Lay Analysis 5e Solutions - Chapter 6

Lay Analysis 5e Solutions - Chapter 6

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时间:2019-09-21

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1、Section6.1ñTheDerivative64ThisworkisprotectedbyUnitedStatescopyrightlawsandisprovidedsolelyfortheuseofinstructorsinteachingtheircoursesandassessingstudentlearning.Disseminationorsaleofanypartofthiswork(includingontheWorldWideWeb)willdestroytheintegrityoftheworkandisnotpermitted.Thewor

2、kandmaterialsfromitshouldneverbemadeavailabletostudentsexceptbyinstructorsusingtheaccompanyingtextintheirclasses.Allrecipientsofthisworkareexpectedtoabidebytheserestrictionsandtohonortheintendedpedagogicalpurposesandtheneedsofotherinstructorswhorelyonthesematerials.AnalysiswithanIntro

3、ductiontoProof5thEditionbyStevenR.Layslay@leeuniversity.eduChapter6–DifferentiationSolutionstoExercisesThefollowingnotationsareusedthroughoutthisdocument:8=thesetofnaturalnumbers{1,2,3,4,…};=thesetofrationalnumbers<=thesetofrealnumbers"=“forevery”$=“thereexists”§=“suchthat”Section6.1

4、–TheDerivative1.(a)False:Thelimitmustbefinite.(b)False:Practice6.1.5,orconsiderf(x)=

5、x

6、atx=0.(c)True:Theorem6.1.6.Copyright©2014PearsonEducation,Inc.Section6.1ñTheDerivative652.(a)True:Theorem6.1.7(a).(b)True:Theorem6.1.7(b).(c)False:gmustbedifferentiableatf(c).2fxf()--(1)x13.(a)Yes.l

7、im==limlimx+1=2andxxç+11xx--11ç+xç+1fxf()--(1)(21)1x-2(1)x-fxf()-(1)lim==limlim=2,sofÅ(1)==lim2.xxxç-111xxx---111ç-ç-xç1x-1(b)No.limfx()=1andlimfx()=2,sofisnotcontinuousatx=1,andthereforenotdifferentiableatxç+1xç-1x=1.Also,notethattheleft-handlimitforthederivativeisequalto–Ñwhiletheri

8、ght-handlimitis3.2fxf()--(1)x1fxf()--(1)(32)1x-(c)No.lim==limlimx+1=2andlim==limxxç+11xx--11ç+xç+1xxç-11xx--11ç-3(x-1)f()xf-(1)lim=3.Sincetheseone-sidedlimitsarenotequal,limdoesnotexist.xç-1x-1xç1x-14.Thisisroutine.1/31/32/31/31/32/35.(a)Hintinbook:Notethatxcx-=()-cx(+cx+c).6.(a)fÅ(x)

9、=2xsin(1/x)-cos(1/x),forxò0.2xxsin(1/)0-(b)limxxçç00==limxxsin(1/)0,sofÅ(0)=0.x-0(c)fÅisnotcontinuousatx=0sincelimxç0fÅ(x)=limxç0[2xsin(1/x)-cos(1/x)]doesnotexist.22gxg()--(0)x0gxg()--(0)xsin(1/)0x(d)Wehavelim==limlimx=0andlim==limxxç-00xx--00ç-xç-0xxç+00xx--00ç+limxxsin(1/)=0,sogÅ(0)

10、exist

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