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Chapter 3 5 - SECTION 3.1 DEHlVATIVES OF POLYNOMIALS AND...

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Unformatted text preview: SECTION 3.1 DEHlVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS 155 44. (a) 8 (b) From the graph in part (a). it appears that f’ is zero at 9151 R8 0.2 and mg m 2.8. The slopes are positive (so f’ is positive) on (~00. m1) and ($2, 00). The slopes are negative (so f’ is negative) on (2111. 232). (C) 9(53) 2 ex — 3$2 => g'(a:) 2 ea” 2 63c 8 45. The curve y 2 2m3 + 3x2 — 12% + 1 has a horizontal tangent when y’ 2 6m2 —l— 6:1: ~ 12 2 O 42) 6(ac2 + a: 2 2) 2 O 2) 6(a: + 2)(x — 1) 2 0 <=> a: 2 —2 orm 2 1. The points on the curve are (‘2, 21) and (1. —6). 46. flat) 2 $3 + 33:2 + a: + 3 has a horizontal tangent when f’(:v) 2 3m2 + 63: + 1 2 0 <2 : T6iv636h122—1iéx/O. 1' 47.y—6m3l5m 3 2 m—y'218w2+5.but9:220f0rallm.som25forall$. 48. Theslopeofy: 14—261 —3wisgivenbym2y’ 2261 —3. y=1+2e“23x Theslopeof3ar~y25 <:> y23m—5is3. m—3 26x 3—3 (Bx—3 2> $2ln3.This occurs at the point (11137 7 — 31n3) % (1.1.3.7). Let (a. (12) be a point on the parabola at which the tangent line passes through the point (0. —4). The tangent line has slope 2a and equation y 2 (—4) 2 2a(a: — 0) <2) y 2 2am 2 4. Since (IL/(12) also lies on the line. a2 2 2a(a) and (—2, 4). — 4, or a2 2 4. So a 2 i2 and the points are (2, 4) ...
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