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chapter 2 57

# chapter 2 57 - SECTION 2.8 DERIVATIVES 1Z1 9(a Using...

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Unformatted text preview: SECTION 2.8 DERIVATIVES 1Z1 9. (a) Using Deﬁnition 2 with F(:c) = x3 — 593 + 1 and the point (17 —3). we have F(1+h)—F(1) [(1+h)3—5(1+h)+1]—(~3) F (1) _ 1,13%) h 113% h . (1+3h+3h2+h3—5~5h+1)+3 . h3+3h2—2h : 11m _ 11m —— h—>0 h hﬁo h 2 l _ : lim W = lim(h2 +3h — 2): —2 h—vO h h—>O So an equation of the tangent line at (1. ~3) is y — (—3) = —2(:c — 1) 4:) y = —2:1: — 1. (b) 8 a+h a — 1 2 h ’1 2 1o. (a)G/(a):iﬂw:iﬂ + (a+h) + a . a+2a2+h+2ah—a—2a2~2ah , 1 ,2 —l l — 1 2 #3) h(1+2a+2h)(1+2a) hill(1+2a+2h)(1+2a) ( + ‘1) So the slope of the tangent at the point (~ﬁ, —%) is (b) m 2 [1 + 2(—%)] _2 : 4, and thus an equation is y+%:4(m+i) ory:4w+%. , . f(1 + h) — f(1) . 31th — 31 (a) f ( ) hm) h Iii—+1110 h (b) 3.4 31+h — 3 So let F(h) 2 T. We calculate: [ h F(h) F(h) 0 1 3.484 3.121 0.9 0.01 3.314 3.278 2- 0.001 3.298 3.294 From the graph. we estimate that the 00001 3.296 3.296 slope of the tangent is about 3.2—2.8 ‘ 0.4 ~33 We estimate that f’(1) x 3.296. 1.06 ~ 0.94 _ 0.12 N ' ‘ ...
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