高考數(shù)學(xué)導(dǎo)數(shù)題型歸納(文科)

（二次函數(shù)區(qū)間最值的例子）第三種：構(gòu)造函數(shù)求最值題型特征：f(x)g(x)h(x)f(x)g(x)0xP(1,b)3，例f(x)t632g(x)xxtx3t0)322a,bx[f(x)xf(x)g(x)t二、題型一：已知函數(shù)在某個區(qū)間上的單調(diào)性求參數(shù)的范圍解法：轉(zhuǎn)化為在給定區(qū)間上恒成立，回歸基礎(chǔ)題型f(x)或f(x)0''解法：所給區(qū)間是求的增或減區(qū)間的子集；做題時一定要看清楚“在（）上是減函數(shù)”與“函數(shù)的單調(diào)減區(qū)間是（兩句話的區(qū)別：前者是后者的子集1a1Rf(x)xx(4ax．例a32122f(x)f(x)g(x)f(x)是(,)a11x(2a)x(1a)x(a0).例f(x)3232（f(x)（f(x)在，求a子集思想三、題型二：根的個數(shù)問題題1函數(shù)f(x)與g(x)（或與x軸）的交點======即方程根的個數(shù)問題解題步驟主要看極大值和極小值與0的關(guān)系；(k11xxg(x)f(x)(2,)例f(x)，32323（）k（）f(x)與g(x)k1axx2xc例f(x)3221f(x)是f(x)xf(x)1bxxdbg(x)f(x)的g(x)x221b題：切線的條數(shù)問題以切點為未知數(shù)的方程的根的個數(shù)x0axbxcxx0f'(x)0x的例f(x)321,m)yf(x)m為(1,3)）f(x)P(題：已知在給定區(qū)間上的極值點個數(shù)則有導(dǎo)函數(shù)=0的根的個數(shù)f(x)解法：根分布或判別式法例8、a11例9、f(x)xx，(aR,a0)＝＋ff(x)g(x)32324（3a其它例題：（最值問題與主元變更法的例子）ax2ax（）Rf(x)a0在32f(x)t[f(0x（根分布與線性規(guī)劃例子）2（1）f(x)xaxbxc323(Ⅰ)f(x)在xf(x)11)3xy0平行,求1)x,Mba(Ⅱ)當(dāng)f(x)在xL將S,L.：Ⅰ由f(x)2x2axb,f(x)在x1,2b20∴2a1c1∵f(0)∴y0f(x)在(0,1)3x,1∴fb3a故221xx3x1∴f(x).7分3232Ⅱ):由f(x)2x2axb及f(x)在x1)x2),2b0f(0)0xb2∴f(1)0即2ab20令M(x,y),則ya1f(2)04ab80x20ay1x20∴∴2ySMx2b4yx603S2B(CDE(0,)(,,,,,21S3S,∴L:x0kxxLS,Ly它10S,2F:xFykx6G:x由4yx604k1G136112SSS116k2k502∴即224k122k1四邊形151kyx)::k或28210:x或yx分2Ⅱ):由f(x)2x2axb及f(x)在x1)x2),2b0f(0)0xb2∴f(1)0即2ab20令M(x,y),則ya1f(2)04ab80x20ay1x20∴∴2ySMx2b4yx603B(CDE(0,)2(,,,,,S2ABC1S3S,L:x01x:y,與H,2111,)2L與:H(211111112SS212S222,S2,∵SH2222ABCDEC1∴:x0或yx213（根的個數(shù)問題）f(x)xaxx1(aR)32x,xxxx2f(x)在xf(x)a1212115f(x)g(x)x(2ax(2x1)af'(x)xx2a,xx1與2226x2ax121212xx(xx)4xx4a4222121212a0分f(x)x2ax1x122令f(x)0得xx1或令f(x)0得1x1∴f(x)(,1)(1,)，(1,1)分13156g(x)xaxx1x(2axf(x)得32221即x11(a)x2ax032326111x(a)x2ax(2x1)令(x)分32326(x)x(2a1)x2a(x2a)(x1)2令(x)0得x2a或x1分1Qa22a1即時當(dāng)2ax21(x)(x)－92aa98a0a0分，2121a即當(dāng)2a2x21(x)(x)＋—09223168aa2(32)aa21Qa(32a)02,36998a01a即216998a，且a0a0當(dāng)即216129a8a0當(dāng)0分2910aa或1629a016當(dāng)分x210535簡單切線問題）f(x)3a2g(x)f(x)3．a(chǎn)2g(x)在x1g(x)g(x)[[mb2mb4g(x)（根的個數(shù)問題）、f(x)axbx(c3a2b)xd(a0)32c、df(x)f(xx5,0af(x)3ax2bx+c-3a-2b2f(x(0,3f1=0d3d3c0得abc3a2b0f2=–3且f(2)=512abab3