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hw-1-sec-4.9-solns

# hw-1-sec-4.9-solns - 3“ |—‘ Ul’lnrlun-v ru I...

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Unformatted text preview: 3“) |—‘ Ul’lnrlun-v ru I LIVI‘UIVI‘U v: u” . ._.“_... .. ...v.. {E’ 4.9 Antiderivatives 1+1 1. f(:L‘)=:v—3:x1—3 => F(:1c): f+1—3x+C:%x2—3\$+C Check: F'(m) : %(2\$) — 3+0 = a: — 3 : f(m) 3 4 2 ®f(\$)=%\$2—2\$+6 => F(a:):%%——-2%+6m+C:%x3—x2+6x+C } 3. f(x)=%+%:c2—§x3 => F(\$)=%x+—-———33+1+C:%a:+im3—%m4+C ‘ Check: F'(a3) =%+ﬁ(3m2)— %(4x3)+0=%+%x2 — 5:1: =f(x) , .ﬂx) : 8x9 — 3x6 + 123:3 :> F(ac) = 8(%\$1°)— 3 (\$227) + 12 (if!) + C 2 gm“) — \$937 +3304 + C I} Q 5. f(x)=(:c+1)(2ac—1)=2\$2+a:—1 :> F(1:)22%;?)+%\$2—x+C=-§\$3+%w2-m+0 I 6. f(:1:) =x(2—m)2 =x(4—4x+\$2) :4m—4x2+:v3 => mm = 4 (W) — 4 (m + + o = 2x? — + +0 1/4+1 3/4+1 5/4 7/4 7_ = 1/4_ 3/4 F = 53 _733 C: 33 _ x _4 5/4_ 7/4 f(m) 5:1: 755 i (a?) 5i+1 g+1 + 55/4 77/4—l—C as 413 +0 9. f(\$)=6\/§— W=6I1/2'\$1/6 => 1/2+1 1/6+1 3/2 7/6 an _ as + C : 6 £— _ :1: §+1 g+1 3/2 7/6 ' 7 F(1‘)=6 1o. f(x) = \4/x3—1— \3/134 2373/4 +x4/3 :> F(a:) = 7/4 + m +0: \$x7/4+%m7/3+C 10 #8 _ 10 338 Cl_—4—5-8-+C1 lfaﬁ<0 11. f(:c) = F = 1093-9 has domain (—oo,0) U (0, 00), so F016) 2 —5 a: __4m8+C2 ifac>0 See Example 1(b) for a similar problem. _ 3 6 12. 9(23) : w 2 San—6 # 435—3 + 2 has domain (—00.0) U (0, 00), so \$6 *5 —2 2 . 5L‘4L+2\$+C1=—-1—+—+2x+01 1fx<0 _ —5 ——2 1‘5 \$2 GW— 1 2 “7+7+2\$+C2 ifzc>0 I: {E 4 4 1/2 13. f(u):E_—;__:\/a:§:3+3::2 ZU2+3u43/2 ﬁ 3 —3/2+1 1 71/2 1 U U +C_—u3 31L 3 +0 ‘F(“>=€+3T?./2—+—1 ‘3 + ~1/2+C:§”"ﬁ SECTION 4.9 ANTIDERIVATIVES — 327 ®f(t)=3cost—4sint => F(t)=3(sint)—4(-cost)+C:33int+4cost+C 15. g(6):cos6—5sin0 :> C(19)=sin6—5(—cos€)+C=sin6+5c050+0 .f(0) = 602 — 7sec2 6 : F(()) = 263 —— 7tan6 + Cn on the interval (mr — 5,72% +— g.) 17. f(t) = ZSect taut + \$t‘1/2 has domain (O, NH ) and (mr — 3, n7r +— g) for integers n 2 1. The antiderivative is F(t) :- 2sect+t1/2 + Co onthe interval (0, %) orF(t )— — 2sect+tu2 + CT. on the interval (mr — §,n7r+ ) for integers n 2 1. 3/2 18. f(:c):ZVE-l—6cosmz2ml/2—i—6cosm : F(m)=2<:—/2>+65inz+C=§x3/2+6sinm+0 5 6 m a: 5 19. f(:r)=5m4—2\$5 :» F(x)=5-3—2‘F+C=x — éﬂie + C. F(0)=4 :5 05—5-06+C=4 :5 C=4,soF(x) =x5~§x6+4 The graph conﬁrms our answer since f (:3) z 0 when F has a local maximum, f is positive when F is increasing, and f is negative when F is decreasing. 20. f(x)_:c+2sm1: :> F(:c)= 51:2 —2c0sx+C 12 F(0)=—6 : 0—2+C=—6 => C:—4,so F03): 2x2 —2cosx—4. 12 17- The graph conﬁrms our answer since f (cc) : 0 when F has a local minimum, f is positive when F is increasing, and f is negative when F is -12 decreasing. x2 m3 21. f"(a:):6:12+12:c2 :> f’(x) =6.7+12.?+C:3\$2+4x3+0 : x3 1:4 f(a:) : 3 ' 3 + 4' ‘4‘ ‘l‘ 017 + D = \$3 + £164 + 090 + D [C and D are just arbitrary constants] 22. f"<x>=2+x3+x6 :» f’<x>=2w+iw4+%x7+0 a» from" +2~om +- 1890 +C\$+D \$8/3 23.f”(x)_ 2/3 :> f’(x)=§<—5—/33>+C= -x5/3+C => f(x)= 2(873 >+Cx+D= 2—35zB/3+Cm+D 2 24. f”(:r:) = 6m+sinx :> f’(1‘) : 6(E. 2>—cosm+C=3x2—Cosm+0 => 3 f(\$)=3<%> —Sin\$+Cx+D=;r3—sinx+Ca:+D 25. f”’(t) = 60t2 2 f”(t) = 20:3 + C 3 f’(t) 2 5t4 + Ct + D => 1%) = t5 + 5&2 + Dt + E 26. f”’(t) : t — \/Z :> f”(t) = y — gt“ + C => f’(t) = §t3 — list“? + Ct + D :> f(t) = it“ — itm + §Ct2 + Dt + E 105 27. ’(m)=1—6m => f(;c):x—3x2+C.f(0):Candf(0)=8 => 0:8,sof(x)=ac—3x2+8. 328 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 1%.? .f’(m)=8x3+12m+3 2 f(a:)=2m4+6x2+3m+c.f(1)=11+Candf(1)=6 2 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 11+C=6 2 C:~5,sof(m)=2x4+6x2+33:—5. f’(:c) = ﬁ(6+ 5x) = 6331/2 + 5203/2 :> f(a:) : 4333/2 +215/2 + C. f(1) = 6 + C and f(1) = 10 2 C = 4, so f(a:) 2 413/2 + 29:5/2 + 4. f/(aj) : 25v — 3/174 2 2m — 31274 => f(\$) = m2 + \$“3 + C because we’re given that a: > 0. f(1)=2+Candf(1)=3 :> C=1,sof(1:):a:2+1/;r3+1. f’(t)=2cost+sec2t => f(t):2sint+tant+Cbecause—7r/2<t<7r/2. f(§)=2(\/§/2)+\/§+C=2\/§+Candf(§)=4 2 C:4—2x/§,sof(t):2sint+tant+4—2\/3—). gym—ICE ifm>0 f’ 3:) = 304/3 has domain (—00, 0) U (0, 00) => f(:r) : ( %x2/3+C2 ifx<0 f(1) — 3 +C1 andf(1) = 1 9 Cl _§ -%-f(—1)=§+Cgandf(-1):_1 2 022—3. ﬁmZ/S—l ifac>0 Thus, f(:c) : {: 2/3 : 53: — 5 if as < 0 f"(:c) = 24952 +255 + 10 2 f'(a:) 2 8:173 +x2 + 10:3 + C. f’(1)= 8 + 1 + 10+ Cand f’(1)= —3 2 19 + C = —3 : C = —22, so f’(;v) = 8933 —I— x2 + 10x — 22 and hence, f(a:) : 2:54 + %\$3 + 5332 — 22cc —I— D. f1 =2+l+5—22+Dandf1 =5 2 D222—[email protected],sofx =2m4+lx3+5m2—22x+§2. 3 3 3 .3 3 f"(z)=4—6a:——40;v3 2 f’(x)=4m~3m2—1Ox4+C.f’(0)———C’andf’(0)=1 2 C=1,so f’(:£):4x—3zr2«10x4+1andhence,f(:v)=2m2—zr3—2m5+:c+D.f(0):Dandf(0)=2 : D:2,so f(:1:)=2ac2—\$3—2x5+m+2. f"(6):sin6+cos€ => f’(6)=—cos¢9+sin0+C.f’(0)=—1—I—Candf’(0)=4 : 025,30 f’(6)=—C056+sin0+5andhence,f(0):—sin6—cosc9+50+D.f(0)=—1—I-Dandf(0)=3 : D24, sof(6)=—sin6—cos€+56+4. f”(t) 2 3N5 2 3r“? 2 f’(t) 2 6251/2 + C. f’(4) = 12 + C and f’(4) = 7 2 C = 25, so f’(t) = 6231/2 — 5 and hence, m) = 49/2 - 5t + D. f(4) = 32 — 20 + D and f(4) = 20 2 D 2 8, so me) = 45"” — 5t + 8. f"(.r):2—12:r :> f’(:v)=2ct:—6x2—I—C :> f(z)=\$2—2x3+Cm+D. f(0)=Dandf(0)=9 2 D29.f(2)=4—16+20+9=20—3andf(2):15 2 20:18 2 I 0:9,sof(x)=x2—2\$3+9x+9. f”(a?)=20:r3+12x2+4 :> f'(\$)=5\$4+4m3+4m+0 => f(\$):a:5+z4+2x2+Cm—I—D. f(0):Dandf(0):8 2 D28.f(1)=1+1+2+C+8:C+12andf(1)=5 2 C:—7,so f(:r)=:c5+x4+2x2—7\$+8. ...
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