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ch review - CHAPTER4 REVIEW E(b To find the maximum height...

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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: F...
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