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File0020

# File0020 - 86 20 21 22 23 24 25 26 27 28 29 30 31 32...

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Unformatted text preview: 86 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. Chapter 2 Limits and Continuity -h(4 + h) , 2 _ ._ _ _ 2 . Atx—2 y— —3 *> m— hlim 1‘ (“EH “—3) — 11111 W : 11m 11 :,—4 slope h—10 h—>0 11—40 1 l _ _ 1 * - (3+11171'2 2— (2+h) _h___— _.__1_ A” 3,y 2 A m 1111310 11 _1111.I4no 211(2+11> =h11£n0 2h(2+h)_ 45109e . b—H) . (11-1)+(11+1) - 211 _ _ _ _ h+1 : : : At x — 0, y — 1 :> m — hll_r+n0 ——h hlgn0 h(h+1) hlgn0 h(h+1) 2, slope 2 _ _ 2 _ At a horizontal tangent the slope m: 0 :> 0— — m— — hlim0 (Hm “(High 1 0‘ +4x 1) w» : hlimo WW : hlimo w : hlimo (2x + h + 4) : 2x + 4; 2x + 4 = O :> x z —2. Then f(—2) = 4 — 8 — 1 = —5 => (—2, —-5) is the point on the graph where there is a horizontal tangent. 0: m: hlimo [(x+h)3—3(x+11)]— (x3—3x) _ lim 1x3+3x2h+3xh2+h373x—3h2—(x3—3x) 11—10 h 2hlimoh221%ﬂ‘h—3h:hlim0(3x2 +3xh+h2—3)=3x2 3;3x2 3:O—>x—— lorx—l. Then f(— 1) : 2 and f(l) : —2 :> (—1, 2) and (17 —2) are the points on the graph where a horizontal tangent exists. l l _ _ . —x+ , —;,— __ (x—I)—( h—l) —___. — ”1 -m—h1£,no ( h) 11 l —h1i_‘,n0 h(x—l)(xx4:h»l) =hIEPO h(x-—l)(xh+h—l) ” _(x—11)7 =>(x—1)2=1=> x2~2x=0 => x(x—2)=O :> x:00rx=2. Ifx:0,theny:-1andm=——l => y=—l——(x—0)=—(x+1). lfx=2,theny=1andm‘—1 -> y—1»(x 2): (x 3). l— _ . 1/x+h ___\Z_§—_ 1/x+h—\/ZI 1/x+h+\/§ . (x+h)7x 4 _ m—hlino 11111310 h 1/x+h+1/i _ 1111—190 h11/x4h4ﬁ) 1 ~hInn0 W: m. Thus, 4 57; : ﬂ 2 => x 4 :> y 2. The tangent l1ne IS _ 2 , _ 2 . _ 2 lim f(2+h)—f(2) _ lim £100 49424.11); (100 49(2)) _ 11m 4.9(4+411+11)+4.9(4) 11 —» 0 h 11 —» o 11 a 0 h 2 hlim0 (~ 19.6 — 4.9h) = - 19.6. The minus sign indicates the object is falling downward at a speed of 19.6 m/sec. . 2 2 2 11m 1004.11}: 1(101_ ~ lim 3(10411)h 3(10) _ lim 3901:4112 :60 ft/sec. h —> 0 11 ‘4 0 —h —> 0 . 2__ 2 . ‘2_ . 11m «3+1? f(3)_ _ lim 7r(3+11)h 11(3) 2 11m «9+6hh+11 9 : 11m 7r(6+h):67r h —1 0 h —> 0 h —+ O h —> 0 . _ . ‘1 113—4—"23 . 411211 2 113 . 11m ——f<2+hg “2) = 11m 2—M )1 “ 2 11m 1____1 f" t 1: 11m 4?” [12+611+112] = 167r h ——+ O h —1 0 h —1 0 h —> 0 lim <m<x°+h>+b ‘ "“0“” = lim WW 2 lim M = lim mzm h—10 h h—>0 h h—>0T h—>0 1. E41+1‘2_ . It 44:1.h __ 1 2- 4+11 _ . 12*v4th112+v4+h1_. 4—(4+11 1m h — 11m 11 — 11m — 11m 11m 1140 11—10 11—10 2h 4+h 11-10 2111/4411(2+1/4+11) 11—10 2111/4411(2+1/4+11 —h —1 __ —1 —l =lim =lim————= ——>m=— h—10 2111/4+1112+1/4+11) h—>0 21/4+11 11(2+1/4+11 ) 294+0(2+Q4+0)_16 16 ...
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