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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 yvalues are equal. The yvalues 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 yvalues 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 ﬁInction since for z = 0, i1, and i2, there are inﬁniwa many points on the curve. 18(a) no) (b) no
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60! 50 r
:..r‘+3_2 m+3—2(:c+1)
26. f(3)—f(1)=$+1 = 1+1 __I+3—2m—2
z—l a:1 :c—l _ (a:+1)(:c—1)
= “3+1 —(a:—1) 1 (:1:+1)(a:—1) 2 (I+1)(:1:—1) ="z—ﬁ 23.f(a:)=(5m+4)/(z2+3x+2)isdeﬁnedforallxexceptwhenO:x2+3z+2 41> O=(a:+2)(:r+1) e: 32—2
or —1, so the domain is {:c e R  a: aé —2, 1} = (—00, —2) U (~2, —1) U (—1,oo). 21. f(:1':)= 3x2 — a: + 2.
f(2) =3(2)2 —2+2_—. 12
f(—2) = 3(2)2  (2) + 2
ﬁg) = 3a2 — a. + 2.
f(a) = 3mg: 4 (—a) + 2
f(a.+ 1) = 3(a+ 1)2  01+ 2+2=12.
=12+2+2=16 =302+a+2.
1)+2=3(a2+2a+1)—a—1+2=362+6a+3a+l=3a2+5o+4. 2f(a.) .—. 2_f(a)=2(3o.2a,+2)226.12 —2a+4.
ma) = 3(2a)2 — (2a) + 2 _.= 3(4a2) — 2a + 2 = 12c»2 — 20 + 2. W) = 302?)? — (a2) + 2 = 3(a4)—a2+2=3a‘a2+2 [f(a)1”= [3aa — a + 212 = (3a2 — a+ 2) (3a? — a+ 2) = 9a“ — 3a3 +6a2 — fieHi) =3(a+h)2 —(a+ 35. f (t) = t2 — ﬁt is deﬁned for all real numbers, so the domain is R, or (—00, 00). The graph of f is a parabola opening upward since the coefﬁcient
of t2 is positive. To ﬁnd the tintercepts, let y = D and solve fort. 0=t276t=t(t—6) =, 36. 11(1) = ___4 ‘ t2 = (““2 ‘ ‘} 2—t Zt is {t  t 75 2}. So the graph ofH is the same as the graph ofthe function
f(t) = t + 2 (a line) except for the hole at (2, 4). 23+1 if2x+120
38.F(n=)=2$+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 tcoordinate of the
vertex is halfway between the tintercepts, that is, at t = f(3) = 32  63 = «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 yaxis) 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 oflinear 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 yintercept 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 xaxis. 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 rcaxis, 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$+16)—l=—\/—a:2—5rc—4—1 13. y = 1 + 2 cos 3:: Start with the graph of y = cos :c, stretch vertically by a factor of 2, and then shift 1 unit upward. y=2cosx J 14. y = 4 sin 3x: Start with the graph of y = sin 3:, compress horizontally by a factor of 3, and then stretch vertically by a
factor of 4. 24. y = It2 —221 = lz” —2:c+1  1 = [(a: w 1)2 — 1]: Startwith thegraph ofy = 32, shift 1 unitright, shift 1 unit downward, and reﬂect the portion of the graph below the zaxis about the :caxis. 27. (a) To obtain y = f(’£), the portion of the graph of y = f(a:) to the right of the yaxis is reﬂected about the yaxis.
(b)y=sinlw (C)y=\/rb‘ 33. f (x) = 1 —~ 3x“: g(ac) = cos 2:. D = R for both f and g, and hence for their compositesf
(a) (f 09%?) = f(9(w)) = “00593) = 1 — 300”
(b) (9 0 f)(==) = 9(f(z)) = 9(1  3m) = 0060  335)
(c) (fof)(:t:) =f(f($)) =f(1—3:L') = 1—3(13.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}={:cm$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 yaxis, and then about the
zaxis {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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 Fall '07
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 Calculus

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