函数解析式求法和值域求法总结及练习题

relationship,establishedequivalentrelationship14,andsubject:applicationproblem(4)--scoresandpercentageapplicationproblemreviewcontentoverviewanswersscores,andpercentageapplicationproblemofkeyis:accordingtomeaning,(1)determinestandardvolume(units1)(2)findassociatevolumeratecorrespondstorelationship,Thenin-linesolution.CategoryfractionmultiplicationwordproblemscoreDivisionapplicationsengineeringproblemproblemXV,asubject:reviewofthemeasurementoftheamountofcapacity,measurementandunitsofmeasurementofcommonunitsofmeasurementandtheirsignificanceinrate1,currency,length,area,volume,unitsize,volume,weightandrate.(Omitted)2,commonlyusedtimeunitsandtheirrelationships.(Slightly)withameasurementunitsZhijianofofpoly1,andofmethod2,andpolymethod3,andofmethodandpolymethodofrelationshipmeasurementdistanceofmethod1,andtoolmeasurement2,andestimates16,andsubject:geometrypreliminaryknowledge(1)--lineandanglereviewcontentline,andsegment,andRay,andvertical,andparallel,andangleangleofclassification(slightly)17,andsubject:geometrypreliminaryknowledge(2)--planegraphicsreviewcontenttriangle,andedgesshaped,andround,andfanaxisymmetricgraphicsperimeterandareacombinationgraphicsofareasubject:Preliminaryknowledge(3)-reviewofsolidcontentcategory1-dshapesaredividedinto:cylinderandcone2,columnisdividedinto:cuboid,square3,coneconeofthefeaturesofcuboidsandcubesrelationshipbetweencharacteristicsofcircularconeisslightlysolidsurfaceareaandvolume1,size2,table...和和平与发展是当今世界发展的主题，中国作为屹立在世界东方的大国，要担负起重要的责任。从“亚太自由贸易区”到“亚投行”、“一带一路”，再到G20峰会，都体现出中国一个负责任的大国形象。2[()]()()ffxafxbaaxbbaxabb函数解析式的七种求法一、待定系数法：在已知函数解析式的构造时，可用待定系数法．例1设)(xf是一次函数，且34)]([xxff，求)(xf．解：设baxxf)()0(a，则342baba，3212baba　或　　．32)(12)(xxfxxf　　或　　．二、配凑法：已知复合函数[()]fgx的表达式，求()fx的解析式，[()]fgx的表达式容易配成()gx的运算形式时，常用配凑法．但要注意所求函数()fx的定义域不是原复合函数的定义域，而是()gx的值域．例2已知221)1(xxxxf)0(x，求()fx的解析式．解：2)1()1(2xxxxf，21xx，2)(2xxf)2(x．三、换元法：已知复合函数[()]fgx的表达式时，还可以用换元法求()fx的解析式．与配凑法一样，要注意所换元的定义域的变化．例3已知xxxf2)1(，求)1(xf．解：令1xt，则1t，2)1(tx．xxxf2)1(，,1)1(2)1()(22ttttf1)(2xxf)1(x，xxxxf21)1()1(22)0(x．四、代入法：求已知函数关于某点或者某条直线的对称函数时，一般用代入法．例4已知：函数)(2xgyxxy与的图象关于点)3,2(对称，求)(xg的解析式．解：设),(yxM为)(xgy上任一点，且),(yxM为),(yxM关于点)3,2(的对称点．则3222yyxx，解得：yyxx64，点),(yxM在)(xgy上，xxy2．relationship,establishedequivalentrelationship14,andsubject:applicationproblem(4)--scoresandpercentageapplicationproblemreviewcontentoverviewanswersscores,andpercentageapplicationproblemofkeyis:accordingtomeaning,(1)determinestandardvolume(units1)(2)findassociatevolumeratecorrespondstorelationship,Thenin-linesolution.CategoryfractionmultiplicationwordproblemscoreDivisionapplicationsengineeringproblemproblemXV,asubject:reviewofthemeasurementoftheamountofcapacity,measurementandunitsofmeasurementofcommonunitsofmeasurementandtheirsignificanceinrate1,currency,length,area,volume,unitsize,volume,weightandrate.(Omitted)2,commonlyusedtimeunitsandtheirrelationships.(Slightly)withameasurementunitsZhijianofofpoly1,andofmethod2,andpolymethod3,andofmethodandpolymethodofrelationshipmeasurementdistanceofmethod1,andtoolmeasurement2,andestimates16,andsubject:geometrypreliminaryknowledge(1)--lineandanglereviewcontentline,andsegment,andRay,andvertical,andparallel,andangleangleofclassification(slightly)17,andsubject:geometrypreliminaryknowledge(2)--planegraphicsreviewcontenttriangle,andedgesshaped,andround,andfanaxisymmetricgraphicsperimeterandareacombinationgraphicsofareasubject:Preliminaryknowledge(3)-reviewofsolidcontentcategory1-dshapesaredividedinto:cylinderandcone2,columnisdividedinto:cuboid,square3,coneconeofthefeaturesofcuboidsandcubesrelationshipbetweencharacteristicsofcircularconeisslightlysolidsurfaceareaandvolume1,size2,table...和和平与发展是当今世界发展的主题，中国作为屹立在世界东方的大国，要担负起重要的责任。从“亚太自由贸易区”到“亚投行”、“一带一路”，再到G20峰会，都体现出中国一个负责任的大国形象。2把yyxx64代入得：)4()4(62xxy．整理得672xxy，67)(2xxxg．五、构造方程组法：若已知的函数关系较为抽象简约，则可以对变量进行置换，设法构造方程组，通过解方程组求得函数解析式．例5设,)1(2)()(xxfxfxf满足求)(xf．解xxfxf)1(2)(①显然,0x将x换成x1，得：xxfxf1)(2)1(②解①②联立的方程组，得：xxxf323)(．六、赋值法：当题中所给变量较多，且含有“任意”等条件时，往往可以对具有“任意性”的变量进行赋值，使问题具体化、简单化，从而求得解析式．例7已知：1)0(f，对于任意实数x、y，等式)12()()(yxyxfyxf恒成立，求)(xf．解对于任意实数x、y，等式)12()()(yxyxfyxf恒成立，不妨令0x，则有1)1(1)1()0()(2yyyyyyfyf．再令xy得函数解析式为：1)(2xxxf．七、递推法：若题中所给条件含有某种递进关系，则可以递推得出系列关系式，然后通过迭加、迭乘或者迭代等运算求得函数解析式．例8设)(xf是定义在N上的函数，满足1)1(f，对任意的自然数ba,都有abbafbfaf)()()(，求)(xf．解Nbaabbafbfaf,)()()(，，不妨令1,bxa，得：xxffxf)1()1()(，又1)()1(,1)1(xxfxff故①令①式中的x＝1，2，…，n－1得：(2)(1)2(3)(2)3()(1)fffffnfnn，，，将上述各式相加得：nfnf32)1()(，2)1(321)(nnnnf，Nxxxxf,2121)(2．relationship,establishedequivalentrelationship14,andsubject:applicationproblem(4)--scoresandpercentageapplicationproblemreviewcontentoverviewanswersscores,andpercentageapplicationproblemofkeyis:accordingtomeaning,(1)determinestandardvolume(units1)(2)findassociatevolumeratecorrespondstorelationship,Thenin-linesolution.CategoryfractionmultiplicationwordproblemscoreDivis