高考數(shù)學(xué)總復(fù)習(xí)——導(dǎo)數(shù)的應(yīng)用（理科）-預(yù)覽頁(yè)
【正文】 0000200030002yxyxxyxxyxPxxf或有則設(shè)[解析 ] [課前導(dǎo)引 ] 1. 曲線 f(x)=x3+x?2在點(diǎn) P處的切線平行于直線 y=4x?1,則點(diǎn) P的坐標(biāo)為 ( ) A. (1,0) B. (2,8) C. (1,0)或 (?1,?4) D. (2,8)或 (?1,4) ?????????????????????????41014132),(,13)(39。 ( 1 ) 2.1nnxxCCnn ???.ln1)39。ln)39。s i n)39。,.,11,11,02)1(2,0]2)1(2[,0)(39。0)2()(,0.),(,),()(,0,1,0.,)(2212121????????????xeaxxxfxxxxxfxxaxxxxxf時(shí)而當(dāng)函數(shù)上為增在上為減函數(shù)在時(shí)當(dāng)取到極小值處在處取到極大值在即.)(,11,0)(,02 取得最小值時(shí)所以當(dāng)時(shí)當(dāng)xfaaxxfx??????.)(,11,0)(,02 取得最小值時(shí)所以當(dāng)時(shí)當(dāng)xfaaxxfx??????上為單調(diào)函數(shù)的在綜上解得即單調(diào)函數(shù)的充要條件是上為在時(shí)當(dāng)]1,1[)(,.43:,111,1]1,1[)(,0 )2( 22?????????xfaaaxxfa).,43[.43???的取值范圍是即充要條件是aa).,43[.43???的取值范圍是即充要條件是aa [點(diǎn)評(píng) ] 本題主要考查導(dǎo)數(shù)的概念和計(jì)算，應(yīng)用導(dǎo)數(shù)研究函數(shù)性質(zhì)的方法及推理和運(yùn)算能力 . [在線探究 ] [在線探究 ] .,)1,1()( )2( )1,1()(,41 )1( )(3232)( .1 23的取值范圍求個(gè)極值點(diǎn)內(nèi)有且只有一在若減函數(shù)；內(nèi)是在求證時(shí)當(dāng)已知axfyxfaRaxaxxxf????????.)1,1()(,0)(39。,3232)()1( 223內(nèi)是減函數(shù)在故內(nèi)在的圖象開(kāi)口向上，二次函數(shù)又????????????????????????????xfxfxfafafaaxxxfxaxxxf???[法一 ] .)1,1()(41.0)(39。)1,(,0)(39。,)( 0 hhafhafAafaxxfh??????求極限且處可導(dǎo)在設(shè)函數(shù)[例 1] [方法論壇 ] 1. 應(yīng)用導(dǎo)數(shù)定義的等價(jià)形式解題： .2)()3(lim,)(39。23)()(lim21 3)()3(lim232)()(lim2)()3(lim0000Aafafafhafhafhafhafhafhafhafhafhhhh?????????????????????).(39。,m,:,4)(222233???????????????????????????xxxxyxxyxxxyyxxxxxx即有令則有設(shè)容器的容積為得和由.,.,..1,),0(,)()0(,.039。t an),(.33)(39。t an),(.33)(39。,0,),0[)(,0)(39。.44)2()2(2)4()
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