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SolA1 - MAT A30 F 2007 SOLUTIONS TO ASSIGNMENT 1 1.1 Four...

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Unformatted text preview: MAT A30 F 2007 SOLUTIONS TO ASSIGNMENT # 1 1.1 Four Ways to Represent a Function 2. (a) The point (—4, -2) is on the graph off, so f(—4) = —-2. The point (3, 4) is on the graph ofg, so 9(3) = 4. (b) We are looking for the values of z for which the y-values are equal. The y-values for f and g are equal at the points (—2, 1) and (2, 2), so the desired values ofz are —2 and 2. (c) fix) = -1 is equivalent to y = —1.W'heny = —1, we have: = -3 and: = 4. ((1) As I increases from 0 to 4, y decreases from 3 to --1. Thus, 1' is decreasing on the interval {0,144. (e) The domain off consists of all z—values on the graph off. For this function, the domain is —4 S 1: 5 4, or [-4, 4]. The range off consists of all y-values on the graph of f. For this function, the range is —2 S y g 3, or [—2, 3]. (f) The domain ofg is [—‘i, 3] and the range is [0.5, 4]. 5. No, the curve is not the graph of a function because a vertical line intersects the curve more than once. Hence, the curve fails the Vertical Line Test. 6. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is {—2, 2] and the range is [— 1, 2]. 7. Yes, the curve is the graph of a flinction because it passes the Vertical Line Test The domain is [-3, 2] and the range is [—3, —2) U [—1,_3]. 8- N0: the me is 11015 the graph Ofa fi-Inction since for z = 0, i1, and i2, there are infiniwa many points on the curve. 18-(a) no) (b) no 400 31W feet 50 ‘ 30 60: (6) ground ((0 vem'cal speed velocity 500 miles perheur , 60! 50 r :..r‘+3_2 m+3—2(:c+1) 26. f(3)—f(1)=$+1 = 1+1 __I+3—2m-—2 z—l a:--1 :c—l _ (a:+1)(:c—1) = “3+1 —(a:—1) 1 (:1:+1)(a:—1) 2 (I+1)(:1:—1) ="z—fi 23.f(a:)=(5m+4)/(z2+3x+2)isdefinedforallxexceptwhenO:x2+3z+2 41> O=(a:+2)(:r+1) e: 32—2 or -—1, so the domain is {:c e R | a: aé —2, -1} = (—00, —2) U (~2, -—1) U (—1,oo). 21. f(:1':)= 3x2 — a: + 2. f(2) =3(2)2 —2+2_—. 12- f(-—2) = 3(-2)2 - (-2) + 2 fig) = 3a2 — a. + 2. f(-a) = 3mg: 4 (—a) + 2 f(a.+ 1) = 3(a+ 1)2 - 01+ 2+2=12. =12+2+2=16- =302+a+2. 1)+2=3(a2+2a+1)—a—1+2=362+6a+3-a+l=3a2+5o+4. 2f(a.) .—. 2-_f(a)=2(3o.2---a,+2)226.12 —2a+4. ma) = 3(2a)2 — (2a) + 2 _.= 3(4a2) — 2a + 2 = 12c»2 — 20 + 2. W) = 302?)? — (a2) + 2 = 3(a4)—a2+2=3a‘-a2+2- [f(a)1”= [3aa — a + 212 = (3a2 — a+ 2) (3a? — a+ 2) = 9a“ — 3a3 +6a2 — fie-Hi) =3(a+h)2 —(a+ 35. f (t) = t2 — fit is defined for all real numbers, so the domain is R, or (—00, 00). The graph of f is a parabola opening upward since the coefficient of t2 is positive. To find the t-intercepts, let y = D and solve fort. 0=t276t=t(t—6) =,- 36. 11(1) = ___4 ‘ t2 = (““2 ‘ ‘} 2—t Z-t is {t | t 75 2}. So the graph ofH is the same as the graph ofthe function f(t) = t + 2 (a line) except for the hole at (2, 4). 23+1 if2x+120 38.F(n=)=|2$+1|= _(2m+1) §2x+f<1 . 2:1:+1 fizz—é = —2a:—1 1fx<—-;~ The domain is R, or (—00, 00)- “ () M S‘ [I x ifmZO h . 2:=——-. moex: ,we ave g x” —z ifx<0 z . 1 . -— 1fx>0 — lfa:>0 2:3 I 9(r)= _$ _ = 1 _ F Lfm<0 —_—Ilfz<0 Note thatgis not defined form: t = D andt = 6. The t-coordinate of the vertex is halfway between the t-intercepts, that is, at t = f(3) = 32 -- 6-3 = «9, the vertex is (3, —9). 3a3+a2—2a+6a2—2a+4=9a4—6a3+'13a2—4a+4. h)+2=3(a2+2ah+h2)—a—h+2=3a2+6ah+3h2ma—h+2. 3. Since ,sofort9é2,H(t) =2+t. Thedomain 0. The domain is (—00, O) U (0, oo). 3—%:c'if1:52 2x—5 ife>2 The domain is R. 4:. fix) = { x+% if .7:<-—3 44. f(.7:)= —2z if ImIS3 —6 if 2:)3 Notethatforx= —3, bothx+93nd ~29: areequaltofi; and form =3, both —2.2 and —631eequalto —6. Domainiis. 31;) and ($2,312) is m = :2 _ Lu and an equation ofthe Line connecting those two points is y -— y1 = m(z - 1:1). The slope of this line segment is 7 5— (513) = 3, so an equation is y— (—3): 5(3— 1). Thefimctionisflz) = gx+ %,1 52: 55. 46. The slope of this line segment is ‘10 — 10 5 ——___ -— __... ' ' ._ _-— _§ _. _ ( 5) — 3,soanequationlsy 10 six ( 5)]. The function is f(:1:) = —§:e+ g, —-5 < a: < ?. 47. We need to solve the given equation for y. x + (y —1)2=0 ep(y—1-)‘=—z e» y—1=:l: -a: y = 1 :l: ‘/-—:r:. The expression with the positive radical represents the top half 0 f the parabola, and the one with the negative radical represents the bottom half. Hence, we want f (1:) = 1 — \/—3. Note that the domain is :1: S 0. 43.32+(y_2)2=4 es (y—2)2=4—a:2 (e, y—-2=:L—-\/4—:? e: y=2:|:\/4—-—-F.Thetophaifisgivenby thefimctionflm)=2+v4—m2,—25m52. as. f(x) = 32:]. 554(1): I f(-:I:)= (753% =I;:1 =-—$2fl:"'_1 =—f(:c). f(..x)__fii):___i.__f(m)_ So f is an odd function. 67.f(:c)=$:l,sof(—a:)=_;:1=mi1. 68.f(:c)=a:[z|. Since this is neither f(:z:) nor —f(m), the functionf is “—3) = ('3?) |_I| = (—3) lxl = a—(xlxl) neither even norodd. .= —f(1') So f is an odd function. 1.2 Mathematical Models: A Catalog of Essential Functions 3. We notice from the figure that g and h are even fimctions (symmetric with respect to the y-axis) and that f is an odd function (symmetric with respect to the origin). So (b) [y = 2:5] must be f. Since 9 is flatter than h near the origin, we must have (c) [y = 3:8] matched with g and (a) [y = 3:2] matched with h. 4. (a) The graph of y = 3a: is a line (choice G). (b) y = 3s is an exponential function (choice f). (c) y = 2:3 is an odd polynomial fiinction or power fimction (choice F). (d) y = 3 x = rel/3 is a root function (choice 9). 5. (a) An equation for the family of-linear functions with slope 2 is y = f(a:) = 29: + b, where b is the y—intercept. 9. Since f(—l) = f(0) = f(2) = 0, f has zeros of—l, 0, and 2, so an equation for f is f(::) = a[:c — (—1)](sr. — 0)(:c — 2), or fix) = aa:(n': + 1)(z — 2). Because f(1) = 6, we’ll substitute 1 for wand 6 for flat). 6 = a.(1)(2)(—1) => —2a = 6 => 0. = -—3, so an equation for f is f(cr:) = —3a:(a: + 1)(a: — 2). a. The vertex of the parabola on the left is (3,0), so an equation is y = o(z — 3)2 + 0. Since the point (4, 2) is on the parabola, we’ll substitute 4 for z: and 2 fory to find a. 2 = a.(4 - 3)” => a. = 2, so an equation is ffm) = 2(2 — 3)2. The y-intercept of the parabola on the right is (0, 1), so an equation is y = an” + be + 1. Since the points (-2, 2) and (1, —2.5) are on the parabola, we’ll substitute -2 for as and 2 for y as well as 1 for z and —2.5 for y to obtain two equations - with the unknowns a and 6. (—2,2): 2=4a—2b+1 => 4o—2b=1 (1} (1,—2.5): —2.5=a+b+1 => a+b=-—3.5 (2) 2- (2) + (1) gives us 60. = —6 :- a = —1. From (2), —l + b = -35 => b = —2.5, so an equation is g(m) = —:52 — 2.51: + 1. 1.3 New Functions from Old Functions 3. (a) (graph 3) The graph of f is shified 4 units to the right and has equation yr: f(.r — 4). (b) (graph 1) The graph of f is shifted 3 units upward and has equation y = f (as) + 3. (c) (graph 4) Title graph of f is shrunk vertically by a factor of 3 and has equation y = g f (:r). (d) (graph 5) The graph off is shifted 4 units to the left and reflected about the x-axis. Its equation is y = —f(:c + 4). (e) (graph 2) The graph of f is shifted 6 units to the Ielt and stretched vertically by a factor of 2. [ts equation is y = 2 f (I + 6). G. The graph of y = f(.1:) = v3z — 2:9 has been shifled 2 units to the right and stretched vertically by a factor of 2. Thus, a flinction describing the graph is y=2f(;c—2)=2 3(3—2)—(:r-—2) =2 3:2:—-6—(:t:2—4a:+4)=2v-—x2+7:c—10 7. The graph of y = f (m) = v3x — :52 has been shified 4 units to the left, reflected about the rc-axis, and shined downward 1 unit. Thus, a function describing the graph is y: ‘-1- f(m+4) - 1 V W ‘—«-’ reflect shift shift about z—axis 4 units left 1 unit lefi: This fimction can be written as y=_f(;c+4)—1=—‘/3(m+4)—(5c+4)2—1=— 3w+12-(m2+8$+16)-—l=—\/—a:2—5rc—4-—1 13. y = 1 + 2 cos 3:: Start with the graph of y = cos :c, stretch vertically by a factor of 2, and then shift 1 unit upward. y=2cosx J 14. y = 4 sin 3x: Start with the graph of y = sin 3:, compress horizontally by a factor of 3, and then stretch vertically by a factor of 4. 24. y = It2 —221 = lz” —2:c+1 - 1| = [(a: w 1)2 — 1]: Startwith thegraph ofy = 32, shift 1 unitright, shift 1 unit downward, and reflect the portion of the graph below the z-axis about the :c-axis. 27. (a) To obtain y = f(|-’£|), the portion of the graph of y = f(a:) to the right of the y-axis is reflected about the y-axis. (b)y=sinlw|- (C)y=\/|rb‘| 33. f (x) = 1 —~ 3x“: g(ac) = cos 2:. D = R for both f and g, and hence for their compositesf (a) (f 09%?) = f(9(w)) = “00593) = 1 — 300”- (b) (9 0 f)(==) = 9(f(z)) = 9(1 - 3m) = 0060 - 335)- (c) (fof)(:t:) =f(f($)) =f(1—3:L') = 1—3(1-3.r) = I —3+92=9:c—2. (d) (g og)(:r) = g(g(:c)) = g(cos 1:) = oos(cos:r) [Note that this is not cosz - cos 1.] 34. f(x) = J5, D = [0, oo); g(:r) = {VI—:5, D 2R. (a) (foam) = f(9($)) = f(x71_:i)= \t’m: {VI—:5. Thedomainoffogis{a:| m20}={z]1~x20}={:c|m$1}=(—oo,l]. (b) (90.00?) = 90619)) = 9N5) = v3 1 — fi. Thcdomainofgofis{mIzisinflnedomainoffandf thatis, [0,00). (c) (fof)(a:) = f(f(1:)) = mm = M: ya. ThedornainoffOfis{z t x 2 Oandfiz 0} = {0,00}. (d) (g og)(a:) =g(g(w)) = 9(‘3/1 — m) = \3/1 -— «3/1 — m, and the domain is (—oo,oo). (x) is in the domain ofg}. This is the domain off, 31r+21rn => z¢3fi+m (“fining“) ’ '21— (mounts)=g(t(z))=.c.r(1+m 1+ ( I.) (1 ) 3 +3 1: :1: (c)(f°f)(a=)=f(f(x))=f(1: )=—ii—t—=———1i§——1——)=Hm+z=2m+l I 1+1+I (1+1+:r)'( +3; . = _1 Since fix) is not defined fora: = —1, and f(f(:r:)) isnot defined fora: 2, the domain of(fof)(:s) is D = {35] 2: ¢ —1,-—§}. (d) (g 0 9X9) = 9(9(I)) = 9(8'11123) = sin(2 511122) Domint R 4a. (raga now) = tome») = rows» = f( 3 V5 1) = tan( 3/? 1) h. 49. Let h{x) = J5, 9(1) = sects, and fix) = :4. Then (f as o me) =__ tome)» = flak/5)) = f(secfi) = (secfil‘ = sec“ (fl) = Hos)- 1.5 Exponential Functions 1. (a) f (1:) = a”, a > 0 (b) R (c) (D, 00) (CD See Figures 6(c), 6(1)), and 6(a), respectively. 2. (a) The number e is the value of a. such that the slope of the tangent line at a: = 0 on the graph of y = a” is exactly 1. (b) e 2 2.71828 (c) fix) = 6' 7. We start w‘th the graplfbfy = 4‘ (Figure 3) and then shifi 3 units downward. This shift doesn’t afi'ect the domain, but the range ofy = 4‘ — 3 is (—3,oo). There is a horizontal asymptote of y = —3. B. We start with the graph ofy = 4” (Figure 3) and then shift 3 units to the right. There is a horizontal asymptote of y = 0. 9. We start with the graph ofy = 2I (Figure 2), reflect it about the y-axis, and then about the z-axis {or just rotate 180° to handle both reflections) to obtain the graph of y = *2“. In each graph, y = 0 is the horizontal asymptote. y = —2“ ...
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