Test bank and solution of derivetive (2)

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Test bank and solution of derivetive (2)

The DerivativeExercise Set 2.11.(a) ni,.,n = (50 - 10)/(15 - 5) = 10/10 = 1 m/s.3 5£ 4I ĩ(b)0 5 10 15 20Time (s)2.At I. = 4 8, mlan « (90 - 0)/(10 - 2

Test bank and solution of derivetive (2)2) = 90/8 = 11.25 m/s. At t = 8 8, wlnn = (140 - 0)/{ 10 - 4) = 140/6 « 23.33 m/s.3.(a) (10 - 10)/(3 - 0) = 0 ctn/8.(b)I = 0. I = 2. t = 4.2. and ! =

8 (horizontal tangent line).(c)maximum: t = 1 (slope > 0), minimum: t = 3 (slo|M? < 0).(d)(3 - 18)/(4 - 2) = -7.5 cm/s (slope of estimated tangent lin Test bank and solution of derivetive (2)

e to curve at t = 3).1. (a) decreasing (slope of tangent line decreases wit h increasing time)(b)increasing (slope of tangent line increases with incr

Test bank and solution of derivetive (2)

easing time)(c)increasing (slope of tangent line increases with increasing time)(d)decreasing (slope of tangent line decreases with increasing time)5.

The DerivativeExercise Set 2.11.(a) ni,.,n = (50 - 10)/(15 - 5) = 10/10 = 1 m/s.3 5£ 4I ĩ(b)0 5 10 15 20Time (s)2.At I. = 4 8, mlan « (90 - 0)/(10 - 2

Test bank and solution of derivetive (2)at time. From t ime to time ti, the velocity, and the slope, decrease. At time Cl, the velocity, anil hence the slope, instantaneously drop to zero, s

o there is a sharp bend in the curve at that point.60Chapter 2(b)mllM> = Um= litn ~J‘= lim 2xi = 0*1"»*!-()*1-4I)X1-O*“-40(c)mtan = Jini—7^-—^ = lim= Test bank and solution of derivetive (2)

Um (2x1 + 2x(i) = 4xo*1-4X0 Xl — X|)*1-4*0 Xl — Xd *1-»Xo12. (a) ,„„ = fíM = ỉù2! = 7"» m“° = A ii^rn = Ã. M ^-1>W4X.+1) = (XJ + Z1->1a!| — I «|-»1 X1

Test bank and solution of derivetive (2)

— l Z1-»1X| — 1Z1->1Exercise Set 2.161(c) nr,aa = lim= lim -y—yi = lim (xf + xxx0 + J-„) = 3xj$X|-»Xo Xj - Xy«|-MT* X, — Zy X|->z<,(d)ia) m /(*)-/(2)

The DerivativeExercise Set 2.11.(a) ni,.,n = (50 - 10)/(15 - 5) = 10/10 = 1 m/s.3 5£ 4I ĩ(b)0 5 10 15 20Time (s)2.At I. = 4 8, mlan « (90 - 0)/(10 - 2