函數(shù)與導(dǎo)數(shù)解題方法知識(shí)點(diǎn)技巧總結(jié)-

....函數(shù)與導(dǎo)數(shù)解題方法知識(shí)點(diǎn)技巧總結(jié)高考試題中，關(guān)于函數(shù)與導(dǎo)數(shù)的解答題(從宏觀上)有以下題型：〔1〕求曲線yf(x)在某點(diǎn)出的切線的方程2〕求函數(shù)的剖析式3〕談?wù)摵瘮?shù)的單調(diào)性，求單調(diào)區(qū)間4〕求函數(shù)的極值點(diǎn)和極值5〕求函數(shù)的最值或值域6〕求參數(shù)的取值圍7〕證明不等式8〕函數(shù)應(yīng)用問(wèn)題2.在解題中常用的有關(guān)結(jié)論〔需要熟記〕：〔1〕曲線yf(x)在xx0處的切線的斜率等于f(x0)，且切線方程為yf(x0)(xx0)f(x0)?！?〕假設(shè)可導(dǎo)函數(shù)yf(x)在xx處獲取極值，那么f(x)0。反之不成立。00〔3〕關(guān)于可導(dǎo)函數(shù)f(x)，不等式f(x)0(0)的解是函數(shù)f(x)的遞加〔減〕區(qū)間?！?〕函數(shù)f(x)在區(qū)間I上遞加〔減〕的充要條件是：xI,f(x)0(0)恒成立〔f(x)不恒為0〕.〔5〕假設(shè)函數(shù)f(x)在區(qū)間I上有極值，那么方程f(x)0在區(qū)間I上有實(shí)根且非二重根?！布僭O(shè)f(x)為二次函數(shù)且IR，那么有0〕?！?〕假設(shè)函數(shù)f(x)在區(qū)間I上不只一且不為常量函數(shù),那么f(x)在I上有極值?！?〕假設(shè)xI,f(x)0恒成立，那么f(x)min0；假設(shè)xI,f(x)0恒成立，那么f(x)max0〔8〕假設(shè)x0I使得f(x0)0，那么f(x)max0；假設(shè)x0I使得f(x0)0，那么f(x)min0.〔9〕設(shè)f(x)與g(x)的定義域的交集為I，假設(shè)xI,f(x)g(x)恒成立，那么有[f(x)g(x)]min0.〔10〕假設(shè)對(duì)x1I1,x2I2,f(x1)g(x2)恒成立，那么f(x)ming(x)max.假設(shè)對(duì)x1I1,x2I2，使得f(x1)g(x2)，那么f(x)ming(x)min.假設(shè)對(duì)x1I1,x2I2，使得f(x1)g(x2)，那么f(x)maxg(x)max.〔11〕f(x)在區(qū)間I1上的值域?yàn)锳,g(x)在區(qū)間I2上值域?yàn)锽，假設(shè)對(duì)x1I1,x2I2使得f(x1)g(x2)成立，那么AB?！?2〕假設(shè)三次函數(shù)f(x)有三個(gè)零點(diǎn)，那么方程f(x)0有兩個(gè)不等實(shí)根x1,x2且f(x1)f(x2)0〔13〕證題中常用的不等式:①lnxx1(x0)〔僅當(dāng)x1時(shí)取“〞〕1/16(x0,y0)，求出切線方程....②ln(x1)x(x1)〔僅當(dāng)x0時(shí)取“=〞〕③ln(1x2)x(x0)④lnxx1(x1)x12⑤lnx11(x0)x222x2ex1xex1x函數(shù)與導(dǎo)數(shù)解答題常有題型的解法1〕曲線yf(x)〔含參數(shù)〕的切線方程為【解法】先設(shè)切點(diǎn)坐標(biāo)為

kxb，求參數(shù)的值yf(x0)(xx0)f(x0)f(x0)k再與切線方程比較系數(shù)得:,解此方程組可求參數(shù)的值xf(x0)f(x0)b〔2〕函數(shù)yf(x)〔含參數(shù)〕，談?wù)摵瘮?shù)的單調(diào)性【解法】先確定f(x)的定義域，并求出f(x)，觀察f(x)可否恒大于或等于〔恒小于或等于〕0，若是能，那么求參數(shù)的圍，談?wù)摫銖倪@里開(kāi)始，當(dāng)參數(shù)在上述圍以外取值時(shí)，令f(x)0，求根x1,x2.再分層談?wù)?，可否在定義域或談?wù)搙1,x2的大小關(guān)系，再列表談?wù)?，確定f(x)的單調(diào)區(qū)間?！泊蠖鄶?shù)函數(shù)的導(dǎo)函數(shù)都可以轉(zhuǎn)變成一個(gè)二次函數(shù)，因此談?wù)摵瘮?shù)單調(diào)性問(wèn)題又經(jīng)常是談?wù)摱魏瘮?shù)在某一區(qū)間上的符號(hào)問(wèn)題〕〔3〕函數(shù)yf(x)〔含參數(shù)〕在區(qū)間I上有極值，求參數(shù)的取值圍.【解法】函數(shù)f(x)在區(qū)間I上有極值，可轉(zhuǎn)變成方程f(x)0在區(qū)間I上有實(shí)根，且為非二重根。進(jìn)而確定參數(shù)〔或其取值圍〕?！?〕可導(dǎo)函數(shù)f(x)〔含參數(shù)〕在區(qū)間I上無(wú)極值，求參數(shù)的取值圍【解法】f(x)在區(qū)間I上無(wú)極值等價(jià)于f(x)在區(qū)間在上是單調(diào)函數(shù)，進(jìn)而獲取f(x)0或f(x)0在I上恒成立〔5〕函數(shù)f(x)〔含單個(gè)或多個(gè)參數(shù)〕僅在xx0時(shí)獲取極值，求參數(shù)的圍【解法】先由f(x)0，求參數(shù)間的關(guān)系，再將f(x)表示成f(x)=(xx0)g(x)，再由g(x)0(0)2/16....恒成立，求參數(shù)的圍?！泊祟?lèi)問(wèn)題中f(x)一般為三次多項(xiàng)式函數(shù)〕〔6〕函數(shù)f(x)〔含參數(shù)〕在區(qū)間I上不只一，求參數(shù)的取值圍【解法一】轉(zhuǎn)變成f(x)在I上有極值?！布磃(x)0在區(qū)間I上有實(shí)根且為非二重根〕?！窘夥ǘ繌姆疵婵紤]：假設(shè)f(x)在I上單調(diào)那么f(x)0(0)在I上恒成立，求出參數(shù)的取值圍，再求參數(shù)的取值圍的補(bǔ)集〔7〕函數(shù)f(x)〔含參數(shù)〕，假設(shè)x0I，使得f(x0)0（0）成立，求參數(shù)的取值圍.【解法一】轉(zhuǎn)變成f(x)在I上的最大值大于0〔最小值小于0〕【解法二】從反面考慮：假設(shè)對(duì)xI，f(x)0(0)恒成立那么f(x)max0〔f(x)min0〕,求參數(shù)的取值圍，再求參數(shù)的取值圍的補(bǔ)集8〕含參數(shù)的不等式恒成立，求參數(shù)的取值圍解法一】分別參數(shù)求最值【解法二】構(gòu)造函數(shù)用圖像注：關(guān)于多變量不等式恒成立，先將不等式變形，利用函數(shù)的最值消變?cè)?，轉(zhuǎn)變成單變量不等式恒成立問(wèn)題〔9〕可導(dǎo)函數(shù)f(x)〔含參數(shù)〕在定義域上存在單調(diào)遞加(減)區(qū)間，求參數(shù)的圍.【解法】等價(jià)轉(zhuǎn)變成f(x)0（0）在定義域上有解即x0I使f(x0)0（0）成立〔1〕可用分別參數(shù)法〔2〕利用圖像與性質(zhì)〔10〕證明不等式【解法】構(gòu)造函數(shù)f(x)并確定定義域I，觀察在I上的單調(diào)性〔注意區(qū)間端點(diǎn)的函數(shù)值〕也許求f(x)在上的最值注：關(guān)于含有正整數(shù)n的帶省略號(hào)的不定式的證明，先觀察通項(xiàng)，聯(lián)想根本不定式，確定要證明的函數(shù)不定式，再對(duì)自變量x賦值，令x分別等于1,2,,n,把這些不定式累加，可得要證的不定式?！?.函數(shù)f(x)4x2x，實(shí)數(shù)s,t滿(mǎn)足f(s)f(t)0，設(shè)a2s2t,b2st.〔1〕當(dāng)函數(shù)f(x)的定義域?yàn)?,1時(shí)，求f(x)的值域；〔2〕求函數(shù)關(guān)系式bg(a)，并求函數(shù)g(a)的定義域；〔3〕求8s8t的取值圍.3/16....1x[1,1]m2x[1,2]12f(x)l(m)m2m(m1)21[1,2]2242f(x)minl(m)minl(1)1f(x)maxl(m)maxl(2)2324f(x)[14,2].42s,tf(s)f(t)04s2s4t2t0st22st(2st06(22)22)a2s2tb2sta22ba0bg(a)1(a2a)72b0,a012a)0a18(a22s2t4s4t2(2s2t)222aaa2st921a2.1038s8t(2s2t)(4s2s2t4t)a(ab)a(a1a21a)1a33a2a(1,2]122222h(a)133a2(1,2]2a2,ah'(a)3a23a3a(a2)0a(1,2]1422h(a)a(1,2]h(a)(h(1),h(2)]8s8t(1,2].162.

f(x)

ex,g(x)

ax2

bx

c1

fx

gx

y4/16....線互相垂直，求b和c的值?！?〕假設(shè)a＝c＝1，b＝0，試比較f〔x〕與g〔x〕的大小，并說(shuō)明原由；〔3〕假設(shè)b＝c＝0，證明：對(duì)任意給定的正數(shù)a，總存在正數(shù)m，使適合x(chóng)(m,)時(shí)，恒有f〔x〕＞g〔x〕成立。解:ac1，b0時(shí)，g(x)x21，5分①x0時(shí)，f(0)1，g(0)1，即f(x)g(x)②x0時(shí)，f(x)1，g(x)1，即f(x)g(x)③x0時(shí)，令h(x)f(x)g(x)exx21,那么h'(x)ex2x.設(shè)()'()=x2，那么x，kxhxexk'(x)=e2當(dāng)xln2時(shí),k'(x)0,k(x)單調(diào)遞減;當(dāng)xln2時(shí),k'(x)0,k(x)單調(diào)遞加.因此當(dāng)xln2時(shí),k(x)獲取極小值,且極小值為k(ln2)eln22ln22ln40即()'()=x20恒成立，故h(x)在R上單調(diào)遞加,又h(0)0,kxhxex因此,當(dāng)x0時(shí),h(x)h(0)0,即f(x)g(x).9分綜上，當(dāng)x0時(shí),f(x)g(x)；當(dāng)x0時(shí),f(x)g(x)；當(dāng)x0時(shí),f(x)g(x)．10分⑶證法一：①假設(shè)0a1，由⑵知，當(dāng)x0時(shí),exx21.即exx2ax2，因此，0a1時(shí)，取m0，即有當(dāng)xm，，恒有exax2.②假設(shè)a1,f(x)g(x)即exax2，等價(jià)于xln(ax2)即x2lnxlna令t(x)x2lnxlna,那么t'(x)12x22時(shí),t'(x)0,t(x)在(2,)單調(diào)遞xx.當(dāng)x增.取x0ae2,那么x0e22，因此t(x)在(x0,)單調(diào)遞加.又t(x0)e2a2lne2alnae2a43lna7a43lna4(a1)3(alna)0即存在mae2,當(dāng)xm，時(shí)，恒有f(x)g(x).15分綜上，對(duì)任意給定的正數(shù)a，總存在正數(shù)m，使適合x(chóng)m，，恒有f(x)g(x).16分證法二：設(shè)h(x)exex(x2)x2，那么h'(x)x3，5/16....當(dāng)x(0,2)時(shí)，h'(x)0，h(x)單調(diào)減，當(dāng)x(2,)時(shí)，h'(x)0，h(x)單調(diào)增，故h(x)在(0,)上有最小值，h(2)e2，12分4①假設(shè)ae2，那么h(x)2在(0,)上恒成立，4即當(dāng)ae2時(shí)，存在m0，使當(dāng)x(m,)時(shí),恒有f(x)g(x)；4②假設(shè)ae2，存在m2，使當(dāng)x(m,)時(shí),恒有f(x)g(x)；4③假設(shè)ae2，同證明一的②，15分4綜上可得，對(duì)任意給定的正數(shù)a，總存在m,當(dāng)x(m,)時(shí),恒有f(x)g(x).16分設(shè)函數(shù)fxx2lnx-ax2+b在點(diǎn)x0,fx0處的切線方程為yxb.數(shù)a與x0的值；(2)求證：對(duì)任意實(shí)數(shù)，函數(shù)fx有且僅有兩個(gè)零點(diǎn).6/16....4.函數(shù)f(x)lnx1b；〔取e為2.8，取ln2為0.7，取21.4〕，g(x)axx〔1〕假設(shè)函數(shù)h(x)f(x)g(x)在(0,)上單調(diào)遞加，數(shù)a的取值圍；〔2〕假設(shè)直線g(x)axb是函數(shù)f(x)1lnx圖象的切線，求ab的最小值；x7/16....3b0f(x)g(x)A(x1,y1)B(x2,y2)2122exx1h(x)f(x)g(x)lnx1axbh(x)11axxx2h(x)f(x)g(x)(0,)x0h(x)11a02xx2x0a11110a0xx2xx2a(,0]412(x0,lnx01)y(lnx01)(11)(xx0)x0x0x0x02y(11)x(11)x0(lnx01)y(11)x(lnx021)xx2x0x2x0xx2x0000001t0a11tt2,blnx021lnt2t17x0x0x02x0ab(t)lntt2t1(t)12t1(2t1)(t1)ttt(0,1)(t)0(t)(0,1)t(1,)(t)0(t)(1,)ab(t)(1)1ab1103lnx11ax1lnx21ax2lnx1x2x1x2a(x1x2)x1x2x1x2x2x1x2lnx21a(x2x1)x1alnx1x1x2x2x1x1x2x1x2lnx212(x1x2)x1x2x2lnx1x2(x1ln12x1x2x2x1)(x1x2)lnx1x2x1x2x2x1x1x1x20x1x2tx21F(t)lnt2(t1)1)F(t)(t1)20x1t(tt(t1)1F(t)lnt2(t1)(1,)F(t)lnt2(t1)F(1)0t1t1lnt2(t1)lnx22(x2x1)lnx1x22(x1x2)x1x2lnx22t1x1x1x2x1x2x2x1x1lnx1x22(x1x2)lnx1x24x1x2lnx1x242lnx1x24x1x2x1x2x1x2x1x22lnx1x242lnx1x221x1x2x1x2G(x)lnx2x0G(x)120G(x)(0,)xxx2ln2e21ln2120.851G(x1x2)lnx1x221ln2e22e2ex1x22ex1x22ex1x22e2■168/16....x1)xeaxaR,e.( )(1a1f(x)(1,f(1))2f(x)3bRf(x)bxRab.1a1f'xex1f'1e1f1e2fx1,f1yee1x1ye1x142f'xexaa≤0f'x0fxR6a0f'xexa0xlnax,lnaf'x0fxxlna,f'x0fxa≤0fx(,)a0fxlna,,lna932a0fxRfx≥b10a0b≤0ab011a0fx≥bxRb≤fminxfminxflna2aalnab≤2aalna13ab≤2a2a2lnaga2a2a2lnaa0g'a4a2alnaa3a2alnaa0g'a0lna332ae233a0,e2g'a0gaae2,g'a0ga9/16....gmaxae3abe32233ae2,b1e21625.yf(x)xx0x0yf(x).f(x)ax33xlnx1(aR).1a0f(x)2f(x)(e,1)a.e10/16....函數(shù)f(x)ex，g(x)mxn.1〕設(shè)h(x)f(x)g(x).①假設(shè)函數(shù)h(x)在x0處的切線過(guò)點(diǎn)(1,0)，求mn的值；②當(dāng)n0時(shí)，假設(shè)函數(shù)h(x)在(1,)上沒(méi)有零點(diǎn)，求m的取值圍；〔2〕設(shè)函數(shù)r(x)1nx4m(m0)，求證：當(dāng)x0時(shí)，r(x)1.f(x)，且ng(x)解：〔1〕由題意，得h(x)(f(x)g(x))(exmxn)exm，因此函數(shù)h(x)在x0處的切線斜率k1m，又h(0)1n，因此函數(shù)h(x)在x0處的切線方程y(1n)(1m)x，將點(diǎn)(1,0)代入，得mn2.〔2〕方法一：當(dāng)n0，可得h(x)(exmx)exm，因?yàn)閤1，因此ex1e

2分4分，①當(dāng)m1時(shí)，h(x)exm0，函數(shù)h(x)在(1,)上單調(diào)遞加，而h(0)1，e1111因此只需h(1)0，解得m6分m，進(jìn)而m.1eeee②當(dāng)m時(shí)，由h(x)exm0，解得xlnm(1,)，e當(dāng)x(1,lnm)時(shí)，h(x)0，h(x)單調(diào)遞減；當(dāng)x(lnm,)時(shí)，h(x)0，h(x)單調(diào)遞加.因此函數(shù)h(x)在(1,)上有最小值為h(lnm)mmlnm，令mmlnm0，解得me，因此1me.e11/16....m[1,e).10en0exmxx0x1x0mexexyexxexexx11x0yxx2x2xy0yex0x1y0yex1y0xxexxyyx1yx1em[1,e).1xenxe1nx114x3r(x)mf(x)g(x)exnexxx4m14xr(x)1ex(3x4)x40exx4F(x)ex(3x4)x412F(0)0F(x)ex(3x1)1F(0)0G(x)F(x)G( )x(3x2)xex0G(x)014F(x)[0,)F(x)F(0)0F(x)[0,)F(x)F(0)0.16f(x)lnx1ax2x,aR2(1)f(1)0f(x);(2)xf(x)ax1a:(3)a2x1,x2f(x1)f(x2)x1x20:x151x221f(1)1a0a212f(x)lnxx2x,x0f12x12x2x10)2xxf(x)02x2x10x0x1f(x)(1,)412/16....2g(x)f(x)-(ax1)lnx12(1a)x12g(x)1(1a)ax2(1a)x1axxxa≤0x0g(x)0g(x)(0,)g(1)ln11a12(1a)13a2022xf(x)≤ax161a0g(x)ax2(1a)x1a(xa)(x1)xxg(x)0x1ax(0,1)g(x)0x(1,)g(x)0aag(x)x(0,11,))x(aag(x)g(1)ln11a(1)2(1a)111lnaaa2aa2a8h(a)1lna2a1h(2)10h(a)a(0,)h(1)0ln224a≥2h(a)0a2102f(x)≤ax1lnx1ax2x≤ax1(0,)2a≥lnxx11x2x(0,)213/16....g(x)lnxx11x2xa≥g(x)6max2(x1)(1xlnx)1xlnx0g(x)12g(x)0x2x)22(2h(x)1xlnxh(x)110h(x)(0,)22x1xlnx0x02x(0,x0)g(x)0x(x0,)g(x)0g(x)x(0,x0)x(x0,)lnx0x0111x1g(x)maxg(x0)201182x02x0x0x0(12x0)h(1)ln210h(1)102241x01112g(x)max(1,2)x02a≥2a21032f(x)lnxx2x,x0af(x1)f(x2)x1x20lnx1x12x1lnx2x22x2x1x20(xx)2(xx)xx2ln(xx)131212112tx1x2(t)tlnt(t)t1t(t)(0,1)(1,)(t)≥(1)115(x1x2)2(x1x2)≥1x1x2≥5116214/16....a，b為實(shí)數(shù)，函數(shù)f(x)1b，函數(shù)g(x)lnx．xa〔1〕當(dāng)ab0時(shí)，令F(x)f(x)g(x)，求函數(shù)F(x)的極值；〔2〕當(dāng)a1時(shí)，令G(x)f(x)g(x)，可否存在實(shí)數(shù)b，使得關(guān)于函數(shù)yG(x)定義域中的任意實(shí)數(shù)x1，均存在實(shí)數(shù)x2[1,)，有G(x1)x20成立，假設(shè)存在，求出實(shí)數(shù)b的取值會(huì)集；假設(shè)不存在，請(qǐng)說(shuō)明原由．解：〔1〕F(x)1lnx，xF(x)x21，令F(x)0，得x1．1分x列表：x(0,1)1(1,)F(x)0+F(x)↘極小值↗因此F(x)的極小值為F(1)1，無(wú)極大值．4分〔2〕當(dāng)a1時(shí)，假設(shè)存在實(shí)數(shù)b滿(mǎn)足條件，那么G(x)(1b)lnx≥1在x(0,1)(1,)上恒成x1立．5分1〕當(dāng)x(0,1)時(shí)，G(x)(1b)lnx≥1可化為(bx1b)lnxx1≤0，x1令H(x)(bx1b)lnxx1,x(0,1)，問(wèn)題轉(zhuǎn)變成：H(x)≤0對(duì)任意x(0,1)恒成立；〔*〕那么H(1)0，H(x)blnx1bb1，H(1)0．x令Q(x)blnx1bbb(x1)1x1，那么Q(x)2．x①b≤1時(shí)，因?yàn)閎(x1)1≤1(x1)11210，222故Q(x)0，因此函數(shù)yQ(x)在x(0,1)時(shí)單調(diào)遞減，Q(x)Q(1)0，即H(x)0，進(jìn)而函數(shù)yH(x)在x(0,1)時(shí)單調(diào)遞加，故H(x)H(1)0，因此〔*〕成