02_R - 136 I: CHAPTERZ LIMITS AND DERIVATIVES 48. (a) T (c)...

Info iconThis preview shows pages 1–11. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11
This is the end of the preview. Sign up to access the rest of the document.

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 f’(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) : tan’1 2 3 63°. 2 Review CONCEPT CHECK ____———————— 1. (a) lim : L: See Definition 2.2.] and Figures 1 and 2 in Section 2.2. I—‘Hl (b) lim : L: See the paragraph after Definition 2.2.2 and Figure 9(b) in Section 2.2. CE—HZ+ (c) lim f(m) : L: See Definition 2.2.2 and Figure 9(a) in Section 2.2. 12—»(1’ (d) lim flan) : 00: See Definition 2.2.4 and Figure 12 in Section 2.2. I—‘a (e) lim flat) 2 L: See Definition 2.6.] and Figure 2 in Section 2.6. z—mo 2. In general. the limit of a function fails to exist when the function does not approach a fixed number. For each of the following functions. the limit fails to exist at a: : 2. The left— and right—hand There is an There are an infinite limits are not equal. infinite discontinuity. number of oscillations. 3. (a)—(g) See the statements of Limit Laws 1—6 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 Definition 2.2.6 and Figures 12—14 in Section 2.2. (b) See Definition 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 = tan‘1 z: Horizontal asymptotes y : :: Ml: n an integer (e) y : 6”: Horizontal asymptote y = 0 ( lim er : 0) xfl—OO (f) y 2 ln 1': Vertical asymptote an : 0 ( lim Ina; : ~00) :c—AO‘l' (g) y : l/x: Vertical asymptote :c : O. (h) y = fl: 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 Definition 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 — 1 . , . . _ 33 ~ 93 (b) The instantaneous rate of change ofy With respect to a: at m = x1 15 hm See Definition 2.8.2. The pages following the definition discuss interpretations of f’(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 f’ 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 TRUE-FALSE QUIZ \ 1. False. Limit Law 2 applies only ifthe individual limits exist (these don‘t). 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 doesn’t exist since doesn‘t approach any real number as x approaches 5. (The denominator approaches 0 and the numerator doesn‘t.) 138 C CHAPTERZ LlMlTS AND DERIVATIVES 5. False. Consider lim M or hm §m(_:E_—i) mas x _ 5 was :6 _ . The first 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 $— 6. False. Consider : lim 7 6) m—>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). x—vb . . . . 1 1 . . . . . . . 8. False. Consrder llni) [f i : 1111i) g2— — E; . This limit is —00 (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 ¢ :Eoo and f has no horizontal asymptote. 15—100 1 x — 1 if a: 1 11. False. Consider f(m) = ) 7é 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. —1 if :c = 3 13. True. Use Theorem 2.5.8 with a = 2. b : 5, and g(m) : 4:32 — 11. Note that f(4) = 3 is not needed. 14. True. Use the Intermediate Value Theorem with a : —1. b : 1. and N : 7r. since 3 < 7r < 4. 15. True. by the definition of a limit with 5 = 1. x2 + 1 if :c gé 0 16. False. For example. let f = . 2 if :c : 0 Then f(9:) > 1 for all cc. but lirrbflx) : lim (m2 + 1) : 1. tic—>0 17. False. See the note after Theorem 4 in Section 2.9. 18. True. f’(r) exists => f is differentiable atr :> f is continuous atr => lim f(ac) : f(r). I—PT ____"—’—— EXERCISES / ' ' : ” 1' = 0 1. (a) (1) £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 —2.) (v) lim = 00 (vi) lim f(a:) : 700 E—'0 w—‘r2' (vii) lim : 4 (viii) lim : —1 :t—roo 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 —3. 0, 2. and 4. The discontinuities are jump, infinite. infinite, and removable. respectively. 10. 11. 12. 13. 14. 15. 16. 3 $—>1 32—9 0 2 . . . . , av — 9 Since rational functlons are continuous. 11m 2 was x + 29c — 3 rm x24) _ hm (m+3)(m—3)_ hm x—3_73—3_—_6 z—1»73:c2+2:1: 3—z>~3(:c:3)(a: 1)_r>3x 17 3 1— 4 . x279 . 2 + 502—9 mgrn+x2+2m_3——m31ncex +2z—3—>0asz—>1 21mm 1 (h_1)3+1 . (h3—3h2+3h—1)+1 , h3—3h2+3h 11m ~11m —11m% h—‘O h hflO h h—~>0 Another solution: Factor the numerator as a sum of two cubes and then simplify. 232+2(3)—3_12 : lim h—vO CHAPTER 2 REVIEW lim ex '2 : 61—1 = 60 =1. ‘0. <0f0r1<x<3. — 3 ~ 3 3 h—l 1 h~12—1h—1 12 limW:limmeth ) ( )+ ] h—>O h_,0 hfio h :lim[(h—1)2~h+2]:1_0+2:3 h—>O limtgc‘lehm (t+2)(t—2) _lim t+2 _ 2+2 _3 H“3‘8_‘*2(t*2)(t2+2t+4)_t~2t2+2t+4_4+4+4‘12 x/F W 3. Since the exponential function is continuous. (h2—3h+3) =3 3 lg(r_g)4zoosince(r—9)4—10asr—»9and(T~9)4>0forr¢9. 4—21 4—1) 1 .1311 |4~v| viii—(4%) .1311 —1 1 lim4_\/E:lim 4—“; :lim _1 2—51:—1 Hles—w 5—.16(\/§+4)(\/§—4) Salem/5+4 \/16+4 8 2 2 _ _ hm'u —: v 8_1im (v+4)(v 2) _lim n+4 2 2+4 :3 H2 11—16 v—»2(v+2)(v—2)(v2+4) v_.2(v+2)(v2+4) (2+2)(22+4) 16 m—S . |$Hsla x_8 1fm—8>0 {1 ifx>8 ‘8 _ _ ¥ : _ .. as (w 8) if$_8<0 11tx<8 x—8 . [re—8| . Th ,1 :1 — 2—. "8.33...-8 .331 1) 1 “13+ (x/m—9+flac+1]]): lirgi+\/x—9+ lim+[[x+1]]:\/9—9+10210 m—i ac—> x—i9 hml—x/l—ccz 1+x/1—x2 l. 1—(1—a22) 1, m2 l. x k'xfi 1m 1 — M m hut/11302 m-*°2:(1+\/1~x2) z$x(1+\/1_x2) I%1+\/_—1—x2 hm v$+2~ x/Zm x/cc+2+\/2x hm —(w—2) 1. —1 -K— ‘ 1m :— 1—0 33(15—2) x/m+2+\/§x z*2m(x—2)(\/x+2+\/2x) I—Qx x+2+\/2a:) 139 0 140 3 CHAPTERZ LIMITS AND DERIVATIVES , 2 1 _ 2 2 2 A 17. liml—fl-Sc—i22hmfi273—iflzhm 1/30 +2/a: 1_0+071__1 mew—934F296 Wo<>(1—:c+2mz)/vc2 Wool/whimssz 0—0+2 2 3_ 2 3_ 2 3 3 18. um L13: m M: hm Sal/mm _5—0+0w 5 x—Poo 2x3+x—3 za—oo (2m3+w—3)/m3 z—v—oo2-i-1/m2—3/333—“2—h0—0fli 19. Since a: is positive. Vzc : 2 cc. hm x/xQ—Q _ hm x/mz—g/x/P hm x/1—9/m2 x/l—O 7 1 woo 2x—6 woo (2x—6)/a: Us.» 2—6/zc ’ 2—0 ‘2 20. lirlnr ln(10() — x2) : #00 since as a: —> 10—. (100 — x2) ——> 0+. 21. lim e_3”” = 0 since —3:c —> ’00 as :c —> 00 and lim eIt 2 0. mace t—t—oo 22. Ify = x3 — :1: = $(sc2 — 1). then as a: —+ 00,1; —> 00. lim arctan(x3 — cc) 2 lim arctany : g by (2.6.4). x—roo y—aoo 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 . :> ——3 3— :> OS 3—.But 11m 0:0and :62 £62 $2 x2 x2 m—vz—oo , 1 11m -—— : 0, so by the Squeeze Theorem. I—>ioo x2 _l cos2 3: cos2 3: lim : 0. Thus. y 2 0 is the horizontal asymptote. lim 2 00 because cos2 (B —+ 1 and x2 —> O as z—rioo 362 1'40 1'2 a: —> 0, so a: : 0 is the vertical asymptote. 24. From the graph of y = f : V132 + w + — V152 — x, it appears that there are 2 horizontal asymptotes and possibly 2 vertical asymptotes. To obtain a different form for f . let’s multiply and divide it by its conjugate. ./2 1/2_ f1(w)—(\/x2I$+1 $2 at) x+$+1+ m m x/x2+ac+1+\/m2—m (x2+$+1)—($2—LE) 2m+1 Jz2+x+1+fi2—m \/$2+m+1+\/x2—ar Now 1. f ( ) hm 2:0 + 1 1m :3 fl 3*” 1 m—'°°\/:132+:c+1+\/ac2~11: — lim 2 + (since V562 2 ac form > 0) Ho \/1 +(1/m) +(1/w2)+ \/1 -(1/:c) 2 : —— : 1‘ 1+ 1 so y = 1 is a horizontal asymptote. For x < 0, we have V732 : |$| : ~33, so when we divide the denominator by :c, with so < 0, we get 2 2_ 2 2— 1 1 \/:c +zc+1+\/:C m_ x/x +m+1+\/; $__[I1+l+_§+./1__] a: «$2 :3 x a: 25. 26. 27. 28. 29. 30. 31. CHAPTER 2 REVIEW 141 Therefore. 1. f ( ) 1. 2:1: -l- 1 90—12100 1 a: 12200 x/x2+m+1+\/m2 —x 2 1 x 2 —lim +(/) 2¥(1+1)_—1. H” ~ [\/1+(1/$)+(1/$2)+ [1 ~<1/x>] so y 2 —1 is a horizontal asymptote. As a: —> 0‘. f —> 1. so a: 2 0 is not a vertical asymptote. As m —> 1+, f —; so :1: 2 1 is not a vertical asymptote and hence there are no vertical asymptotes. 4 ,1010 —4 Since 2m — 1 g f(m) g 132 forO < x < 3and lim (2x ~1)212 lim :32. we have lim f($) 2 1by the z—tl m—bl ac—>1 Squeeze Theorem. Let flan) 2 —$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 < lav—5| <6.then [(7:17~ 27) *8| <5 <=> [7.x— [z‘ — 5! < €/7.Sotake6 25/7.Then0 < Ix —5| < 6 2> |(7:L’— 27) —8| < 5. Thus. lim m—s5 35|<e <:> (7x—27)28 by the definition of a limit. Givens > Owemustfind6> OsothatifO < lav—0| < 6.then [\3/_—0| <6. Now|\3/_—0| 2 <5 2 [ml 2 <53.Sotake62€3. ThenO < [x—Ol 2 <63 2 IVE—OI 2 2 \3/lml < \3/6—28. Therefore. by the definition ofalimit. lim \3/_ 2 0. 2—0 Given5>0. we need6>0sothatif0< lan—2l <6. then [:62 —3m—(—2)[ then~1<x—2<1.soO<m—1<2 => :> [$223x2(22)| : |(m22)(m— 1) Thus, gig; (:2 — 33:) < 5. First. note that if |m — 2' < 1. [:L' — 1] < 2. Now let 6 2 min {5/2, 1}. Then 0 < [9: — 2] < 6 |2 lac — 2| [x — 1| < (€/2)(2) 2 a. —2 by the definition of a limit. Given M > 0, we need 6 > 0 such that ifO < a: — 4 < 6. then 2/\/:c — 4 > M. This is true <=> Mas—4 < 2/M <:> $24 < 4/M2. Soifwechooseé = 4/M2.then0 < m—4 < 6 :> 2/y/x—4 > M. So by the definition ofa limit. lim (2AM — 4 W“ ) 2 00. (a) f(:c)2Hifx<0.f(w)23—xif0Sx<3.f(m)2(z—3)2ifx>3. (i) “151+ f($) 2 llI(I)1 (3 — z) 2 3 (ii) lim 2 lim V—x 2 0 z__, I__, z—yO— z—VOT (iii) Because of (i) and (ii). lirnJ does not exist (iv) lim 2 Iim (3 ~ in) 2 0 3* z—b3‘ IHB‘ (v) Iim z—>3+ lim (9:232 :0 3+ (vi) Because of (iv) and (v), lint;g flat) 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):230‘302ifOSxS2.g(z):2—mif2<x§3. g(ac) : ac — 4 if3 < a: < 4. g(:c) = 7r ifs: Z 4. Therefore. lim g(z) : lim_ (2x — x2) : 0 and :t—r2‘ ac—vZ $133+ g(z) = $133+ (2 — ac) : 0. Thus. 9(20) : O : g so g is continuous at 2. lim g(:c) = lim (2 7 m) : —1 and x—rB— z—r3— lim+ g(ac) : lim — 4) = —1.Thus. ling 9(13) : —1 = 9(3). so 9 is continuous at 3. $43 $423+ lim 9(22) : lim (cc — 4) : 0 and lim 9(33) : lim 7r : 7r. Thus. lim 9(23) does not exist. so 9 is x—r4’ x—»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. x—v4 33. sin .7: is continuous on IR by Theorem 7 in Section 2.5. Since ea” is continuous on IR. e“ x is continuous on R by Theorem 9 in Section 2.5. Lastly, cc is continuous on R since it‘s a polynomial and the product mesi“ m is continuous on its domain R by Theorem 4 in Section 2.5. 34. x2 — 9 is continuous on IR since it is a polynomial and fl is continuous on [0. 00). so the composition V362 — 9 is continuous on {cc l :02 — 9 Z 0} = (—00. *3] U [37 00). Note that x2 i 2 ¢ 0 on this set and so the quotient 532—9 function g(m) : is continuous on its domain. (—00. —3] U [37 $2 7 35. f(:c) : 2:63 + $2 —I— 2 is a polynomial. so it is continuous on [—2. 71] and f(~2) = #10 < 0 < 1 : f(—1).So by the Intermediate Value Theorem there is a number 6 in (—2. —1) such that f (c) : 0. that is, the equation 2$3 + x2 + 2 = 0 has aroot in (#2., #1). 36. f(:c) = 6—952 # 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. 6” — cc 2 0. or e‘” : x. has a root in (0. 1). 37. (a) The slope of the tangent line at (2, 1) is — A 2— 7 3 —2 2#4 _ —2 *2 2 limwzlimw:lim8 21 :lim—E—thm (a: )(x+ ) 50-3 12—2 1H2 $*2 I—>2 (13—2 1H2 x72 z—r2 33—2 =1im2[-2(m+2)]:A2-4:~8 (b) An equation of this tangent line is y — 1 : #8(sc # 2) or y z #836 + 17. 38. For a general point with :c—coordinate a. we have 2/(1—3w) —2/(1;3a) 2(1 #3a) -2(1—3:c) # . _ 1. _,_______ m :13: aka $13}. (1—3a)(1—3m)(wra) i hm 5(56 a) 7 lim —————d ‘ I a(]. 3a)(1 3x)(x a) m»a(1 3a)(1 33:)_(1—3a)2 CHAPTER 2 REVIEW 143 —0)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 — 2 2 6(:c §(m+ )ory2 gml— g. a 2 71. m 2 g and f(—1) 2 so an equation of the tangent line is y — § 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) — 1 h 4h 4 ' So for the following intervals the average velocities are: (i) [1,3]: h : 2.7;” = (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 — 1 1 h 10 (b) When t 2 1. the instantaneous velocity is lim w 2 lim 0—+ 2 — 2 2.5 m/s i120 h h—>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 3—28 — 3—33 2 3.2 — 4 2 —0.81b/in2. So the average rate of change is AP —0.8 Ib/m2 — 2 — 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) — V (P) _ . 800/(P + h) — 800/P limo E V1335 h V135 h _ hm 800 [P 2 (P + h)] _ mm —800 _ _@ * h—»0 h(P + h)P a 1:20 (P + h)P _ P2 which is inversely proportional to the square of P. , _. f(a:)—f(2)_l $3—2JJ—4 “Wflm—ii’éfi—mfi . (m—2)(:c2+2a:+2) 211m (b)y—4210($—2)ory210m—16 42. 26 2 64, so 2 $6 and a 2 2. 43. (a) f’(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 $1200. (c) As r increases. C increases. So f’(r) will always be positive. 144 3 CHAPTERZ LIMITS AND DERIVATIVES For Exercises 44—46, see the hints before Exercise 5 in Section 2.9, 44- 45. 47 (a) f/(aj) — hm Milt—“£1 rm \/3—5($+h) — x/3—5m \/3—5(m+h) +‘/—3_5x ' _ M h _ bio h x/s—sec +7) + m ! hm [3 7 5(30 + h)] — (3 — 5:0) lim —5 _ —5 h*°h( 3—5(x+h)+\/BT5§) h"°\/3—5(x+h)+\/3—5m Mia—sac (b) Domain of f: (the radicand must be nonnegative) 3 — 5:0 2 0 :> 5x 3 3 => as 6 (A00, Domain of f' : exclude % because it makes the denominator zero; :3 E (—00, (c) Our answer to part (a) is reasonable because f l(sic) is always negative and f is always decreasing. 48. (a) Asw —> :00, fine) : (4 # $)/(3+m) —+ —1. so there isa horizontal asymptote at y = #1. As a: —> —3+, f(a:) —> oo. and as :c —> -3’, f(a:) —> —00. Thus, there is a vertical asymptote at a: : —3. CHAPTERZ REVIEW U 145 (b) Note that f is decreasing on (700. *3) and (—3, 00). so f' is negative on those intervals. As :0 —> 5:00, f’ —+ 0. As at —> 73‘ and asx —> —3+. f’ —> —oo. 4~(m+h)_4—:L' 3+(x+h) 3+:c f(x+h) -f(w) <c>f'<w>=m h h _ hm (3+:c)[4—(x+h)]—(4~$)[3+(a:+h)] —h——>O h[3+(m+h)](3+x) _1, (12—3x—3h+4$7m2ihw)—(12+4m+4h—3m—m2—hx) 113% h[3+(x+h)](3+x) —7h —7 7 :lmmufihnom‘i%[3+<x+h)i<3+m)= may (d) The graphing device confirms our graph in part (b). 49. f is not differentiable: at x : —4 because f is not continuous. at m = —1 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: F’(1950) is, Q 0.11, 10 : F'(1965) w $06 : ~0.16. and F'(1987) m 01+; : 0.02. (b) The rate of change of the average number of children born to each woman was increasing by 0.11 in 1950, decreasing by 0.16 in 1965. and increasing by 0.02 in 1987. (c) There are many possible reasons: 0 In the baby-boom era (post-WWII). there was optimism about the economy and family size was rising. o In the baby—bust era. there was less economic optimism. and it was considered less socially responsible to have a large family. 0 In the baby—boomlet era. there was increased economic optimism and a return to more conservative attitudes 51. B’(1990) is the rate at which the total value of US. banknotes in circulation is changing in billions of dollars per year. To estimate the value of B’(1990). we will average the difference quotients obtained using the times t : 1985 andt : 1995' Let A I B(1985) ~ B(1990) I 1820 — 268.2 1985 ~ 1990 _5 = 17.24 and 3(1995) — 3(1990) 401.5 a 268.2 C 2 x : I . . 1995 — 1990 5 26 66 Then 8,0990) : hm B(t) — B(1990) A + C 17.24 + 26.66 ~ _ ~ __ t—>1990 t — 1990 2 2 : 21.95 billions of dollars/year. 146 C CHAPTERZ LIMITS AND DERIVATIVES 52. The slope of the tangent toy : 13+: is x, ($+h)+1#$+1 hm (m+h)71 22—1_lim($—1)(m+h+1)—(ac+1)(a:+h—1) h—+O h 11—.0 h(:c—1)(a:+hi1) lim _2h — 2 —hH0h($*1)($+h*1)— (93—1)2 2 Soat(2,3).m— (2_1)2* 2 —5 y—3—*2(sc#2) => —72a:+7 At(—10)m———2————l => y_ ‘ " 7 (~1—1)Qfl 2 yz—ém—I—l) => y=*%z—%. 53. S g(:c) <=> 79(03) 3 f(:c) S g(x) and lim 9(50) 2 0 = lim —g(m). Thus, by the Squeeze Theorem, (II—’11 (t—Va lim : O. x-wt 54. (a) Note that f is an even function since f(x) : f(—;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]] : + (— A 1) = #1. So lim f(.’t) exists EH04 (and is equal to —1) for all values of a. (b) f is discontinuous at all integers. ...
View Full Document

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.

Page1 / 11

02_R - 136 I: CHAPTERZ LIMITS AND DERIVATIVES 48. (a) T (c)...

This preview shows document pages 1 - 11. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online