ch review - CHAPTER4 REVIEW E 399 17 22 (b) To find the...

Info iconThis preview shows pages 1–24. 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

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

View Full DocumentRight Arrow Icon
Background image of page 12
Background image of page 13

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

View Full DocumentRight Arrow Icon
Background image of page 14
Background image of page 15

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

View Full DocumentRight Arrow Icon
Background image of page 16
Background image of page 17

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

View Full DocumentRight Arrow Icon
Background image of page 18
Background image of page 19

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

View Full DocumentRight Arrow Icon
Background image of page 20
Background image of page 21

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

View Full DocumentRight Arrow Icon
Background image of page 22
Background image of page 23

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

View Full DocumentRight Arrow Icon
Background image of page 24
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHAPTER4 REVIEW E 399 17 22 (b) To find the maximum height. set v(t) on 3 < t S 17 equal to 0. —32(t — 3) + 270 : 0 => t1 = 11.4375 s and the maximum height is 5(t1) = —16(t1 — 3)2 + 270(t1 — + 270 : 14090625 ft. (c) To find the time to land. set 5(t) = —18(t V 22) + 424 = 0. Then t e 22 : 4—125 : 23.5. sot e. 45.6 s. 79. (a) First note that 90 mi/h : 90 X 3% ft/s 2 132 ft/s. Then a(t) = 4 ft/s2 :> v(t) 2 4t + C. but 11(0) : 0 => C : 0. Now 4t : 132 when t : ¥ : 33 s. so it takes 33 s to reach 132 ft/s. Therefore, taking 3(0) : 0. we have 5(t) : 2752. 0 S t S 33. So 5(33) 2 2178 ft. 15 minutes : 15(60) : 900 S. so for 33 < t g 933 we have up) : 132 ft/s :> 3(933) : 132(900) + 2178 : 120,978 ft = 22.9125 mi. (b) As in part (a). the train accelerates for 33 s and travels 2178 ft while doing so. Similarly. it decelerates for 33 s and travels 2178 ft at the end of its trip. During the remaining 900 — 66 = 834 s it travels at 132 ft/s. so the distance traveled is 132 ~ 834 = 110.088 ft. Thus. the total distance is 2178 + 110.088 + 2178 2114.444 ft: 21.675 mi. (c) 45 mi 2 45(5280) : 237.600 ft. Subtract 2(2178) to take care of the speeding up and slowing down. and we have 233.244 ft at 132 ft/s for a trip of 233.244/132 : 1767 s at 90 mi/h. The total time is 1767 + 2(33) : 1833 s = 30 min 33 s : 30.55 min. ((1) 37.5(60) : 2250 s. 2250 7 2(33) 2 2184 s at maximum speed. 2184(132) + 2(2178) : 292.644 total feet or 292.644/5280 : 55.425 mi. 4 Review CONCEPT CHECK 1. A function f has an absolute maximum at ac : c if f (c) is the largest function value on the entire domain of f. whereas f has a local maximum at c if f (c) is the largest function value when x is near c. See Figure 4 in Section 4.1. 2. (a) See Theorem 41.3. (b) See the Closed Interval Method before Example 8 in Section 4.1. 3. (a) See Theorem 41.4. (b) See Definition 4.1.6. 4. (a) See Rolle’s Theorem at the beginning of Section 4.2. (b) See the Mean Value Theorem in Section 4.2. Geometric interpretation—there is some point P on the graph of a function f [on the interval (a. b)] where the tangent line is parallel to the secant line that connects (a. flu» and (b. 5. (a) See the VB Test before Example I in Section 4.3. (b) See the Concavity Test before Example 4 in Section 4.3. 10. C CHAPTER 4 APPLICATIONS OF DIFFERENTIATION (a) See the First Derivative Test after Example I in Section 4.3. (b) See the Second Derivative Test before Example 6 in Section 4.3. (C) See the note before Example 7 in Section 4.3. (a) See I‘Hospital‘s Rule and the three notes that follow it in Section 4.4. f g — or —. 1/9 1/f (c) Convert the difference into a quotient using a common denominator. rationalizing. factoring. or some other method. (b) Write fg as (d) Convert the power to a product by taking the natural logarithm of both sides of y = f9 or by writing f9 as eg I" f . Without calculus you could get misleading graphs that fail to show the most interesting features of a function. See the discussion following Figure 3 in Section 4.5 and the first paragraph in Section 4.6. (a) See Figure 3 in Section 4.9. _ f0“) (W172 # $1 — fwd“) (C) $n+1 = $1; — (d) Newton‘s method is likely to fail or to work very slowly when f'(:r1) is close to 0. (a) See the definition at the beginning of Section 4.10. (b) If F1 and F2 are both antiderivatives of f on an interval I. then they differ by a constant. TRUE-FALSE QUIZ . False. For example. take f(:c) : m3. then f’(x) : 3:02 and f/(O) : 0. but f(0) : O is not a maximum or minimum; (0. 0) is an inflection point. False. For example. f(:c) : has an absolute minimum at 0. but f’(0) does not exist. False. For example. f(a:) = as is continuous on (0, 1) but attains neither a maximum nor a minimum value on (0. 1). Don‘t confuse this with f being continuous on the closed interval [(1. b]. which would make the statement true. True. By the Mean Value Theorem. f'(c) : W = g 2 0. Note that |c| < 1 <=> c E (71.1). . True. This is an example of part (b) of the [/D Test. . False. For example. the curve y = f(a:) : 1 has no inflection points but f”(c) : 0 for all c. False. f’(:c) : g'(:c) :> flat) : g(a:) + C. For example. if : 9: + 2 and g(;E) : x + 1. then f'(96) : 9'(I) :1-butf($)75 9W)- False. Assume there is a function f such that f (1) = —2 and f (3) : 0. Then by the Mean Value Theorem there f(3) — f0) 7 w 3*1 2 :1,Butf'(m)>1forallx.a exists a number C E (1, 3) such that f’(c) = contradiction. CHAPTER4 REVIEW 3 401 9. True. The graph of one such function is sketched. 10. False. At any point (a. f(a)). we know that f'(a) < 0. So since the tangent line at (a. f(a)) is not horizontal. it must cross the x—axis—at ac = b. say. But since f”(:c) > O for all m. the graph of f must lie above all ofits tangents; in particular. f(b) > 0. But this is a contradiction. since we are given that flat) < 0 for all m. 11. True. Let 361 < 1:2 where 9:1.x2 E I. Then f(w1) < flu) and g(m1) < 9(12) (since f and g are increasing on D- 30 (f +9)($1) I f(561) +9061) < f(i€2) +9062) = (f + 9)($2)- 12. False. f(:c) : ac and 51(1) : 2:5 are both increasing on (0. 1). but f(:c) — g(:c) : ix is not increasing on (0.1). 13. False. Take f(:c) : x and g(m) : :c — 1. Then both f and g are increasing on (0. 1). But f(z)g(:c) : 93(33 7 1) is not increasing on (0.1). 14. True. Let 251 < 102 where $1,322 6 I. Then 0 < f(;c1) < flu) and O < g(m1) < 9(552) (since f and g are both positive and increasing). Hence. f(;c1)g(:v1) < f(12)g(z1) < f(a:2)g($2). So fg is increasing on I. 15. True. Letz1.:52 E I and £1 < $2. Then f(x1) < f(3:2) (f is increasing) :> “in > fab) (f is positive) : g(z1) > 9(a) => g(m) : 1/f(:c) is decreasing on I. 16. False. The most general antiderivative is : —1/:c + 01 for in < 0 and : —1/x + 02 for as > 0 (see Example 1 in Section 4.10). 17. True. By the Mean Value Theorem. there exists a number 0 in (0, 1) such that f(1) — f(0) : f’(c)(1 — 0) : f’(c). Since f’(c) is nonzero. f(1) i f(0) # 0. so f(1) % f(0). 18. False. iii?) 6% = : (13 = 0. not 1. $—>0 EXERCISES 1. f(:c) : 10 + 27x — x3. 0 g m g 4, f’(m) = 27 — 3x2 : —3(m2 — 9) : —3(:c + 3)(:1: — 3) : 0 only when x : 3 (since —3 is not in the domain). f'(ac) > 0 for at < 3 and f/(m) < 0 for x > 3. so f(3) : 64 is a local maximum value. Checking the endpoints. we find f(0) : 10 and f(4) : 54. Thus. f(0) = 10 is the absolute minimum value and f(3) : 64 is the absolute maximum value. f(:c) :$— fil) S at £4. f’(x) =1—1/(2fi) :0 <:> 2\/_=1 => 1:; f’(:c)doesnotexist <:> ac : 0. f’(z) < 0f0r0 < e < i and f’(m) > Ofor 41 < :c < 4. so f6) ~71 isalocal and absolute minimum value. f(0) : 0 and f(4) : 2. so f(4) : 2 is the absolute maximum value. z (m2+x+1)(1)—x(2$+1) 17;: W‘ ($2 + at + n2 T m a: : —1 (since 1 is not in the domain). f’(:v) < 0 for —2 < x < —1 and f’(a:) > 0 for —1 < x < 0. so f(—1) : 71 is a local and absolute minimum value. f(72) = ~§ and f(0) : 0.80 f(0) : 0 is an absolute maximum value. 2 f(;t:): —2§x§0.f’(:c): :0 <=> 402 CI CHAPTEFM APPLICATIONS OF DIFFERENTIATION 4. f(w) = (m2 + 29:)3. [—2. 1]. f'(a:) : 3(x2 + 2$)2(2x + 2) : 6(x +1):c2($ + 2)2. so the only critical numbers in 10. 11. 12. 13. 14. the interior ofthe domain are a: = —1,0. f’(33) < 0 for —2 < x < 71 and f’(:c) > Ofor —1 < a: < Oand 0 < 20 < 1. so f is decreasing on (—2, —1) and increasing on (—1,1). Thus. f(—1) : 71 is a local minimum value. f(—2) = 0 and f(1) : 27. so the local minimum value is the absolute minimum value and f(1) = 27 is the absolute maximum value. f(w)=:c+sin2m.[0,7r].f’(a:):1+2cos2m:0 <=> cos2xi % 5 2$a2gor4gr < x— 7r 3 or f"(a:) : —4sin2x, so f"(§) : —4sin2§—r = —2\/§ < Oand f”(%") : v4sin4§ : 2\/§ > 0~ so I g + g B 1.91 is a local maximum value and = 2?" — A? m 1.23 is a local minimum value. Also f(0) = O and f(7r) = 71’. so f(0) : 0 is the absolute minimum value and f(7r) = 71 is the absolute maximum value. may , m2~l—(ln:c)(2x) m—Zmlnm 1721119? 1 m : 61/2 : \/E m 1.65. f’(;c) > 0 form < fiand f'(x) < 0 fora: > J5, so f is increasing on (1, V5) and 1 , . l 3 . decreasing on (\/E, 3) . Hence, f(\/E) = 2— IS a local max1mum value. f(1) = 0 and f(3) : HT 2 0.12. S1nce 6 2; m 0.18. f(\/E) : 2i is the absolute maximum value and f(1) : 0 is the absolute minimum value. 6 e 1, tanwz H hm 7rsec2 71m 7r~12 7T 1m —— : —— : : za01n(1+:c) 1—901/(14-37) 1/1 lim1—Cosm 5 hm smx _ 9 _0 x—»0 m2+x _z—>02:L'+1 _ 1 7 4m 4m 41 — — a 1 hing—#3211111“ 4 211m 66 : lim8e4zz8-128 z—>O SC z—rO 212 95—0 2 z—rO 4m _ _ 4:: _ 1 4x lim S—ug 2 11111 46 4 = lim 6e 2 11111 86439 : oo I—roo (132 :c—>oo 2m 1—700 2 m—roo 3 2 6 . 6 lim m3e*m:1im 3”— £11m 35” £11m 1’ 211m — =0 iii—>00 z-MX) em z—>oo ex ac—voo ez z—voo e _ . lnzv H . 1/9: , 1 2 2 : — 1 = l *- = 0 213%,?” mm $133+ 1/2:2 1331+ —2/ac3 1351+< 2m ) l, a: 1 rm xlnm—m—I—l H hm x-(l/x)+lnw71 A — : 1 —— 2 —————— xinil+ 36— 1 Ina: x_.1+ (.7:— 1)lna: x_.1+ (m—1)-(1/:c)+ln:c _ 1. Inst 5 lim l/x : 1 :1 _ x31 1 — l/x—I—lnx _ mam 1/:c2 + 1/m 1 + 1 2 y : (tan zero” => lny : cosmlntanm, so 2 lim lny : hm lntanw E lim (1/ tanm) sec :0 : hm sec2z 2—»(7r/2r :c—>(7r/2)’ secsc x—>(7r/2)— secmtanw 2—.(7r/2)— tan :1: : lim 6,0? = % =0.so z—>(7r/2)’ sm cc 1 1im (tan 1:)“3” = lim elny 2 60 : 1. z—>(7r/2)_ $H(7r/2>7 CHAPTER4 REVIEW 3 403 15. f(()) : O. f'(72) : f'(1) = f'(9) : 0. lim : 0. (II—>00 aligns f(m) : foo. f’(:c) < Oon (700, —2). (1,6). and (9.00). f’(:c) > 0 on (—2. 1) and (6.9). f"(:c) > 0 on (—00.0) and (12. oo). f”(z) < 0 on (0.6) and (6.12) 16. ForO < x < 1. f/(ac) : 2x. so f(2:) : $2 + C. Since f(0) : O. f(ac) : x2 on [0.1]. Forl < ac < 3. f’(w) : *1. so f($)=—$+D.1:f(1):71+D => D:2.so f(:c):Zix.Forx>3.f’(n:)=1.sof(ac):a:+E. —1:f(3):3+E :> E:—4.sof(z):x—4.Sincef is even. its graph is symmetric about the y—axis. 17. f is odd. f’(m) < 0 forO < :1: < 2. f’(a:) > 0 fora: > 2. f"(x) > 0for0 < at < 3. f”(w) < 0 form > 3. limgc_.oO : —2 18. (a) Using the Test for Monotonic Functions we know that f is increasing on (—2, 0) and (4. 00) because f’ > 0 on (—2.0) and (4.00). and that f is decreasing on (—00. —2) and (0.4) because f’ < 0 on (~00. i2) and (0. 4). (b) Using the First Derivative Test. we know that f has a local maximum at m = 0 because f’ changes from positive to negative at a: = 0. and that f has a local minimum at an : 4 because f’ changes from negative to positive atx=4. (d) possible graph of f ><V 19. y : : 2 — 2x ~ 173 A. D : R B. y—intercept: f(0) = 2. The m—intercept (approximately 0.770917) can be found using Newton‘s Method. C. No symmetry D. No asymptote E. f’(:zr) : ~2 7 3x2 : —(3m2 + 2) < 0. so f is decreasing on IR. F. No extreme value G. f”(a:) : —61: < 0 on (0.00) and f”(z) > 0 on (—007 0). so f is CD on (0.00) and CU on (—007 0). There is an IP at (0.2). 4M 20. 21. 22. 23. 3 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION y=f(;v):m376m2—15sc+4 A. DzR B. y—intercept: f(0) : 4; :c—intercepts: f(m) = 0 => x m 72.09. 0.24. 7.85 C. No symmetry D. No asymptote E. f’(:c) — 3x2 15 * 3(ac2 5) : 3(x + 1)(:r i 5). so f is increasing on (700. —1). decreasing on (71. 5). and increasing 12w 4x on (5. 00). F. Local maximum value f(71) = 12. local minimum value f(5) : 796. G. f"(:c) : 6m — 12 : 6(50 — 2). so f is CD on (700. 2) and CU on (2. There is an [P at (2. —42). y : : 2:4 — 3:83 + 3x2 # cc = 93(m — 1)3 A. D : IR B. y—intercept: f(0) = 0; x‘intercepts: f(ac) : 0 <=> 9: = 0 or m = 1 C. No symmetry D. f is a polynomial function and hence. it has no asymptote. E. f’(a:) : 4mg — 9x2 + 6w — 1. Since the sum of the coefficients is 0. 1 is a root of f’. so f’(:c) : (3c —1)(49v2 7 5w +1) 2 (x 71)2(4a: — 1). f'(ac) < 0 :> m < i. so f is decreasing on (700. :11) and f is increasing on (i, 00). F. f'(:c) does not change sign at ac : 1. H. so there is not a local extremum there. f6) : —% is a local minimum value. G. f”(:c):12;t2 —189: + 6 2 6(22? —1)(x — 1). f"(:c):0 <=> :c:%or1. f"(:c)<0 <¢ %<:c<1=> f is CD on 1) and CU on (700. and (1. 00). There are inflection points at —%5) and (1,0). 1 1 T 1—12 7 (1+$)(1~x) C. f(—x) = so f is even and the graph of f is symmetric about the y—axis. D. Vertical asymptotes: 2m (1* $2)2 A. D : {ac | :c 79 :I:1} B. y—intercept: f(0) = 1; no sic—intercept ac : i1. Horizontal asymptote: y : 0 E. y’ : 0 <=> x : 0. so f is decreasing on (—00. —1) and (—1.0). and increasing 0n (0, 1) and (1. 00). F. Local minimum value f(0) = 1; no local maximum (1 7 m2)2 - 2 7 2a: - 2(1 # (1:2)(—2;L‘) G' f”($) i (1_x2)4 21_ 2 82 2 :(__$)’+3::_6w_+_2§<0 s, .2» (l—mz) 0—1-2) so f is CD on (A00. —1) and (1,00). and CU on (-1. 1). No [P y : f(:c) = fl); A. D = {2: I m ¢ 03} : (700,0) U (0.3) U (3,00) B. No intercepts. ac an - C N D l' 1 0 0 ‘s aHA I‘m 1 - —— = . : I. . 1 ————— : . 0 symmetry. $11100 mm 7 3>2 so y mam mm 7 3)2 . 1 . 1 _ :oo.sox:Oand93—3areVA. (a: 7 3)2 + 29m — 3) : 3(1 — x) m2 (cc — 3)4 122(36 ~ 3)3 E. f’($)— 24. 25. 26. 405 CHAPTER 4 REVIEW f/(cc) > 0 <=> 1 < m < 3. so f is increasing on (1.3) and decreasing H. on (—00.0). (0, 1). and (3. oo). 6 232274 +3 (if/(m): (36%;)4 ) since it has negative discriminant. So f"($) > 0 <=> .z' > 0 => f is CU on (O, 3) and (3. 00) and CD on (—00, 0). No IP F. Local minimum value f(1) : i . Note that 2w2 7 4m + 3 > 0 for all a: : : — : —— A. D : 0. —1 B. No -mterce t. m—lnterce t : —— y f(w) L6er+1 WWI) {$Im# } y p p 2 2m+1 . 2x+1 ‘ = = ' . ‘ —— : . l * : — . C. No symmetry D. 0011,2300 f(ac) 0. soy 0 IS a HA 331—13511L Mac + 1) oo Stirs: mm + 1) oo lim gig». lim 2—361:—oo.soac=0.x=~lareVA. x—vil+ $(m ‘i’ 1—r—1’ $(x + 1 1 . E. / : —— i — < 0. so is decreasm on ~00. —1 . H. No y a“? f g< ,) (~1, 0) and (0, 00). F. No extreme values I, _ 2 2 ‘ 2(2$+1)($2+m+1) Gfmcg+afi$*—*Efim?—‘ f"(x) > 0 <=> as > Oor~1<x < —%.sofisCUon (0.oo)and (—1,—§)andCDon(—oo.—1)and(—51.0). lPat(—%.0) E. f’(x):1 >0 <:> m>00rzc<~16. _ (:1:+8)2 : (m+8)2 so f is increasing on (—00, —16) and (0. 00) and decreasing on (~16. —8) and (—8. 0). F. Local maximum value f(—16) —32. G.f"(:l;)=128/(:1: + 8)3 > 0 <=> $ > —8. so f is CU on (—8. 00) and CD on (—00. 78). No [P 2 3/ ~ f(w) — x: 8 — m — 8 : iii: A. D : {at | an 75 —8} B. Intercepts areO C. No symmetry D li $2 00 butf(;c) (ac 8) 64 Oasx 00 so ac Sisaslant as m tote . m = , — _ : _, __) ~ : fl . ' zflwx+ x+8 y y p x2 x2 lim 2 co and lim : —00. so 2: = 78 is aVA. H. ; xA—8+ mgr—8‘ E 64 x($ +16) Eo (—16. —32) y : f($) : a: + \/1——a_: A. D 2 {z I m S 1} : (~oo.1] B. y-intercept : 1; x—intercepts occur when $+m=0 => m2—m :1—2321:2 => x2+w—1:O : 5c::1%@.but the larger root is extraneous. so the only m-intercept is :1%5 . C. No symmetry D. No asymptote E. f’(m):1~1/(2fl):0 <=> 2W:1 <=> H. 1—w: local minimum value f(0) : 0 % <=> ac = E and f’(z) > O <:> a: < %. sof is increasing on (—00, decreasing on 1). F. Local maximum value f6) 1 G. " :_‘ . i _ ‘ . f (x) 4(l_ac)3/2<0 <=> £13<180flSCDOn(OOV1) NOIP E 4 406 27. 28. 29. 3 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION y : f(m) : cc #2 + x A. D : [—2, 00) B. y—intercept: f(0) = 0; :c—intercepts: —2 and 0 C. No x 1 3m+4 — +\/2+Cc:—— 2\/2+a: 2x/2—I—m 2 «2 + :0 2 when m 2 —§. so f is decreasing on (—2., —§) and increasing on (—g F. Local minimum value 3» symmetry D. No asymptote E. f'(1:) [x + 2(2 + = f(—§) : igfi : 7% R1 i109. no local maximum H. Y 1 2x/2—I—zr-3A 3m+4 II ( ) 2+1? G. f (as) — V 4(2+:I:) _ 6(2+m)—(3x+4) i 3m+8 4(2 +m)3/2 # 4(2+m)3/2 f”(w) > 0 form > ~2. so f is CU on (—2.00). NO [P y:f(x):\/—, 3/; A. D:[0,oo) D. (331/2 * 351/3) _2/3 _ 3531/6 — 2 B. y—intercept 0: sic—intercepts 0. 1 : lim Sic—Poo lirri 1—»00 C. No symmetry [931/3 (301/6 — 1)] : 00. no asymptote E. f’(m) = %x’1/2 i 1:1: > 0 <=> 3301/6 > 2 <=> a: > (96. sof is increasing on 3 6032/3 ((§)6 .00) and decreasing on (O, (§)6> . F. : —% is a local minimum value. <=> .1: < (3)6. so f is CU on (O. (3)6) and CD on ((3)6 i 64 729 IPat ((3)6. ) y : flan) : sin2 :c — 200520 A. D : IR B. y—intercept: f(0) —2 C. f(—;E) : so f is symmetric with respect to the y—axis. f has period 271'. D. No asymptote E. y’ : 2sinwcosa: + 23mm : QSiIlJC(COS$ + 1). y' = O <:> sinx : 0 or cosa: : —1 <=> a: : nvr or as : (2n + 1)7r. y' > 0 when since > 0. since (3089: + 1 2 0 for all :0. Therefore. y' > 0 (and so f is increasing) on (27m. (271 + 1)7r); y’ < 0 (and so f is decreasing) on ((2n — 1)7r, 27m). F. Local maximum values are f((2n + 1)7r) = 2; local minimum values are f(2n7r) —2. G. 3/ : sin 256 + 2 SlHZL’ :> y”:2c032m+2cosw 22(2cos2x—1)+2cos;c :4COSZ$+2COSZL'*2 22(2cos2x+cos:c — 1) : 2(2cossc i 1)(cos:c-l- 1) y" :0 <:> cosm: % 0r—1 <:> a: :27L7r:l: germ: (2n+1)7r. y" > 0(andsofisCU)on (2n7r — g, 2mr + y” g 0 (and so f is CD) on (27177 + §,27m + There are inflection points 1 _l) at (2n7ri 3. 4. 0 30. 31. 32. 33. :1 CHAPTER 4 REVIEW 407 71' 7r 2‘2 ). B. y—intercept : f(0) = 0 C. f(—:c) : if(:c (4m— 1321mm): 00. 5020 : y _ i 42: tanx. so the curve is symmetric about (0. 0) . g<z<g A.D—( ). D. lim g (436 ~ tan :3) 2 —oo. {11—b—7r/2+ lim Ian/2* andm: 7% areVA. E. f’(w) :4isec2$>0 <=> seczr<2 <=> cosm >§ <=> —§ < 50 < §.sofisincreasing on decreasing on (7%, —§) and . F. : 47" — x/gisa local maximum value. : \/§ — 4?” is a local minimum value. G. f"(w):i2seC2:rtancc>0 <:> tanm<0 <=> sofisCUon (750) and CD on (0. g). [P at (0.0) —§<m<0. y : f(:c) : sin’1(1/x) A. D ={m| —1§1/x S 1}: (—oo,—1]U[1.oo). B. No intercept C. f(~ac) : symmetric about the origin D. lilil sin’1(1/x) = sin’1 (0) : 0. so y : 0 is a HA. E. f’(:c) : 1 1 = 71 < 0. so f is decreasing on (—00, —1) and (1. oo). y/1~(1/ac)2 $2 $4~$2 F. No local extreme value. but f(1) : g is the absolute maximum value H. and f(—1) ~§ is the absolute minimum value. 4553 — 211'; 13(21'2 — 1) 2(354 E $2)3/2 ($4 7 m2)3/2 f"(2:) < Oforx < *1. so f is CU on (1.00) and CD on (—00. —1). No IP G. f"(ac) : > 0fora: > land 2 :0. 2m~m y: f(:v) nm e x—vioo A. D : R B. y-intercept 1: no x—intercept C. No symmetry D. soy : 0 is 21 HA. E. y : f(m) = 62””2 : f’(:c) : 2(1— flew—’2 > 0 <:> a: < 1. so f is increasing on (~oo.1) and decreasing on (1. 00). F. f(1) : e is a local and absolute maximum value. G. f”(:c) : 2(23L’2 i 41' + 1)eh’22 : 0 <=> m 21¢ “75. f"(:c) >0 <=> $<1—3§0ra:>1+‘/7§.sofisCUon g) and (1+§.oo).andcoon (1_ 391+?) 1Pat(1i§,\/E) (—w y : : ex + 6‘“ A. D : R B. y—intercept 2: no z—intercept C. No symmetry D. lim :E—vioo f/(x) : ex 7 36‘” : 6‘31 (en ~ 3) > O <=> e41 > 3 <=> 4:0 > ln3 <=> .1“ > filn3 z 0.27. so f is increasing on (i ln3. 00) (6“ + e‘am) : 00. no asymptote E. y : : ex + 6*?“ :> H. and decreasing on (—00. £1113). F. Absolute minimum value f(;111n3) : 31/4 + 373/4 z 1.75. G. f”(:13): ea” + 98—3z > 0. so f is CU on (~00. 00). No IP 408 C CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 34. y : f(m) = ln(ac2 — 1) A. D = (—00, ~1) U (1, 00) B. No y—intercept; x—intercepts ix/i C. Symmetric 35. about the y-axis D. lim ln(;c2 ~ 1) = 00, lim ln(;c2 ~ 1) 2 —oo. lim ln(at2 * 1) : —oo. son; : 1 z—mkcx: $_,1+ xail— andsc : 71 are VA. E. y = f($):1n(;€2 — 1) 2 f’(:c) : 22$ 37 _ for x < *1. so f is increasing on (1, 00) and decreasing on (—007 71). Note that the domain of f is > 1. F. No extreme value ,, 2:2 +1 . G. f : —2———2 < 0. so f 1S CD on (00, 71) and (1,00). (m2 — 1) N0 IP 3:2 — 1 I $3995) — (x2 A1)3$2 3 g x2 : $3 :> f A $6 7 x4 2) 1 H x4(—2$) — (3 7 2234353 7 2x2 a 12 f m $8 ’ 3135 f Estimates: From the graphs of f ’ and f ". it appears that f is increasing on -5 V. '— 5 (—1.730) and (0,1.73) and decreasing on (—00, -1.73) and (1.73, 00); #012 f has a local maximum of about f(1.73) : 0.38 and a local minimum of 0'2 about f(—1.7) : —0.38; f is CU on (—2.45, 0) and (2.45., 00). and CD on (—00, —2.45) and (0, 2.45); and f has inflection points at about (—2.45, 70.34) and (2.45., 0.34). 3A3:2 m4 Exact: Now f’(m) : is positive for 0 < :02 < 3. that is. f is increasing on (—\/§, 0) and (0, and fl is negative (and so f is decreasing) on (#00, ix/g) and 00). f’ : 0 when m : :tx/g. f' goes from positive to negative at w : so f has a local maximum of 71,5 2 7 -0.25 : (:33 1 : z—g/‘E’; and since f is odd. we know that maxima on the interval (0, oo) correspond to minima on (7007 0). so f has a local 2 2 — 12 . . , minimum of flex/23>) = #2459. Also. f” : £55— 13 posrtive (so 13 —0.4 f is CU) on (—x/6,0) and 5/6, 00), and negative (so f is CD) on (_00, _\/6) and (0, There are IP at 5%?) and («-g). CHAPTER 4 REVIEW 409 2 3 {1:22/3 1 2.73 f’(;z:)i$1/3( 1)(1 m)’2( 1)+(1 93)—1(%)w7/: 3 (ac—:1)? fuw : x—2/3W + 1+ 29; (225/3) 2954/3 5x? + 5x — 1 9 (cc — 1)3 3 (x — 1>4 (x 71? —0.6 0 0‘8 From the graphs. it appears that f is increasing on (A060. 1) and (1. 00). with a vertical asymptote at as : 1. and decreasing on (7007 —0.50); f has no local maximum. but a local minimum of about f(70.50) = 70.53; f is CU on (71.110) and (017,1) and CD on (—00. —1.17). (0‘ 0.17) and (1, 00); and f has inflection points at about (—1.17,—0.49). (0,0) and (0.17, 0.67). Note also that lim f(m) : 0. so y : 0 is a horizontal asymptote. r—vioo 37. my) : 31:6 g 5x5 + x4 i 5273 i 22:2 + 2 => f’(;r) : 18w5 ‘ 25204 + 4$3 — 15:2:2 — 43: => f"(:(:) i 903:4 100x3 l 12562 30m 4 From the graphs of f’ and f”. it 100 2.5 appears that f is increasing on - (70.2310) and (1.62. 00) and decreasing on (—00. —0.23) and (0, 1.62); f has a local maximum of 71's 25 _0,5 05 about f (0) = 2 and local minima 725 048 of about f(70.23) : 1.96 and f(1.62) : —19.2: f is CU on (—00, —0.12) and (1.24., 00) and CD on (—0.12, 1.24); and f has inflection points at about (—0.12 1.98) and (1.24, ~12.1). 410 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 38. f(m) : sinxcos2sc => f'(x) : cossw — 2sin2x cossc => f"(x) = —7sin:c c0522 + 25in3x aw i” From the graphs of f’ and f”, it appears that f is increasing on (0, 0.62). (1.57, 2.53). (3.76. 4.71) and (5.677 271') and decreasing on (0.62157), (2.53, 3.76) and (4.71, 5.67); f has local maxima of about f(0.62) : f(2.53) = 0.38 and f(4.71) : 0 and local minima ofabout f(1.57) = 0 and f(3.76) : f(5.67) : —0.38; f is CU on (1.08, 2.06). (314,422) and (5.20, 27r) and CD on (0.1.08), (206, 3.14) and (4.22. 5.20); and f has inflection points at about (0, 0). (1.08, 0.20). (2.06, 0.20). (3.14., 0). (4.22, 70.20). (5.20, —0.20) and (271'. 0). 39. | From the graph. we estimate the points of inflection to be about (i082. 0.22). f(:c) : (fl/$2 :> f’(;c) = 2.5351022 3 f”($) : 2 [:c’?’(2zr_3)e’1/x2 + 6'1/22 (73174)] 2 235—661”2 (2 — 3x2). —5 5 0 This is 0 when 2 7 3:02 : 0 <:> x : :: so the inflection points are (:: %.e_3/2). 1 , 1 _ 1 40- (a) H (1)) fix) — my zlglgoflx) — 1 + 1 — 2’ . 1 1 . 1 _ 1 ‘1010 11m f(_qj):——:]_ v0.1 250— 1 +0 (c) From the graph of f. estimates for the IP are (—0.4. 0.9) and (0.4, 0.08). 61/“ [cal/W250 i 1) + 2x + 1] 354031” +1)3 (e) From the graph. we see that f” changes sign at x : $0.417 (d) f"(I) — — (cc : 0 is not in the domain of f). 1P are approximately (0.417. 0.083) and (70.417. 0.917). 41. 42. 43. CHAPTEI'M REVIEW E 411 flat) 2 arctan(cos(3arcsin We use a CAS to compute f’ and f". and to graph f, f’. and f”: From the graph of f’. it appears that the only maximum occurs at w = 0 and there are minima at w : i087. From the graph of f ". it appears that there are inflection points at a: : i052. f(:c) : ln(Zw + as sin We use the CAS to calculate 4 2+sinm+$cosm I _ d f 2m + xsinm an f,,( ) 2x25inz+4sinar—c032x+x2+5 x : —-————— x2(cos2 a: — 4sinm i 5) From the graphs. it seems that f’ > 0 (and so f is increasing) on approximately the intervals (0. 2.7). (4.5. 8.2) and (10.9. 14.3). It seems that f" changes sign (indicating inflection points) at a: z 3.8. 5.7. 10.0 and 120. Looking back at the graph of f. this implies that the inflection points have approximate coordinates (3.8. 1.7). (5.7, 2.1). (10.0, 2.7), and (12.0, 2.9). The family of functions f(m) : ln(sin x + C) all have the same period and all have maximum values at x : g + 27m. Since the domain of In is (0.00). f has a graph only if sinac + C > 0 somewhere. Since —1 S sinzz: g 1. this happens ifC > *1. that is. f has no graph ifC E —1. Similarly. ifC' > 1. then since + C > 0 and f is continuous on (700. 00). As C increases. ,5 / —0.5 0 l the graph of f is shifted vertically upward and flattens out. If—l < C’ S 1. f is defined where sinac + C > 0 42> sinm > —C’ <=> sin’1(—C) < m < 7r ~ sin—1(~C’). Since the period is 27r. the domain of f is (27m + sin’1(—C), (2n + 1)7r — sin‘1(—C’)). n an integer. . We exclude the case 0 : 0. since in that case f(93) : 0 for all :c. To find the maxima and minima. we differentiate: 2 = ave—m” :> f’(;c) = c[9:e““2(—20m) + e_"12(1)] : ce’cmg (720222 + 1). This is 0 where —2c:c2 + 1 = 0 <:> ac : i1/V2c. So if c > 0. there are two maxima or minima. whose git—coordinates approach 0 as c increases. The negative root gives a minimum and the positive root gives a maximum. by the First Derivative Test. By substituting back into the equation. we see that f(:t1/\/2_c) : c(:l:1/x/2_c) ¢2_”(i1/‘/§E)2 2 :l: c/Ze. So as c increases. the extreme points become more pronounced. Note that if c > 0. then lirin f(x) : 0. If c < 0. x‘) 00 Iim 14>:oo then there are no extreme values. and (as) : :Foo. 412 El CHAPTER4 APPLICATIONS OF DIFFERENTIATION 45. 46. 47. 48. 49. To find the points of inflection. we differentiate again: f’(m) (re—“2 (—26232 + 1) => , 2 f"(2:) = c[e"‘I (74cm) + (—20:52 +1)<—2cxe_"m2>] = —2c2:2:e_””2 (3 — 20:32). This is 0 at :c : 0 and where3 ~ 262:2 : 0 <:> 5c : ::\/3/(2c) => [Pat (i 3/(20), is/30/2 673/2). Ifc > 0 there are three inflection points. and as c increases. the sic—coordinates of the nonzero inflection points approach 0. If c < 0. there is only one inflection point. the origin. 0.75 -0.75 f(ac) = $101 + x51 + :c i 1 : 0. Since f is continuous and f(O) : 71 and f(l) : 2. the equation has at least one root in (0. 1). by the Intermediate Value Theorem. Suppose the equation has two roots. (1 and b. with a < b. Then f(a) = 0 : f(b). so by the Mean Value Theorem. there is a number an in (a, b) such that W) : 1%) A M) _ _o_ b—a ibia f’(z) = 10155100 + 519550 +1 2 1 for all as. : 0. so f’ has a root in (a. b). But this is impossible since By the Mean Value Theorem. f’(c) : w <:> 4f’(c) : f(4) i 1 for some 6 with 0 < c < 4. Since 2 g f’(c) g 5. we have 4(2) 3 4f’(c) 5 4(5) <:> 4(2) 3 f(4) — 1 3 4(5) <:> 8 S f(4) A 1 g 20 <2; 9 s f(4) : 21, Since f is continuous on [32, 33] and differentiable on (32, 33). then by the Mean Value Theorem there exists a 5 5 l -4/5:_\/§—__@: 5 3_ 5‘3 33—32 ‘/3— => {3/33 > 2. Also f’ is decreasing. so that f’(c) < f’(32) : §(32)"V5 = 0.0125 => 0.0125 > f’(c) : \5/33 7 2 :> {3/33 < 2.0125. Therefore. 2 < \5/33 < 2.0125. number 6 in (32.33) such that f’(c) : 2. but écT4/5 >0 => 3/33 — 2 > 0 For(1.6)tobeonthecurvey :m3+a502 +bm+ 1.wehavethat6:1+a+b+ 1 :> b: 4711. Now y' : 3.762 + 2am + b and y” : 6m + 2a. Also. for (1. 6) to be an inflection point it must be true that y"(1) i6(1) I 2a 0 > a -3 > b y" : 62: ~ 6 : 6(22 7 1). so y” changes sign at z : 1. proving that (1., 6) is a point of inflection. [This does not 4 — ( 3) = 7. Note that with a. : 73. we have follow from the fact that yl/(l) : 0.] (:1) 9(2) : f(sc2) :> g'(a:) : 2xf'(302) by the Chain Rule. Since f’(a:) > 0 for all z ¢ 0. we must have f’(a:2) > O for ac % 0. so g'(a:) : 0 <=> x : 0. Now g'(:c) changes sign (from negative to positive) at 50. 51. 52. 53. CHAPTEIM REVIEW II 413 :c : 0. since one of its factors. f' is positive for all x. and its other factor. 2m. changes from negative to positive at this point. so by the First Derivative Test. f has a local and absolute minimum at at : 0. (b) g’(:c) : 25cf’($2) => g”(z) = 2 [Ifl/(IIIQ) (21:) + f’(:c2)] = 4w2f”(m2) + 2f’(m2) by the Product Rule and the Chain Rule. But 3:2 > 0 for all cc 75 0. f”(m2) > 0 (since f is CU form > 0). and f’(a:2) > 0 for all :c 75 0. so since all of its factors are positive. g"($) > 0 for as 76 0‘ Whether g"(0) is positive or 0 doesn’t matter (since the sign of 9” does not change there); g is concave upward on R. Call the two integers :c and y. Then as + 4y 2 1000. so as = 1000 — 4y. Their product is P : my 2 (1000 * 4y)y. so our problem is to maximize the function P(y) = 1000y * 4y2. where O < y < 250 and y is an integer. P’(y) : 1000 — 8y. so P/ : 0 <:> y = 125. P”(y) : *8 < 0. so P(125) : 62.500 is an absolute maximum. Since the optimal y turned out to be an integer. we have found the desired pair of numbers. namely :6 : 1000 — 4(125) : 500 and y :125. If B = 0. the line is vertical and the distance from x : —g to IASCl + Byl + 0| A ————.so W assume B 75 0. The square of the distance from ($1411) to the line is : (m 7 x1)2 —I— (y i y1)2 where (x1.y1) is m+gi 1A7 2 Am + By + C : 0. so we minimize f(93) : (a: — x1)2 + (—330 — g — y1> => airs-w . r . , A . . . . . a minimum smce f '(x) : 2 <1 + > 0. Substituting this value of so into and simplifying gives f’(.1;) :2(z—e1)+2<_§ma % , x 32m — ABy1 i AC _ A2 + B2 and this gives (A151 + Byi -l- C)2 _|A1‘1 +By1 +C| A2+B2 ———- W On the hyperbola my : 8. if d(;c) is the distance from the point (33,31) 2 8/x) to the point (3i, 0). then [d($)]2 (a: i 3)2 + We2 : f(:v). f’(:c) : 2(2: — 3) — 128/233 : 0 :5 m4 7 31:3 — 64 : 0 => (:2: — 4)(m3 + 1:2 + 426 +16) 2 0 => :3 = 4 since the solution must have cc > 0. Then y = g = 2. so the point is (4‘ 2). f($) : . so the minimum distance is f (96) . , . r By Similar triangles. g : —. so the area of the triangle is LE 352 — 27m 2 A(:c) — = my : NC :> $2 — 2m: rx2 (ac e 31") (:02 * 27:0)3/2 A'($) < 0 when 21" < a: < 37'. A'(x) > 0 when :0 > 37180.7: : 3r gives a minimum and A(3¢) : r(9r2)/(\/§r) : 3 V373. : 0 when at : 3r. 414 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 54. The volume of the cone is V : $712120" + 1:) = %7r(7'2 — $2) (7" + 77‘ S :c < 7'. V’(:c) : g [(1"2 ~ x2)(1) + (r + $)(~2m)] g [(1’ + :c)(r — cc 7 2:0] 2 £0 + 1‘)(r ~ 32:) V xv :OWhCn$:*7‘O]‘$=T/3. A Now V(r) : 0 : V(—r), so the maximum occurs at as = r/3 Qv—’ 2 3 v and the volume is V(:) = E T2 _ L. g : 327W ' i 3 3 9 3 81 I | 55. We minimize L(m) : |PA| + IPBI + |PC| = 2x/232 + 16 + (5 — C T 0£m§5iL’(w)=2m/\/$2+16—1:O 4=> T 2m:\/m2+16 <:> 4x2:x2+16 <=> m:%. A B i L(0) : 13. 4%) m 11.9. L(5) m 12.8. so the minimum 4 D 4 4 occurs when a: = W ~ 2.3. 56. C If |CD| : 2. the last part of L(m) changes from (5 — ac) to (2 — ac) with 0 g m g 2. But we still get L’(x) 2 O <=> a. as : %. which isn’t in the interval [0, 2].Now L(0) : 10 I A 4 D 4 B and L(2) : 2 \/2_ : 4 x/g % 8.9. The minimum occurs wheanC. L C’ dv K 1 C) 1 C 2 2 57. = —+— —:——-— —i— —0 <—> —*— <-3 L *C A v C L dL 2 (L/0)+(C/L)(C L2 C L2 L = C. This gives the minimum velocity since 11’ < 0 for 0 < L < C and 11' > 0 for L > C. 58. We minimize the surface area S : 71'7'2 + 27r7‘h + %(47rr2) : 37rr2 + 27r7‘h. V — 271'1‘3 V I 3 _ 3 _ 2 Solvmg V : 7rr2h + gm“ for h. we get It i —7r—T2— i W A 37". so V 2V 2 i 2 5(r) : 37TT‘ + 27T7’[m — gr] — gm" + , 2V EM"3 — 2V S(r):~—73+1—§7rT:————3 T2 :0 <=> 1397rr3:2V 3V 42> r3 : fl <=> r = {3/ This gives an absolute minimum since S’(r) < 0 for 0 < r < 3 — and 5m SW 571’ 2 3V l / 3 3V V’ 57?“ 57 (V# gv) 3 (5702 3VE/(57r)2 _ 3 3V g S(r)>0forr> ——.Thus.h: 7 3 2 3 2— 5 —r 5" 7r 3 (gr/)2 7r \/ (3V) 571' \/(3V) 7r CHAPTER 4 REVIEW 415 59. Let 3: denote the number of $1 decreases in ticket price. Then the ticket price is $12 — $1(m). and the average attendance is 11.000 + 1000(w). Now the revenue per game is R(a:) : (price per person) x (number of people per game) = (12 i w)(11.000 + 10003:) : 71000352 + 1000w + 132,000 for 0 S as S 4 (since the seating capacity is 15.000) => R’(93) : —2000x + 1000 = 0 <=> :c = 0.5. This is a maximum since R”(m) : —2000 < 0 for all :0. Now we must check the value of R(x) : (12 — x)(11.000 + 100035) at :c : 0.5 and at the endpoints of the domain to see which value of ac gives the maximum value of R. 3(0) : (12)(11.000) : 132.000. R(0.5) : (11.5)(11.500) : 132.250, and R(4) : (8)(15.000) : 120.000. Thus. the maximum revenue of $132250 per game occurs when the average attendance is 11.500 and the ticket price is $11.50. 60. (a) C(x) : 1800 + 2533 e 0.2m2 + 0.00m3 and 12000 R(:I:) : 3313(33) : 48.256 * 0.03x2. The profit is maximized when C(30) : R’(m). From the figure. we estimate that the tangents are parallel ( 300 0 when x m 160. (b) C'(z) : 25 e 0.4a: + 0.003%2 and R’(3c) : 48.2 — 0.06m. 0'05) : R’(a:) :5 0.0032:2 7 0.34m — 23.2 : 0 :> :61 x 161.3 (x > 0). 3%) : 70.06 and 0"(33) = —0.4 + 0.00613. so R”(m1) : 70.06 < C"(m1) R“ 0.57 => profit is maximized by producing 161 units. ‘ C(zc) _ 1800 (C) C(30) : 200 m m + 25 — 0.220 + 0.00m2 is the average cost. Since the average cost is minimized when the marginal cost equals the average cost. we graph c(:v) and C/(zc) and estimate the point of intersection. From the figure. C’(m) : C(av) 42> :c m 144. ‘4 300 0 61. :xs—x4+3x2—3xi2 => f’(a:)=5:c4i4x3+6x73.so xi—mi—l—3mi—3xni2 5m$—4m%+6$n~3 xn+1 1% .Now .271 : 1 => $2 = 1.5 :> $3 8 1.343860 => :34 m 1.300320 => m5 z 1.297396 => 9:6 3 1.297383 m 2:7. so the root in [1.2] is 1.297383. to six decimal places. 416 I: CHAPTEH4 APPLICATIONS OF DIFFERENTIATION 62. Graphing y : sinac and y : $2 7 3x + 1 shows that there are two roots. one about 0.3 and the other about 2.8. f(;c) : sinm — 202 +33} — 1 :> f’(m) : cosm — 256 +3 => ~ 2 smx —:c +350 —1 $n+1:xn—#——n—.Nowx1:03 => cosmn—2mn—I—3 $2 % 0268552 :> :63 % 0.268881% :64 and mi : 2.8 i 122 z 2.770354 :> 1:3 2 2.770058 a“ 3:4. so to six decimal places. the roots are 0.268881 and 2.770058. 63. f(t) 2 cost + t i t2 :> f’(t) = — sint + 1 — 2t. f’(t) exists for all If. so to find the maximum of f. we can examine the zeros of f’. From the graph of f’. we see that a good choice for 151 is t; : 0.3. Use g(t) : — sint —I— 1 — 2t and g'(t) = — cost ~ 2 to obtain 2:2 e 033535293. 253 z 033541803 a #4. Since f"(t) : 7 cost 7 2 < 0 for all t. f(0.33541803) % 1.157185571s the absolute maximum. 64. y : f(x) : rsin ac. 0 S :c S 271'. A. D = [0. 271'] B. y—intercept: f(0) : O; m—intercepts: f(:c) : 0 <=> ac : 0 or sinw = 0 <=> :c : 0. 7r. or 277. C. There is no symmetry on D. but if f is defined for all real numbers cc. then f is an even function. D. No asymptote E. f’(ac) : mcos m + sin :c. To find critical numbers in (0. 27r). we graph f’ and see that there are two critical numbers. about 2 and 4.9. To find them more precisely. we use Newton‘s method. setting g(;v) : f’(:c) = accosx —I— sin 3:. so that g’(:c) : f”(x) = 2cosm — xsinx and 3:” cos zen + sin 3:” $7,,“ = :6” 7 $1 = 2 => :02 3 2.029048. x3 % 2.028758 % m4 and 1'1 2 4.9 :> 2cos mu — xn sin xn' $2 % 4.913214. 1133 2: 4.913180 w :64. so the critical numbers. to six decimal places. are n = 2.028758 and r2 : 4.913180, By checking sample values of f’ in (0.11). (7"1.T‘2). and (r2, 2%). we see that f is increasing on (0.?1). decreasing on (T1.T‘2). and increasing on (7‘2.271‘). F. Local maximum value f(?"1) % 1.819706. local minimum value f(r2) 2 —4.814470. G. f”(m) = 2COSZL’ — msin as. To find points where f”($) : 0. we graph f" and find that f”(x) : O at about 1 and 3.6. To find the values more precisely. we use Newton’s method. Set 2 cos mu ~ urn sin mn : f”(m) = 20059: A msinx. Then h'(:c) = —3sinm — :ccosx. so acn+1 : In # —*——‘_3Sin$n i m" cosmn- $1 : 1 => £132 % 1.078028. 333 2 1.076874 % m4 and $1 : 3.6 => 1'2 z 3.643996. mg % 3.643597 % $4. 65 65. 67. 68 69. 70. 71. f”(:c) = 1 — 6m + 482:2 => f’(m) : x 7 3w2 +16m3 + C. f’(0) : c and Ho) : 2 :5 C 2 2. so 72. f"(;c) = 2303 + 3362 ~ 493+ 5 :> f’(;c) : so the zeros of f". to six decimal places. are 713 = 1.076874 and r4 : 3.643597 By checking sample values off" in (0,7'3). (73,721). and (721.2%). we see that f is CU on (0,?3). CD on (73.72;), and CU 0n (721.270. f has inflection points at (r3, fang) z 0.948166) and (7’4, f(r4) z 4.753240). 7 02,, —4 was) : —4/e/5 : x5” Him-“5 :» we) 2 ~4(%x“/5) +0 = 7 5954/5 +0 f'(w)=8$—3sec2$ => f($):8(%$2)#3tanm+0 :417273tansc+Cnontheinterval (nvrigmmfi—g). 2 ex 7 : eI —2$_1/2 => f 2 $-1/2+1 C SCI/2 w _ :r _ x_2 : _ mi () _1/2+1+ 6 1/2 C e 4\/E+C .f’(:c)=2/(1+ac2) => f(:c)=2arctanw+C. f(0)=2arctan0+C:0+C:Candf(0):—1 => C:—1.Sof(x):2arctanx—1. f’(t):2t735int => f(t)=t2+3cost+C. f(0)=3+Candf(0):5 => 0:2.sof(t)=t2+3cost+2. 2 f’(u):%\/E:u+u~l/2 :> f(u):%u2+2u1/2+C. f(1):%+2+0andf(1):3 : 02%.sof(u):%u2+2\/E+%. f'(:c) = a: — 3x2 + 16233 + 2 and hence. = D:1.s0f(m):§$2—x3+4x4+2$+1. éxQ—m3+4m4+2x+D. f(0):Dandf(0):l é CHAPTER 4 REVIEW H. §z4+x3—2x2+5x+0 => f(w)=$x5+iw4—§m3+gfi+0$+n f(0):Dandf(0):2 ;> 0:2. f(1):fi+fi—§+E+C+2andf(1)—O A> 0— 6 f(fc) : fix5+ 1 438 4 ,2 3 E 2_E 3$+2£C 60x+2. 150 60 120 15 40 + 60 7 _ 251 60 60 ~ 60 SO 60 417 418 D CHAPTER4 APPLICATIONS OF DIFFERENTIATION 73. (a) Since f is Ojust to the left of the y—axis. 5 we must have a minimum of F at the same place since we are increasing through (0, (l) on F. There must be a ,4 L 4 H- 72 local maximum to the left ofx = —3. since f changes from positive to negative there. (b) : 0.16m +sina: => = 0.1ex — cosm + C. (c) 5 F(()):O => (11714—020 :> C=O.9.so : 0.16I i cosx + 0.9. _4 4 __ -—l 74.: x4 + 3:3 + 0102 => : 4x3 + 3302 + 2630‘ This is 0 when $(4w2 —I— 3:1: + 2c) : 0 <=> z : 0 or 4:62 + 3x + 2c : 0. Using the quadratic formula. we find that the roots of this last equation are i 734; t/9 — 32c _ 8 :c . Now if9 — 320 < 0 ¢> c > 3—92. then (0, 0) is the only critical point. a minimum. lfc : 395. then there are two critical points (a minimum at m : 0. and a horizontal tangent with no maximum or minimum at LL’ 2 —§) and ifc < 3—92. then there are three critical points except when G : 0, in which case the root with the + sign coincides with the critical point at a: : 0. For 0 < c < 3—92. there is a minimum at 3 v9 — 32C 3 V9 ~ 326 .r I —§ — a maximum at a: : —g + 8 . and a minimum at 1: =0. For c : 0. there is a minimum at as : —% and a horizontal tangent with no extremum at a: : 0, and for c < 0. there is a maximum at m : 0. and there are minima at J? : 7% :: Now we calculate f”(rc) : 12:52 + 630 + 2c. The roots of this equation are ac : So if 36 i 960 S 0 4:) c 2 3. then there is no inflection point. If c < 3. then there are two inflection points at as = —i i Value ofc No, of CP No. of IP 75. Choosing the positive direction to be upward. we have a(t) 76. 77. CHAPTEIM REVIEW E] 419 :UO —9.8 => 0(t) = —9.8t + 110, but0(0) = 0 110) —9.8t 5'0) > s(t) 4.9:2 + so. but 3(0) : so = 500 :> 3(t) _ —4.92:2 + 500. When 3 : 0. —4.9t‘2 + 500 : 0 => 251 = % m 10.1 => 0(t1) : 79.8 3% x —98.995 Since the canister has been designed to withstand an impact velocity of 100 m/s. the canister will not burst. Let sA(t) and 53(t) be the position functions for cars A and B and let f(t) : sA(t) — 5(t). Since A passed B twice. there must be three values of t such that f(t) : 0. Then by three applications of Rolle’s Theorem (see Exercise 4.2.22). there is a number 6 such that f"(c) : 0. So 513(0) : 3% that is. A and B had equal accelerations at t = c. We assume that f is continuous on [0. T] and twice differentiable on (0. T). where T is the total time of the race The cross—sectional area of the rectangular beam is (a) A:2m-2y=4my=4$\/100—x2.03a:g10.50 1/2. — = 42%) (100 — x2)71/2(*2m) + (100 4 m2) 4 —4:2:2 (100 i 952)”? 4 [—x2 + (100 — fl] (100 i 3201/2 ‘ +4 (100 — 952)”? dA dcc y : i/ 100 — (v50 )2 : x/SO. Since A(0) : A(10) = 0. the rectangle of maximum area is a square. 0when—x2+(100—x2):0 :2 42:50 :~ x:\/50%7.07 => (b) The cross—sectional area of each rectangular plank (shaded in the figure) is A : 2z(y — m) : 2$[\/100 — $2 — x/50].0 g $ : t/50. so dA , E : 20/100 7 2:2 ~ «50) + 2m(%) (100 — .262) 1/2 (7220) 2 _ 2(100 we 2 x/50 2"" (100 — x2)1/2 % 7 , 2 2 1/2 2 2 2 1/2 Set dm _0. (1()0—m)7\/%(100—x) —m :0 :> 100—2m : \/50(100—m) : 10.000 — 400172 + 41:4 : 50(100 4 2:2) :> 4m4 — 35022 + 5000 : 0 :> . 175 :t ‘/—10,625 2264 — 1752:2 + 2500 : 0 :> x2 : —4— m 69.52 or 17.98 => w m 8.34 or 424. But 8.34 > V50. so 331 m 4.24 => y i v50 : \/ 100 i 2:? * v50 % 1.99. Each plank should have dimensions about 8% inches by 2 inches. 420 C CHAPTER4 APPLICATIONS OF DIFFERENTIATION (c) From the figure in part (a). the width is 2m and the depth is 2y. so the strength is s : k(2:c)(2y)2 : 8k$y2 : 8ke(100 — 2:2) : 800m i 81663.0 3 x g 10. dS/dz = 800k — 24kw2 : 0 when24kac2 800k > .722 1—30 > m 37% > y M? 13/35 Since 5(0) : 3(10) : 0. the maximum strength occurs when m = %. The dimensions should be % m 11.55 inches by M m 16.33 inches. \/§ 78. (a) .V y = (tan t9):c — —g—;cZ. The parabola intersects the 2112 cos2 0 line when (tan (1)0: : (tan 60:3 — may => (tanfi — tan 002122 cos2 6 x : ————— :> 9 12(0) V x _ sint9 * sina 212200826 7 sin6 _ sina (€080 COSQ)2v2c080 icosa _ cos€ cosa gcosa _ c086 cosoz gcos2a 2 2 2 2 = (sin0 cosa i sina cos QM : sin(0 — 00w 9 cos2 a 9 cos2 a 2 2 2 (b) R'(0) : 930:2 a [c050 - cos(0 i a) + sin(0 — oz)(— sin 6)] : 903:2 a cos[0 + ((9 i (1)] 2 : 202 cos(26*a) :0whencos(26—Ct) :0 => 20—0: => gcos oz 7r/2 + (1 7r (1 . . . . 1 . 0 2 —— : -— + —. The First Derivative Test shows that this gives a max1mum value for R(0). 2 4 2 [This could be done without calculus by applying the formula for sin 2' cos y to R(6).] 2v2 cost9 sin(t9 + a) (c) Replacmg oz by —a 1n part (a). we get R(6) : g CO82 a . Proceeding as in part (b). or simply by replacing a by —a in the result of part (b). we see that 12(0) is maximized when 0 : g i gt kcos 9 k(h/d) h h h (a) d2 d2 d3 (1/402 + [12)3 (1600 + hl2)3/2 d1 k (1600 + h2)3/2 e hg (1600 + h2)1/2 ‘ 2h _ k (1600 + h2)1/2 (1600 + h? — 3h2) — : —-————— 3 k(1600 i 2h2) : ———— [k is the constant of proportionality] (1600 + m5” Set d1 /dh : 0: 1600 2h2 0 > h2 800 > h #800 20 J2. By the First Derivative Test. I has a local maximum at h : 20 x/2 m 28 ft. CHAPTER 4 REVIEW 421 03) d2: a kcosG _ k[(h—4)/d] _ k(hi4) A k(h—4) _ i g 2 362 -3/2 1: d2 _Ti—d3 _———[(h_4)2+$2]3/2 7km 4)[(h, 4) + ] d1 d1 drc 3 2 2-5/2_ xi? E:d—I~E:k(h—4)(—§)[(hi4) +m] 2 dt _ _ _ _ 2 2—5/2. : —12wk(h—4) Jam 4)( 3x)[(h 4) +x] 4 [(h [gums/2 dI 480k(h—4) E 98:40 [(h — 4)2 +160015/2 80. (a) V’(t) is the rate of change of the volume of the water with respect to time. H’(t) is the rate of change of the height of the water with respect to time. Since the volume and the height are increasing. V’(t) and H'(t) are positive. (b) V’(t) is constant. so V”(t) is zero (the slope of a constant function is 0). (c) At first. the height H of the water increases quickly because the tank is narrow. But as the sphere widens. the rate ofincrease of the height slows down. reaching a minimum at t = t2. Thus. the height is increasing at a decreasing rate on (0,152). so its graph is concave downward and H "(751) < 0. As the sphere narrows fort > 752. the rate of increase of the height begins to increase. and the graph of H is concave upward. Therefore. LII/(t2) 2 0 and HI/(tg) > 0. w 81. We first show that 1:332 < tan‘1 1' for a: > 0. Let f(;13) = tan’1 m — 1+ $2. Then 1 1(1 + m2) * z(2m) (1 +102) — (1— 232) 28 f’x: . — _ >0forx>0.So mis ( ) 1+x2 (1+$2)2 (l—l—av2)2 (ll-cc?)2 f( ) increasing on (0.00). Hence. 0 < :16 => 0 : f(0) < f(m) : tan‘1 as 96 So x < tan’1 ac _ 1+ x? 1 + m2 for 0 < m. We next show that tan‘1 m < as for a: > 0. Let h(ac) : 2: — tan‘1 up. Then 1 $2 h' :1— (1) 1—l—sc2 1+:102 > 0. Hence. h(m) is increasing on (0, 00). So for 0 < :27. <tan_1$<ar 0 = h((]) < h(:c) : w — tan‘1 m. Hence. tan‘1 at < ac form > 0. and we conclude that + 2 :c for a: > 0. 422 D CHAPTER4 APPLICATIONS OF DlFFERENT’ATlON 82.1ff’(m)< 0for all 9:. f"(:c) > 0for>1.f"(:v)< 0 for < 1‘ and lim + ac] = 0. then f is decreasing everywhere. concave up on maicx) (—oo. —1) and (1. oo). concave down on (—1. 1). and approaches the line y : —ac as x —> :too‘ An example of such a graph is sketched. ...
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 / 24

ch review - CHAPTER4 REVIEW E 399 17 22 (b) To find the...

This preview shows document pages 1 - 24. Sign up to view the full document.

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