ma1012_hw12_201012

ma1012_hw12_201012 - hÄ! h hÄ! h hÄ! h at (2ß 5): y c 5...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: hÄ! h hÄ! h hÄ! h at (2ß 5): y c 5 œ 4(x c 2), tangent line c) c(" b h) c 2(1 b h) d c( (c1) a 8 c ( c b h) 4h œ lim c1 b hc#2 c h c 2h b b 1 c12. hm 1)1b limc(" c "lim1d ( h h) (1 b c œ 0. d c œ ) b 1m h œ lim h h h ! h Ä ! œ lim h Ä hÄ !! c8h(c# b h) Ä h ! hÄ! at ("ß c"): œ bim œ c a3(xcc 1),htangent line 12 c 6h b h y l 1 c 12h 6h b b œ lim im h œ 0; at ("ß "): y œÄ ! 0(x8h(c1) h) y œh1, ! 8(c2 b h) 1 b c c #b Ê h Ä m Mathematics I: Homework 12. Tangents and derivatives œ lim h(c3hc 2h) œ c3; hÄ! Ä! 3 œ 38("h28)cœ c 16 ; t line 3 c h) c 3(h c2h 13. c h œb 3 m b lim (3 h) h 2 h(4 b h) lim (3 b h(h b 1) b 1) œ lim h(h b 1) œ c2; œ a1 c4 c4h h Äat ˆim c " ‰ : problemsd 3 (x3) (c2)) Ä !1 c4 c4h c h b b 3 Ä ! " h) c h œ! l c#ß c 8 h7 yœ1c4,bc 16 (c c cc a Total: lim h œ c (28 slope At x œ h y œ c3 h Ä ! œ 2, Êm œ lim œ lim h h at ($ß $): Êcy3œ c h2(x !c 3), tangent line h Ä ! y œ c 3Ä c " , tangent line hÄ! x# 16 h) c 8 œ lim 2h(2 h" œ c " c 2lim )At xhœ 3, y œ b h)Ê (2mh)4 , slope œœ Äm 14. ! œ lim # h hÄ! Thomas’ Calculus Chapter 2 Limits and Continuity 11th Edition, Chapter 2, Section 7: 5, 7, 13, 17, 21, 23, and 25 ch(4 b h) h œ c4, slope 2c (3 1c c " r 2 Limits and Continuity lim h) 8 c 2(2œ h)lim lim(2 b8h) 2œ4 blimb h b œh lim c 2h(4slopeœ c8 œ c2; b œ c 4 , b h) œ 2h(2 c a b4h h h(2 h) 4 NOTE: Use !the db h) h Ä ! h Ä ! bderivative 2h(2 bholve th(2 bproblems above: efinition of h) h(2h Ä ! to s h)Ä ! he h) hÄ! hÄ tinuity ß c c2(x c 2) b b 3, P 5 m œ h(4 P (h m at (2" 2): yœ 2 œ c(2 b h) b 11d c 5 œ 4. P" :a5m"4hœ h b c # : œ #lim c3 b h) œ %; # b # : b ")# œ c11. m2h lim h h c (c1) hlim! h 1) Ä! Ä h lim h b 1) x œœ hlÄ !c1 b 1)h m œ slope h 1 At 0, yim h(h Ê œ 2, lim œ œ lim (h ch(hb (h b ") œ Ä ! h(h2h 1) œ 2, slope h b 1) b h 5! hÄ! hÄ! at (2ß 5): y cÄ œ 4(x c 2), tangent line c (c"15. d c a4 c (lim b (2 b h) c 8 œ lim a8 b 12h b 6h b h b c 8 œ lim h a12 b 6h b h b œ 12; b h) m œ c1) h h h hÄ! hÄ! hÄ! h c(x b h) b 4(x b h) c 1d c ax b 4x c 1b c(x b h) b 4(x b h) c 1d c ax b 4x c 1b ( È c 2h bhorizontal tangent cœ 12(t c c œ20 h) d c 0 œ m h b b1t (2ß ) h h( 8the slope m 2(1 b Ê line 2È !At a2È1bah c 12.l):my c #lim2Èc1 " b h) 2), tangent(c1) œ œ hlim 1 b h c 2 c 4h c 2h b b 1 œ lim h(c3 c 2h) œ c3; h 1 c (2 b h) d c (c 3) 1im œim a1 c4 c4h h c 2b b 3 1 b h b 2 ch(4 b h) Ä ! lim a œ œ œim h h) œ b c h † hÈ 2; m l h h h Ä !œ c4, slope hÄ! 1 !h Ä Ä h bh b h h a2xh b h b 4hb ! 4h œ hax bh ÄhlÄ bhbb ! œ12x ax b 4xœ1hlim# ! 4) c hh c b 4;2 c b h b ! bÄ lim 42xh bb h 4x 4h b (2x h ! h œ $lb #(x c (c1)) "ß c"): y2x b œ c3(x c 1), tangent line œ im h Äc at ( cÊ h) hœ3(12b1 d5, 4 œ lim 3h b 3h b h b 3 bœ b c 4 (2x b h hb 4)3h b 2x b 4; lim œhb 4(1 b h) ! h h) 3h h Ä ! im h16.Èm œ limœ (1 b y b 1 bh b 1cœ œ lim h a1 b lim 1; Ä! Ä! lim a6 b œ 6; b‹ Š Ä ! (c2ß c5) is 1the!point ! Èh b graph where there is a Ä h Ê c2h4 œ1 b hÊ Ä œhc2.on the f(c2) œ 4 c 8 c 1 œ c5 Ê (c2ß c5) œ the point onhthe graph where hx hÄ! hÄ! 2x b 0 is 2 c (2 b h) Then h) 1 " h œatlim %c2h(2 ch) œ œlim h 1), tangent line , slope œ 6(t 3c 1, cb h) œ line c4 ):h y œ 2 b tangent. b 4 y h (3 b 22h(2 1(x 1) x ! c tangent 3 h) c 3(h h ("ß c2h horizontal Ä ! ):myœÊ 13. lim Ä h) h œ lim (3 b h(h b 1) b 1) œ lim h(h b 1) œ c2; hÄ! hÄ! hÄ! Answers: 5. 7. there is a h 3xh b h c 3x c 3hb c ax c 3xbb h c 2 1 È (h c 1) b 3(x " c(x b h) 4c(h bb)h)d c ax h 1 c (c1) ! h Ä !1 "1 (x c 1) (x ch lim ch(x 2hœchlim " cœh) Êc xc(2 bc5) is the c 1) c (x8b h12h b œ b b c 8 there h 12 b c h " 8 c! m hb 2) 1 œacc8bcb hœ1) œœc" ;2(c 1) ßœ3):c y!c 3lim1)(xaon c8),graphh where c 1)(x bis c a1) œ6h b (x b œ 12; 2 h) lim point b the 6h hlim! h(x h c 1) Äœ 4 c 1)(x15. cmlim lim at h ß œ Ä œ h(x c " (x c 1) tangent line lim h a œ œ1È! b 35 23.(x c (8 hh 2; œ6 Ä im h(c1 b h)h Ä 9 œ h h 1 h Ä ! b h 6c! b h) hÄ hÄ! Äœ 0 or x œ#2. If h Ä !0,( then y œ c1 and m œ c1 ! x œ x# c 2x œ 0 Ê x(x c 2) œ 0 Ê x œ 0 or x œ 2. If x œ 0, then y œ c1 and m œ c1 Ê (x c 1) œ 1 t Êß )): y c 8 œ 12(t ßnd):myœ c1bÊ yc (21 1))(x c 2)œ 2x (x c 2), tangent line " œ 1 2(x a œ c ( Êy b 3, Ê y œ c1 c (x c 0)c c(x b œ ,c xc 3).cthen y œ 1 and m œ1c1 b h b y 5 1 c (x c 2) 2 b c(x c 3). œ 1). If œ 2, Ê cœ œ 25. œ c1, b 3xh h 3x œ c ax 5( " b h) c 5 œ lim 5 a c 2hh œ lim 5h(ch h) œ c10, slope t 3xb c line19. At x ax b 3x hy œ 5bÊc m c 3hblim c 3xb h œ lim hÄ! hÄ! hÄ! h h Ä ! 16. m œ lim c(1 b h) b 3(1 b h)d c 4 œ lim a1 b 3h b 3h b h b 3 b 3hb c 4 œ lim h a6 b 3h b h b œ 6; x" Èx ! hÈx b h c Èx h Äb h b Èx h h (x b Èx b h c È! h) c x h Ä x (x b h) c x h Ä ! # im œ Ä ! h ŠÈ h aœ # lb 3xhÄ x # c È ‹ b 3; 3x# c 3 œ 0 †Ê xx œ c1 or xlim! h Then h b Èx‹ œ È b h b Èx œ h Ä 1. ŠÈx b x4 3x hm œ hlim! b h b3bxœ 3xœclim! h h at ("ß %): y c h Ä 6(t c 1), tangent line 4œ " " 2) andÊ x œ 4 Ê ypoints The tangent line is "a horizontalœ 2 Ê exists. Ê y œ 2. The tangent line is the œœ l("ß c2) are h œ 2. on the Thus, whereÈ Ê Èx tangent x œ 4 2 im œ #Èx . graph 4 œ # x È h Ä ! h Š x b h b È x‹ È È È œ œ #lim3; œ h b at3(%ß h0): Ê c 82lim2" aor cb 3xhThen h) 3b œ 3x# 8 c 3;43x4h ch3b œ 0 Ê xcœ c1h) x c8 1. Then 3x c 3x # ;c œ # c 3h œ h) œ 1 (x # œ 1. cb h# line 3x " 3xh b c c 2a b # b or œ y x œ c 4 3x x 4), tangent c 2(2 2h(4 4 œ lim (2 b 4(x b h) 1d c 8 4x cbb œ lim œ lim h(2 b b œ 4 œ c2; h Ä ! 14. m h 17. b h) hhÄ ! œ clim ax bh(2 b h) c(x 1 h(2 b h) h) hÄ! hÄ! hÄ! 1 b 0cÊ b0 dœc(" cœ where a!horizontal tangent exists. h on 1 œ c1) œtheand f(1)b 1dhœ lÊ (c"ß 2) and ("ß c2) are the points on the graph where a horizontal tangent exists. m" œ Ä oints 1) 2 cgraph) hlimc2 im h h f( Ä! h Ä 2œ at (2ß 2): hyb 1!c 3h c2(x c È9 b h c 3 È9 b h b 3 c 2) h b 4x c 1b a2xh È(8 b  ­ b( /4 c œ ) lim b h b1h)1c1 b h) 21. 4hcœ lim y(2x b h b 4) œ† 2x b 4; œ lim (9 b h) c 9 œ lim œœ 1 È9 b h b 3 œ 0;18.(h)m"l): y œ 1im 0(xh 1) Ê lim 1, h "ß m ( at œ (im œ l b c È È h ! h h Äh Ä ! at ($ß $): y c 3 œ c3x È42h 3), 2 aÈ4b 3xb 2 c line bh (4 b c 4 c c 3h h) cx 13. œ lim2(xœbhh c tangent h b b œ b h mœ œ 0 œ 17.œ limhlim! h(h bh1) h œ hlim! bh(h b 1) œ †2,xslopeh 2 3xhhlimhc 3xŠÈ4 bb hab #c 3xb hlim! lim È4 b h mh œ lim Ä ‹ Ä Ä! h Ä Ä hhÄ !! hÄ! hÄ! h h ŠÈ4 b h b #‹ œ " È4 b # h h h(c1bh) Ä ! Ä ! h Ä ! h Š 9 b h b 3‹ h Ä ! h Š 9 b h b 3‹ 4bhc2 4bhc2 4h2 (4 b h) c 4 h œ lim † È4 b h b 2 œ lim œ lim œ " 17. m œ lim h h bb y œ 2 b c (x b c 4) œ x b 1. h hÄ! h Ä ! h ŠÈ4 b h b #‹ h Ä ! h ŠÈ4 b h b #‹ (x c 1) 4 (x h c 1) ch " 4 Ä! m h(x c 1)(x b h c 1) œ lim h(x c 1)(x b h c 1) œ c (x c 1) ! " œ h"Äa! (%ß #): y c 2 œ 4 (x c 4), tangent line 4; t È 1 b h c 2È 1 Ê xc4.9cb2)4h bim Ê2È1œ h0100 †x2È12. h Ifb2x aœ 0, 4.9(2) b œ c1 and 4.9 aœ c1 h b b 4.9(4) (xf(2a4 b c f(2) b b 4.9(4) b a or c 4.9(#b h) c 100 c then y x œb m b h) œl 0 œ h œ limh c 2 2È1 b h b # imim lh œ lim c 4 b 4h h bh hÄ! Ä f!x!œ 2, then y œ 1hand m œ c1 Ê y œ 1 c (x c 2) œ c(x c 3). hÄ Ä! hÄ! I b h) c 4 h 4(1 (9 b h) c 9 œ lim 4.9h)2 È 19.6. c 3 œ lim È9 b h c 3 † È h b 3 œ falling downward at alim of h œ at tes the object19.6m œ downward1; bThe minus sign indicates È9 bobject is lim œ lim (c ishfalling 1 b œ c(8 b h) a1speed of c the œ speed h ŠÈ1 b h b 1‹ 18. Ä ! Èlimh b 1 h h Èx b h c Èx 19.6 m/sec. † Èx b h y Èx xœ 1, tangent b h) c x 2 b 1(x c 1)ÈÊ b œ "b lim " ; at (x line y c 3 œ œ œ (8ß 3): h x b h b ÈÈ9 b 3 Ä ! h ŠÈx b h b Èx‹ x 6 h " È4 b # hÄ! hÄ! hÄ! 9bhb3 h Ä ! h ŠÈ9 b h b 3‹ h Ä ! h ŠÈ9 b h b 3‹ " " hb hus, 460 ft/sec.f(10) Èx lim2 Ê b h) c 3(10) œ œ # h) lim f(10 bÈhxc Ê œ œ 3(10 x œ 4 Ê h hÄ! hÄ! " 6 (x c 8), tangent line 5h(c2 b h) h y œlim The20h b h b œ 60 is œ 2. 3 a tangent line ft/sec. h hÄ 19. At x œ c1, y œ 5 Ê m œ lim ! 5(c" bhh) c 5 œ lim 5 a1 c 2hhb h b c 5 œ lim hÄ! hÄ! hÄ! hÄ! c4.9 a4 b 4h b h b b 4.9(4) h 1 c9 b 6h bh c 9d h hÄ! œ c10, slope h sin ˆ h ‰ œ 1 b ˆ c (c1)) Ê y œ 2x f(0)3, 2(x cb Slope at " ‰ œ 0œÊlim f(0 b h)does havelim origin œ a tangent at œ lim h sin ˆ " ‰ œ 0 Ê yes, f(x) does have a tangent at h sin h yes, f(x) h h h 3 ah Äb h b 20h ! c 3(10) hÄ! ! œ lim œ 60 ft/sec. h Ä ! h h Ä ! slopeh0. the origin with hÄ! 1 h) ) œ lim 1 c (c1 b h) b c 4 (2) 4 4 c12h b a b h d of h 6h heh minush) c f(2) ! Tim d f(2 b lim Ä41œ12 c1bthe (2 h#h) œis falling downward at 6h speedœ lim 41 sign c h( b h) object 3 l b œ h h indicates6h3b d h 161 œ lim 3 lim h 3 hc2h ! h b h Ä ! hÄ hÄ! hÄ! 3 a Äb 2ch ! œ lim (c1 b h) œ 2; h(c1 b h) hÄ! h hÄ! hÄ! b h) b c a100 c 4.9(2) b c 9d limœ f(3imh) c (6 b h)lim611(3 b h)h c 1(3) œ lim l b 1 f(3) œ œ h c( œ lim 1(6 b h) œ 61 hÄ! œ lim c12 b 6h b h# d œ 161 1 c9 b 6h bh c 9d h sin ˆ ‰ 1(3) g(0 b 1 h h imœ " lim h) c g(0) œ lim œ hlim œ (6 b h) œ "61 mlÄ ! h Ä !h not exist,Ä ! has Ä !tangent! at h . Since hlim! sin sin h does f(x) h no hlim sin h h Ä Ä ! " h does not exist, f(x) has no tangent at the origin. 4 3 (2) œ lim 4 3 hÄ! c12h b 6h b h d h œ lim hÄ! 41 3 c12 b 6h b h# d œ 161 œ lim hÄ! h sin ˆ h ‰ h œ lim h sin ˆ " ‰ œ 0 Ê yes, f(x) does have a tangent at h fpm@itesm.mx hÄ! ...
View Full Document

Ask a homework question - tutors are online