HW 5 - Tran Tony Homework 5 Due 3:00 am Inst Samuels This...

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Tran, Tony – Homework 5 – Due: Sep 25 2007, 3:00 am – Inst: Samuels1Thisprint-outshouldhave24questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.The due time is Centraltime.001(part 1 of 1) 10 pointsFind the slope of the secant line passingthrough the points(-2, f(-2)),(-2 +h, f(-2 +h))whenf(x) = 2x2+x+ 3.1.slope = 4h-92.slope = 4h+ 73.slope = 4h+ 94.slope = 2h+ 75.slope = 2h-7correct6.slope = 2h-9Explanation:Since the secant line passes through thepoints(-2, f(-2)),(-2 +h, f(-2 +h)),its slope is given byf(-2 +h)-f(-2)h={2(-2 +h)2+ (-2 +h) + 3} -9h=2h2-7hh=2h-7.keywords: slope, secant line002(part 1 of 1) 10 pointsIfP(a, f(a)) is the point on the graph off(x) =x2+ 3x+ 7at which the tangent line is parallel to the liney= 6x+ 6,determinea.1.a= 02.a=-123.a=124.a=32correct5.a= 1Explanation:The slope of the tangent line at the pointP(a, f(a)) on the graph offis the valuef0(a) =limh0f(a+h)-f(a)hof the derivative offatx=a. To computethe value off0(a), note thatf(a+h) = (a+h)2+ 3(a+h) + 7=a2+h(2a+ 3) +h2+ 3a+ 7,whilef(a) =a2+ 3a+ 7.Thusf(a+h)-f(a) =h{(2a+ 3) +h},in which casef0(a) =limh0{(2a+ 3) +h}= 2a+ 3.If the tangent line atPis parallel to the liney= 6x+ 6,
Tran, Tony – Homework 5 – Due: Sep 25 2007, 3:00 am – Inst: Samuels2then they have the same slopes, sof0(a) = 2a+ 3 = 6.Consequently,a=32.keywords: tangent line, parallel, slope, deriva-tive003(part 1 of 1) 10 pointsFind thex-intercept of the tangent line atthe pointP(2, f(2)) on the graph offwhenfis defined byf(x) =x2+ 2x+ 1.1.x-intercept = 22.x-intercept =-33.x-intercept =12correct4.x-intercept =-25.x-intercept =-126.x-intercept = 3Explanation:The slope,m, of the tangent line at thepointP(2, f(2)) on the graph offis thevalue of the derivativef0(x) = 2x+ 2atx= 2,i.e.,m= 6.On the other hand,f(2) = 9.Thus by the point-slope formulaan equation for the tangent line atP(2, f(2))isy-9 = 6(x-2),i.e., y= 6x-3.Consequently,x-intercept =12.keywords: tangent line, x-intercept, slope004(part 1 of 1) 10 pointsFind an equation for the tangent line to thegraph ofgat the pointP(1, g(1)) wheng(x) = 5-4x3.1.y+ 13x+ 12 = 02.y= 13x-123.y+ 12x+ 13 = 04.y= 12x+ 135.y+ 12x= 13correctExplanation:Ifx= 1, theng(1) = 1. Thus the Newto-nian different quotient forg(x) = 5-4x3at the point (1,1) becomesg(1 +h)-g(1)h=h5-4 (1 +h)3i-1h=5-4h3-12h2-12h-5h.Thusg0(1) =limh0(-4h2-12h-12)=-12.Consequently, by the point-slope formula, anequation for the tangent line to the graph ofgatPisy-1 =-12(x-1)
Tran, Tony – Homework 5 – Due: Sep 25 2007, 3:00 am – Inst: Samuels3which after simplification becomesy+ 12x= 13.keywords: tangent line, slope, equation005(part 1 of 3) 10 pointsA Calculus student leaves the RLM build-ing and walks in a straight line to the PCL Li-brary. His distance (in multiples of 40 yards)from RLM aftertminutes is given by thegraph-101234567891011246810246810tdistancei) What is his speed after 3 minutes, and inwhat direction is he heading at that time?

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