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Unformatted text preview: 136 I: CHAPTERZ LIMITS AND DERIVATIVES 48. (a) T (c) dT/dl 0 t (b) The initial temperature of the water is close to room temperature because of the water that was in the pipes.
When the water from the hot water tank starts coming out. dT/dt is large and positive as T increases to the
temperature of the water in the tank. In the next phase. dT/dt : 0 as the water comes out at a constant. high
temperature. After some time. dT/dt becomes small and negative as the contents of the hot water tank are
exhausted. Finally. when the hot water has nm out. dT/dt is once again 0 as the water maintains its (cold)
temperature. 49. In the right triangle in the diagram. let Ay be the side opposite angle ()5 and Arc the side adjacent angle (15. Then the slope of the tangent line 6 is
m : Ay/Asc : tan d). Note that 0 < qb < We know (see Exercise 19)
that the derivative of f(:c) = m2 is fβ(ac) : 230. So the slope of the tangent to the curve at the point (1, 1) is 2. Thus. (15 is the angle between 0 and g whose tangent is 2; that is. d) : tanβ1 2 3 63Β°. 2 Review
CONCEPT CHECK ____ββββββββ 1. (a) lim : L: See Deο¬nition 2.2.] and Figures 1 and 2 in Section 2.2. IββHl (b) lim : L: See the paragraph after Deο¬nition 2.2.2 and Figure 9(b) in Section 2.2. CEβHZ+ (c) lim f(m) : L: See Deο¬nition 2.2.2 and Figure 9(a) in Section 2.2. 12βΒ»(1β (d) lim ο¬an) : 00: See Deο¬nition 2.2.4 and Figure 12 in Section 2.2. Iββa (e) lim ο¬at) 2 L: See Deο¬nition 2.6.] and Figure 2 in Section 2.6. zβmo 2. In general. the limit of a function fails to exist when the function does not approach a ο¬xed number. For each of the following functions. the limit fails to exist at a: : 2. The leftβ and rightβhand There is an There are an inο¬nite limits are not equal. inο¬nite discontinuity. number of oscillations. 3. (a)β(g) See the statements of Limit Laws 1β6 and l l in Section 2.3. 4.
5. 10. 11. 12. 13. CHAPTERZ REVIEW 3 137 See Theorem 3 in Section 2.3.
(a) See Deο¬nition 2.2.6 and Figures 12β14 in Section 2.2.
(b) See Deο¬nition 2.6.3 and Figures 3 and 4 in Section 2.6. (a) y I 1843 N0 aSymPtOIC (b) y : sin at: No asymptote (c) y = tan :3: Vertical asymptotes a: = g + 7m. (d) y = tanβ1 z: Horizontal asymptotes y : :: Ml: n an integer
(e) y : 6β: Horizontal asymptote y = 0 ( lim er : 0) xο¬βOO (f) y 2 ln 1': Vertical asymptote an : 0 ( lim Ina; : ~00)
:cβAOβl'
(g) y : l/x: Vertical asymptote :c : O. (h) y = ο¬: No asymptote horizontal asymptote y = 0 (a) A function f is continuous at a number a if f(m) approaches f(a) as a: approaches a; that is.
gig; am) 2 f(a)~ (b) A function f is continuous on the interval (~00, 00) if f is continuous at every real number a. The graph of
such a function has no breaks and every vertical line crosses it. See Theorem 2.5.10.
See Deο¬nition 2.7.]. See the paragraph containing Formula 3 in Section 2.7. (a) The average rate of change of y with respect to :0 over the interval [231. m2] is 2 β 1 . , . . _ 33 ~ 93
(b) The instantaneous rate of change ofy With respect to a: at m = x1 15 hm See Deο¬nition 2.8.2. The pages following the deο¬nition discuss interpretations of fβ(a)
line to the graph off at x 2 a and as an instantaneous rate of change of f($) as the slope of a tangent
with respect to a: when m = a. (a) A function f is differentiable at a number a if its derivative fβ exists at x = a; that is. if f/(a) exists. (b) See Theorem 2.9.4. This
(C) theorem also tells us that if f is not continuous at a. then f is nor differentiable at a. 14. See the discussion and Figure 8 on page 172. a TRUEFALSE QUIZ \ 1. False. Limit Law 2 applies only ifthe individual limits exist (these donβt). 2. False. Limit Law 5 cannot be applied if the limit of the denominator is 0 (it is).
3. True. Limit Law 5 applies. 4. True. The limit doesnβt exist since doesnβt approach any real number as x approaches 5.
(The denominator approaches 0 and the numerator doesnβt.) 138 C CHAPTERZ LlMlTS AND DERIVATIVES 5. False. Consider lim M or hm Β§m(_:E_βi) mas x _ 5 was :6 _ . The ο¬rst limit ex1sts and 18 equal to 5. By Example 3 in Section 2.2. we know that the latter limit exists (and it is equal to 1). 1 $β 6. False. Consider : lim 7 6) mβ>6 . It exists (its value is 1) but f(6) = 0 and g(6) does not exist. so f(6)g(6) 79 1. 7. True. A polynomial is continuous everywhere, so lim p(z) exists and is equal to p(b). xβvb
. . . . 1 1 . . . . . . .
8. False. Consrder llni) [f i : 1111i) g2β β E; . This limit is β00 (not 0). but each of the 1nd1v1dual functions approaches 00.
9. True. See Figure 4 in Section 2.6. 10. False. Consider f = sinrc for :3 Z 0. lim f Β’ :Eoo and f has no horizontal asymptote. 15β100 1 x β 1 if a: 1
11. False. Consider f(m) = ) 7Γ©
2 if :c : 1 12. False. The function f must be continuous in order to use the Intermediate Value Theorem. For example, let 1 if 0 g at < 3
f(9:) : There is no number c E [0. 3] with f(c) = 0.
β1 if :c = 3 13. True. Use Theorem 2.5.8 with a = 2. b : 5, and g(m) : 4:32 β 11. Note that f(4) = 3 is not needed. 14. True. Use the Intermediate Value Theorem with a : β1. b : 1. and N : 7r. since 3 < 7r < 4. 15. True. by the deο¬nition of a limit with 5 = 1. x2 + 1 if :c gΓ© 0 16. False. For example. let f = .
2 if :c : 0 Then f(9:) > 1 for all cc. but lirrbο¬x) : lim (m2 + 1) : 1. ticβ>0 17. False. See the note after Theorem 4 in Section 2.9. 18. True. fβ(r) exists => f is differentiable atr :> f is continuous atr => lim f(ac) : f(r). IβPT ____"ββββ EXERCISES / ' ' : β 1' = 0
1. (a) (1) Β£13; f(:c) 3 (n) #5113 + f (:16)
(iii) lim3 f does not exist since the left and right (iv) f (:c) = 2 limits are not equal. (The left limit is β2.) (v) lim = 00 (vi) lim f(a:) : 700
Eβ'0 wββr2'
(vii) lim : 4 (viii) lim : β1
:tβroo ma~oo
(b) The equations of the horizontal asymptotes are y = #1 and y : 4. (c) The equations of the vertical asymptotes are a: : 0 and a: = 2.
(d) f is discontinuous at an 2 β3. 0, 2. and 4. The discontinuities are jump, inο¬nite. inο¬nite, and removable. respectively. 10. 11. 12. 13. 14. 15. 16. 3 $β>1 32β9 0 2
. . . . , av β 9
Since rational functlons are continuous. 11m 2
was x + 29c β 3 rm x24) _ hm (m+3)(mβ3)_ hm xβ3_73β3_β_6
zβ1Β»73:c2+2:1: 3βz>~3(:c:3)(a: 1)_r>3x 17 3 1β 4
. x279 . 2 + 502β9
mgrn+x2+2m_3ββm31ncex +2zβ3β>0aszβ>1 21mm
1 (h_1)3+1 . (h3β3h2+3hβ1)+1 , h3β3h2+3h
11m ~11m β11m%
hββO h hο¬O h hβ~>0 Another solution: Factor the numerator as a sum of two cubes and then simplify. 232+2(3)β3_12 : lim
hβvO CHAPTER 2 REVIEW lim ex '2 : 61β1 = 60 =1. β0. <0f0r1<x<3. β 3 ~ 3 3 hβl 1 h~12β1hβ1 12
limW:limmeth ) ( )+ ]
hβ>O h_,0 hο¬o h :lim[(hβ1)2~h+2]:1_0+2:3 hβ>O
limtgcβlehm (t+2)(tβ2) _lim t+2 _ 2+2 _3
Hβ3β8_β*2(t*2)(t2+2t+4)_t~2t2+2t+4_4+4+4β12 x/F W 3. Since the exponential function is continuous. (h2β3h+3) =3 3 lg(r_g)4zoosince(rβ9)4β10asrβΒ»9and(T~9)4>0forrΒ’9.
4β21 4β1) 1
.1311 4~v viiiβ(4%) .1311 β1 1
lim4_\/E:lim 4ββ; :lim _1 2β51:β1
Hlesβw 5β.16(\/Β§+4)(\/Β§β4) Salem/5+4 \/16+4 8
2
2 _ _
hm'u β: v 8_1im (v+4)(v 2) _lim n+4 2 2+4 :3
H2 11β16 vβΒ»2(v+2)(vβ2)(v2+4) v_.2(v+2)(v2+4) (2+2)(22+4) 16
mβS .
$Hsla x_8 1fmβ8>0 {1 ifx>8
β8 _ _ Β₯ : _ ..
as (w 8) if$_8<0 11tx<8
xβ8
. [reβ8 .
Th ,1 :1 β 2β.
"8.33...8 .331 1) 1
β13+ (x/mβ9+ο¬ac+1]]): lirgi+\/xβ9+ lim+[[x+1]]:\/9β9+10210
mβi acβ> xβi9
hmlβx/lβccz 1+x/1βx2 l. 1β(1βa22) 1, m2 l. x
k'xο¬ 1m 1 β
M m hut/11302 m*Β°2:(1+\/1~x2) z$x(1+\/1_x2) I%1+\/_β1βx2
hm v$+2~ x/Zm x/cc+2+\/2x hm β(wβ2) 1. β1
Kβ β 1m :β
1β0 33(15β2) x/m+2+\/Β§x z*2m(xβ2)(\/x+2+\/2x) IβQx x+2+\/2a:) 139 0 140 3 CHAPTERZ LIMITS AND DERIVATIVES , 2 1 _ 2 2 2 A
17. limlβο¬Scβi22hmο¬273βiο¬zhm 1/30 +2/a: 1_0+071__1
mewβ934F296 Wo<>(1β:c+2mz)/vc2 Wool/whimssz 0β0+2 2 3_ 2 3_ 2 3 3
18. um L13: m M: hm Sal/mm _5β0+0w 5
xβPoo 2x3+xβ3 zaβoo (2m3+wβ3)/m3 zβvβoo2i1/m2β3/333ββ2βh0β0ο¬i 19. Since a: is positive. Vzc : 2 cc. hm x/xQβQ _ hm x/mzβg/x/P hm x/1β9/m2 x/lβO 7 1
woo 2xβ6 woo (2xβ6)/a: Us.Β» 2β6/zc β 2β0 β2 20. lirlnr ln(10() β x2) : #00 since as a: β> 10β. (100 β x2) ββ> 0+. 21. lim e_3ββ = 0 since β3:c β> β00 as :c β> 00 and lim eIt 2 0. mace tβtβoo 22. Ify = x3 β :1: = $(sc2 β 1). then as a: β+ 00,1; β> 00. lim arctan(x3 β cc) 2 lim arctany : g by (2.6.4). xβroo yβaoo 23. From the graph of y = (cos2 x) /x2, it appears that y : 0 is the horizontal asymptote and x : 0 is the vertical asymptote. Now 0 3 (cos x)2 S 1 0 cos2 :c 1 cos2 av 1 .
:> ββ3 3β :> OS 3β.But 11m 0:0and
:62 Β£62 $2 x2 x2 mβvzβoo
, 1
11m ββ : 0, so by the Squeeze Theorem.
Iβ>ioo x2 _l
cos2 3: cos2 3:
lim : 0. Thus. y 2 0 is the horizontal asymptote. lim 2 00 because cos2 (B β+ 1 and x2 β> O as
zβrioo 362 1'40 1'2 a: β> 0, so a: : 0 is the vertical asymptote. 24. From the graph of y = f : V132 + w + β V152 β x, it appears that there are 2 horizontal asymptotes and
possibly 2 vertical asymptotes. To obtain a different form for f . letβs multiply and divide it by its conjugate. ./2 1/2_
f1(w)β(\/x2I$+1 $2 at) x+$+1+ m m
x/x2+ac+1+\/m2βm (x2+$+1)β($2βLE) 2m+1 Jz2+x+1+ο¬2βm \/$2+m+1+\/x2βar Now
1. f ( ) hm 2:0 + 1
1m :3 ο¬
3*β 1 mβ'Β°Β°\/:132+:c+1+\/ac2~11:
β lim 2 + (since V562 2 ac form > 0)
Ho \/1 +(1/m) +(1/w2)+ \/1 (1/:c)
2
: ββ : 1β
1+ 1
so y = 1 is a horizontal asymptote. For x < 0, we have V732 : $ : ~33, so when we divide the denominator by :c, with so < 0, we get 2 2_ 2 2β 1 1
\/:c +zc+1+\/:C m_ x/x +m+1+\/; $__[I1+l+_Β§+./1__]
a: Β«$2 :3 x a: 25. 26. 27. 28. 29. 30. 31. CHAPTER 2 REVIEW 141 Therefore.
1. f ( ) 1. 2:1: l 1
90β12100 1 a: 12200 x/x2+m+1+\/m2 βx
2 1 x 2
βlim +(/) 2Β₯(1+1)_β1.
Hβ ~ [\/1+(1/$)+(1/$2)+ [1 ~<1/x>]
so y 2 β1 is a horizontal asymptote. As a: β> 0β. f β> 1. so a: 2 0 is not a vertical asymptote. As m β> 1+, f β; so :1: 2 1 is not a vertical asymptote and hence there are no vertical asymptotes. 4
,1010 β4 Since 2m β 1 g f(m) g 132 forO < x < 3and lim (2x ~1)212 lim :32. we have lim f($) 2 1by the zβtl mβbl acβ>1 Squeeze Theorem. Let ο¬an) 2 β$2. g(x) 2 m2 cos(1/w2) and h(:c) 2 932. Then since lcos(1/rc2)l g lfor x 75 0. we have
f S g(m) g h(;l:) for it 75 0. and so ling) f(z) 2 lirrbhm) 2 O 2> linbgh) 2 0 by the Squeeze Theorem. Given5> 0.weneed6 > 050thatif0 < lavβ5 <6.then [(7:17~ 27) *8 <5 <=> [7.xβ
[zβ β 5! < β¬/7.Sotake6 25/7.Then0 < Ix β5 < 6 2> (7:Lββ 27) β8 < 5. Thus. lim mβs5 35<e <:>
(7xβ27)28 by the deο¬nition of a limit. Givens > Owemustο¬nd6> OsothatifO < lavβ0 < 6.then [\3/_β0 <6. Now\3/_β0 2 <5 2
[ml 2 <53.Sotake62β¬3. ThenO < [xβOl 2 <63 2 IVEβOI 2 2 \3/lml < \3/6β28.
Therefore. by the deο¬nition ofalimit. lim \3/_ 2 0. 2β0 Given5>0. we need6>0sothatif0< lanβ2l <6. then [:62 β3mβ(β2)[
then~1<xβ2<1.soO<mβ1<2 =>
:> [$223x2(22) : (m22)(mβ 1) Thus, gig; (:2 β 33:) < 5. First. note that if m β 2' < 1.
[:L' β 1] < 2. Now let 6 2 min {5/2, 1}. Then 0 < [9: β 2] < 6
2 lac β 2 [x β 1 < (β¬/2)(2) 2 a. β2 by the deο¬nition of a limit. Given M > 0, we need 6 > 0 such that ifO < a: β 4 < 6. then 2/\/:c β 4 > M. This is true <=>
Masβ4 < 2/M <:> $24 < 4/M2. SoifwechooseΓ© = 4/M2.then0 < mβ4 < 6 :> 2/y/xβ4 > M.
So by the deο¬nition ofa limit. lim (2AM β 4 Wβ ) 2 00. (a) f(:c)2Hifx<0.f(w)23βxif0Sx<3.f(m)2(zβ3)2ifx>3. (i) β151+ f($) 2 llI(I)1 (3 β z) 2 3 (ii) lim 2 lim Vβx 2 0
z__, I__, zβyOβ zβVOT
(iii) Because of (i) and (ii). lirnJ does not exist (iv) lim 2 Iim (3 ~ in) 2 0
3* zβb3β IHBβ (v) Iim zβ>3+ lim (9:232 :0 3+ (vi) Because of (iv) and (v), lint;g ο¬at) 2 0. 142 I: CHAPTERZ LIMITS AND DERIVATIVES (b) f is discontinuous at 0 since lim f(9:) does not exist. 1H0 f is discontinuous at 3 since f(3) does not exist. 32. (a)g(ac):230β302ifOSxS2.g(z):2βmif2<xΒ§3.
g(ac) : ac β 4 if3 < a: < 4. g(:c) = 7r ifs: Z 4. Therefore.
lim g(z) : lim_ (2x β x2) : 0 and :tβr2β acβvZ
$133+ g(z) = $133+ (2 β ac) : 0. Thus. 9(20) : O : g so g
is continuous at 2. lim g(:c) = lim (2 7 m) : β1 and xβrBβ zβr3β lim+ g(ac) : lim β 4) = β1.Thus. ling 9(13) : β1 = 9(3). so 9 is continuous at 3. $43 $423+
lim 9(22) : lim (cc β 4) : 0 and lim 9(33) : lim 7r : 7r. Thus. lim 9(23) does not exist. so 9 is
xβr4β xβΒ»4* x_.4+ z_.4+ za4
discontinuous at 4. But lim+ g(x) : 7r : 9(4). so 9 is continuous from the right at 4.
xβv4 33. sin .7: is continuous on IR by Theorem 7 in Section 2.5. Since eaβ is continuous on IR. eβ x is continuous on R by Theorem 9 in Section 2.5. Lastly, cc is continuous on R since itβs a polynomial and the product mesiβ m is continuous
on its domain R by Theorem 4 in Section 2.5. 34. x2 β 9 is continuous on IR since it is a polynomial and ο¬ is continuous on [0. 00). so the composition V362 β 9 is continuous on {cc l :02 β 9 Z 0} = (β00. *3] U [37 00). Note that x2 i 2 Β’ 0 on this set and so the quotient 532β9 function g(m) : is continuous on its domain. (β00. β3] U [37 $2 7
35. f(:c) : 2:63 + $2 βIβ 2 is a polynomial. so it is continuous on [β2. 71] and f(~2) = #10 < 0 < 1 : f(β1).So
by the Intermediate Value Theorem there is a number 6 in (β2. β1) such that f (c) : 0. that is, the equation
2$3 + x2 + 2 = 0 has aroot in (#2., #1). 36. f(:c) = 6β952 # cc is continuous on IR so it is continuous on [0.1]. f(0) :1 > 0 > l/e  1 : f(1). So by the 2 2
Intermediate Value Theorem. there is a number 0 in (0, 1) such that f(c) : 0. Thus. 6β β cc 2 0. or eββ : x. has a root in (0. 1). 37. (a) The slope of the tangent line at (2, 1) is β A 2β 7 3 β2 2#4 _ β2 *2 2 limwzlimw:lim8 21 :limβEβthm (a: )(x+ ) 503 12β2 1H2 $*2 Iβ>2 (13β2 1H2 x72 zβr2 33β2
=1im2[2(m+2)]:A24:~8 (b) An equation of this tangent line is y β 1 : #8(sc # 2) or y z #836 + 17.
38. For a general point with :cβcoordinate a. we have 2/(1β3w) β2/(1;3a) 2(1 #3a) 2(1β3:c) # . _ 1. _,_______
m :13: aka $13}. (1β3a)(1β3m)(wra)
i hm 5(56 a) 7 lim βββββd β I a(]. 3a)(1 3x)(x a) mΒ»a(1 3a)(1 33:)_(1β3a)2 CHAPTER 2 REVIEW 143 β0)ory269:+2.For
1 For a 2 0. m 2 6 and f(0) 2 2. so an equation of the tangent line is y β 2 2 6(:c
Β§(m+ )ory2 gmlβ g. a 2 71. m 2 g and f(β1) 2 so an equation of the tangent line is y β Β§ 2
39. (a) s 2 3(t) 2 1 + 2t + t2/4. The average velocity over the time interval [1. 1 + h] is
_ 8(1+h)_8(1) 1+2(1+h)+ (1+h)2/4213/4 10h+h2 10+h (1 + h) β 1 h 4h 4 '
So for the following intervals the average velocities are: (i) [1,3]: h : 2.7;β = (10 + 2)/4 = 3 m/s (ii) [1. 2]: h 21.2;ave 2 (10 +1)/4 : 2.75 m/s
(iii) [1.1.5]: h = 0.5. vave : (10 + 0.5)/4 = 2625 m/s
(iv) [1.1.1]: It 2 0.1.21ave 2 (10 + 0.1)/4 2 2525 m/s ave 1 β 1 1 h 10
(b) When t 2 1. the instantaneous velocity is lim w 2 lim 0β+ 2 β 2 2.5 m/s
i120 h hβ>0 4 4 40. (a) When V increases from 200 in3 to 250 in3. we have AV 2 250 i 200 2 50 in3. and since P 2 800/V. AP 2 P(250) i P(200) 2 3β28 β 3β33 2 3.2 β 4 2 β0.81b/in2. So the average rate of change is
AP β0.8 Ib/m2 β 2 β 2 ~ . 16 . AV 50 0 0 in3 (b) Since V 2 800/ P. the instantaneous rate of change of V with respect to P is . AV _ . V(P + h) β V (P) _ . 800/(P + h) β 800/P
limo E V1335 h V135 h
_ hm 800 [P 2 (P + h)] _ mm β800 _ _@
* hβΒ»0 h(P + h)P a 1:20 (P + h)P _ P2 which is inversely proportional to the square of P. , _. f(a:)βf(2)_l $3β2JJβ4
βWο¬mβiiβΓ©ο¬βmο¬
. (mβ2)(:c2+2a:+2)
211m (b)yβ4210($β2)ory210mβ16 42. 26 2 64, so 2 $6 and a 2 2. 43. (a) fβ(r) is the rate at which the total cost changes with respect to the interest rate. Its units are
dollars / ( percent per year). (b) The total cost of paying off the loan is increasing by $1200/ (percent per year) as the interest rate reaches 10%.
So if the interest rate goes up from 10% to 11%. the cost goes up approximately $1200. (c) As r increases. C increases. So fβ(r) will always be positive. 144 3 CHAPTERZ LIMITS AND DERIVATIVES For Exercises 44β46, see the hints before Exercise 5 in Section 2.9, 44 45. 47 (a) f/(aj) β hm MiltββΒ£1 rm \/3β5($+h) β x/3β5m \/3β5(m+h) +β/β3_5x ' _ M h _ bio h x/sβsec +7) + m
! hm [3 7 5(30 + h)] β (3 β 5:0) lim β5 _ β5 h*Β°h( 3β5(x+h)+\/BT5Β§) h"Β°\/3β5(x+h)+\/3β5m Miaβsac (b) Domain of f: (the radicand must be nonnegative) 3 β 5:0 2 0 :> 5x 3 3 => as 6 (A00, Domain of f' : exclude % because it makes the denominator zero; :3 E (β00, (c) Our answer to part (a) is reasonable because f l(sic) is always negative and f is always decreasing. 48. (a) Asw β> :00, ο¬ne) : (4 # $)/(3+m) β+ β1. so there isa horizontal asymptote at y = #1. As a: β> β3+, f(a:) β> oo. and as :c β> 3β, f(a:) β> β00. Thus, there is a vertical asymptote at a: : β3. CHAPTERZ REVIEW U 145 (b) Note that f is decreasing on (700. *3) and (β3, 00). so f'
is negative on those intervals. As :0 β> 5:00, fβ β+ 0. As at β> 73β and asx β> β3+. fβ β> βoo. 4~(m+h)_4β:L'
3+(x+h) 3+:c f(x+h) f(w) <c>f'<w>=m h h
_ hm (3+:c)[4β(x+h)]β(4~$)[3+(a:+h)]
βhββ>O h[3+(m+h)](3+x)
_1, (12β3xβ3h+4$7m2ihw)β(12+4m+4hβ3mβm2βhx)
113% h[3+(x+h)](3+x)
β7h β7 7 :lmmuο¬hnomβi%[3+<x+h)i<3+m)= may (d) The graphing device conο¬rms our graph in part (b). 49. f is not differentiable: at x : β4 because f is not continuous. at m = β1 because f has a comer. at m : 2 because
f is not continuous. and at :c = 5 because f has a vertical tangent. 50. (a) Drawing slope triangles, we obtain the following estimates: Fβ(1950) is, Q 0.11, 10 :
F'(1965) w $06 : ~0.16. and F'(1987) m 01+; : 0.02. (b) The rate of change of the average number of children born to each woman was increasing by 0.11 in 1950,
decreasing by 0.16 in 1965. and increasing by 0.02 in 1987. (c) There are many possible reasons: 0 In the babyboom era (postWWII). there was optimism about the economy and family size was rising. o In the babyβbust era. there was less economic optimism. and it was considered less socially responsible to
have a large family. 0 In the babyβboomlet era. there was increased economic optimism and a return to more conservative attitudes 51. Bβ(1990) is the rate at which the total value of US. banknotes in circulation is changing in billions of dollars per
year. To estimate the value of Bβ(1990). we will average the difference quotients obtained using the times t : 1985
andt : 1995' Let A I B(1985) ~ B(1990) I 1820 β 268.2 1985 ~ 1990 _5 = 17.24 and
3(1995) β 3(1990) 401.5 a 268.2
C 2 x : I . .
1995 β 1990 5 26 66 Then 8,0990) : hm B(t) β B(1990) A + C 17.24 + 26.66 ~ _
~ __ tβ>1990 t β 1990 2 2 : 21.95 billions of dollars/year. 146 C CHAPTERZ LIMITS AND DERIVATIVES 52. The slope of the tangent toy : 13+: is
x,
($+h)+1#$+1
hm (m+h)71 22β1_lim($β1)(m+h+1)β(ac+1)(a:+hβ1)
hβ+O h 11β.0 h(:cβ1)(a:+hi1)
lim _2h β 2
βhH0h($*1)($+h*1)β (93β1)2
2
Soat(2,3).mβ (2_1)2* 2 β5 yβ3β*2(sc#2) =>
β72a:+7 At(β10)mβββ2ββββl =>
y_ β " 7 (~1β1)Qο¬ 2 yzβΓ©mβIβl) => y=*%zβ%. 53. S g(:c) <=> 79(03) 3 f(:c) S g(x) and lim 9(50) 2 0 = lim βg(m). Thus, by the Squeeze Theorem, (IIββ11 (tβVa lim : O. xwt 54. (a) Note that f is an even function since f(x) : f(β;v). Now for
any integer n, + II~n]] : n 4 n = 0, and for any real number k which is not an integer,
[[kl] + [[719]] : + (β A 1) = #1. So lim f(.βt) exists EH04 (and is equal to β1) for all values of a. (b) f is discontinuous at all integers. ...
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This note was uploaded on 12/08/2009 for the course MATH 101 taught by Professor Dr.tahir during the Fall '08 term at King Fahd University of Petroleum & Minerals.
 Fall '08
 Dr.Tahir
 Math

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