly to the values of the reciprocal Emotion A model of the form ﬁe) 2: (2/3: seems
appropriate. FUNCTIONS AND MODELS : l3! CHAPTER1 T0“ y z 1  x2 : ~32 + 1: Start with the graph ofy : 2:2, reﬂect about tile m—axis, and then shift 1 unit upward. f1. y m (a: + 1?: Start with the graph efy 2: $2
and shiﬁ 1 unit to ihe leﬁ. 12. y 2 532 % 43: + 3 2 (m2 «— 4x + 4) w 1 2 ~ 2)2  1: Start with éhe graph ofy 2“ $2, shift 2 units to the right, anti than shift 1 unit downward 32V yzcosx y22cosx+i F 2 43in 3m: Start with the graph of y 2 sin compress horizontaiiy by a factor of 3, arrd then stretch vertically by a factor of 4. SECT’ION 1.3 NEW FUNCﬁONS FROM 0%) FUNCTEONS 33 i} 1 1_ 1 ii a:
22;“— + 12$+—+ 2: mw+~w+ a, g; 5:171 11: m‘f}. $(I2+1) I
33+1_1 m+1+1($+2)
, 23+1 $122“ $+2 m+1+$+2 2$+3
[E x 2 :MM—zW: :
(“909)”) MW 9(222) m1” $+E+2($+2) $+1+2x+4 3m+5 $ + 2 11: + 2
Since g(:I:) is not deﬁaed f0; w 2 22 and g{g($)) is not deﬁned fer m m 2%, the domain of(gog)(m} isD : {z 53 ¢ —2, mg , 36. ﬁg): 14:, D:{$im§é~l}; g($}msin2m, 132R. sin 2x (a) (f OQMJ?) m f(9($)) z f{Sin 2$} I m Domaintl+sin2m?é0 22> sin2m%21 ::> 2m$~32£+27m 2:2 32%5Tﬂ2r7r {b) {gof)(m) 2: g(f{$)} z 9(E:I) z sin<1i$m> Domain: {3: i m # —1} :5 w ~(1+$)
55 1+1?) x a:
(c><fof)<w)2f(f<z>>2f( M1” < 22—2—2; 2
1+3?) 1+1:$ (1+1:$)‘(1+x) 1233222: 2x+1 Since ﬁx) is no: éeﬁned for z : ~1, and f{f(3:)) is no? éeﬁned for a: m 22%, n (7: an integer). thedomain of(f o f)(;z;} is D 2 {m 5 m # ~21, —% .
{(1) {g o g){g} :2 2 g(sin 23:) ﬂ sin(25in 2m. Domain: ER __ 37. (f 09 o W) 2 f{9(h(:c))) 2 f(g(a: 2 in 2 mm 21»: 2<x 2 1) + 1 2 2m 2 1 o g 0 mm 2 mm») 2 ma 2 2)) 2 m: 2 2)?) 2 20— as)? 2 1 2 222 2 4m 2 i 39. {f o g o W) 2 f(g(h{m))) 2 MW + 2)) 2 2“in + 2?}
zf(m6+4$3+é) m Wzmﬁ 42 (fogohm 2 f{g(h(w))) 2 f(g($/i"}) 2 f( E) :m( 1) Letg(3:) : $2 2&2 i and f($) m 3:10. Then (f o g)(9:) 2: f(g(:r:)) ‘2 ﬁx? + 1} : ($2 22 U10 z 12 9(2) 2 «E and m 2 2112mm 0 9x2) 2 figirs» 2 f NE) 2 sin was“) 2 Fm. Then (f o gum) 2 mm» 2 mm 2 1 2 Fix). ?m g(:c) w e/Eﬁnd 3 i TS". (I) 2 1' andﬂsc) 2 ﬁmwogm 2 119(2))2 f<1:$)2 13/1; 2 6(2). 1+$ 34 CHAPTER1 FUNCTIONS AND MODEiS 4s. m m 2 cost and ﬁt) 2 v3. Then (f owe) = fee» : ftcost) : we?“ 2 me). 4s. Le{g(t) : tantami m) : ——t~m.Then (fog}(t} : mm) : ﬂtant) = ta“ 1+: 1+:ant2uft)’
47. Let Mac) 2 $2, 51(35} 3 3x, and my) 2 1 w I. Then
(f cg a was} : maze)» a: mm) = 19(9)“) =1~ : Hm.
I " Leth(1:) 3 i931, gm : 2 +$, and ﬁx) : 579?. Then
(f o g e W} x f(g(h(x))) : meme) 3 f(2 + let) = 2 Her).
49. Let M33) 2 ﬁg, 9(3) : sec 3:, and ﬁx) 2 $4. Then
(f 0 go we) a when) = mm» : f(sec x/E) : (sec v45)“ 3 see we) : Hes}. 50. (a) f{g(1)) 3 ftﬁ} : 5
(C) f(f(1)) : f(3) :4 (’0) 96(1)) 2 9(3) 2 2 (d)g(9(1)) = 9(6) 3 3
(‘3) (Q 0 f){3) m Q(f{3}} 3 9(4) : 1 {f} (f 0 WW) : .f{9(6)) 2 “3) t 4
51 (3) 9(2) 2 5, because the point (2, 5) is on the graph ofg. Thus, f{g(2)) 2 f(5) 2 4, because the point (5, 4) is en the
graph off. (b) §{f(0)) * 9(0) 2 3 (C) (f 09W) e f{9(0)) * 1W3): 0
(d) (9 0 f)(5} m 90%)) 3 9(5) This value is not deﬁned, because there is no point or; the graph of g that has
mwcoordinate 6. (e) (g 0 g)(~2) 2 981(4)} : 9(1) = 4
(f) (f 0 M4) 2 f{f(4)) : NZ) 3 ~2 52. To ﬁnd a particuiar value of ff 9(1)}, say for m 2 0, We note from the graph that 9(0) z: 2.8 and f(
f (9 (D)
.r 2.8) m “05. Thus,
) is f(2.8) m ~05. The other values listed in the table were obtained in a simil ar fashion. 53. (22) Using the relationship distance 2 rate  time with the {adius r as the distance; we {save r(t) : 60$. (b) A 2 ‘m'2 :s (A o rm) 3 A{r(t)} 2 7160:}? 2 360077222. This fomml a gives as the extent 9;“ the rippled area
(in cmg} at any time t. SECTION 1.3 NEW FGNCT§ONS FROM OLD FUNCTlONS S 35 1 a) Elle radius r of tile balloon is increasing at a rate cf 2 cm/s, so 7ft} xx (:2 cm/s)(t s) 2 2t (in cm). (b) Using v : gmﬁ, we get (1/ owe) : Vow) 2 W223) : §7r(2t)3 : ergm3. The result, V : §327r£3, gives the volume of the ealioon (in ems) as a functier; of time (in s) 55' (a) From the ﬁgum: we ha“? a right triangie with legs 5 and d, am; hypotaﬂusa s. ____ “““  "
By {he Pythagorean Theosem, d:2 + 62 z: 52 22> s 2 fid) : : xx 8
(b) Using d. 2: wt, we get d 2 (30 km/hr) (t 11;) 2 395 {in my Thus; lighthouse Shﬁreilne d r: 903) :2 3046. (c) (f o g){t) : f(g{t)) m f(30t) : 143%)? + 3 2 «900:2 —i— 36. This function represents the distance betweeo the lighthouse and the ship as a function of the time eiapsed since noon
58. (a) d : rt *4 d(t) :2 350i
(b) There is a Pythagorean} relationship invoiving the iegs with ierigths (1’ anti 1 and the hypotenuse with length 5:
d2 —‘r 12 : 32.T§1us, 5(d) : {e} (s o d)(t) : s(d(t)) m 3(350t) : (3501‘)2 J?" 1 57. (a) (b) V(t) 0 if t < 0 Vet 120H
3 50 Z t .
120 if t 2 o ) ( } Starting with the formuia in part (b), we replace 120 with 240 to reﬂect the
different voltage. Also, because we are starting 5 units to the right of. t 2 £3, we repiace t with t w 5. Thus, the formula is V(t) 2 2430116 ~— 5). 58 (me) we) (eve) D “<0 (we) 6 if“?
, a z: :2 c 3
2t ifogtgeo 4&4) irvgtgss
0 ift<0
“ t éftze soV(t):2tH(t),t§60. sol/(t)24(tw7)H(t—7),tg32.
V V
120
$00
0 60 I 0 7 32‘ 3.59. lff(m) x mm + b; and g($) : 77122: + b2, than
' (JeOQME) 3 : f<m2$ "i" 52) x m: (M23: “in 132) + (31 5 7711771223 + 771152 + 131. _ I So 1’ o g is a linear function witil slope mlmz. ...
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 Spring '08
 Shen
 Calculus, PreCalculus

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