導(dǎo)數(shù)專題(經(jīng)典23題)

1、第1頁(yè)23個(gè)函數(shù)與導(dǎo)函數(shù)類型專題1、函數(shù)第1題已知函數(shù)，若，且，，求ln()x1fxx1x???x0?x1?ln()xkfxx1x???k的取值范圍.解析：⑴將不等式化成模式()()k???由得：，化簡(jiǎn)得：①ln()xkfxx1x???lnlnx1xkx1xx1x?????ln22xxk1x1???⑵構(gòu)建含變量的新函數(shù)()gx構(gòu)建函數(shù)：（，且）ln()22xxgxx1??x0?x1?其導(dǎo)函數(shù)由求得：2uuvuvvv????????()(

2、lnln)()22222gxxxxx1x1?????即：②()[()()ln]()22222gxx1x1xx1?????()ln()222222x1x1xx1x1??????????????⑶確定的增減性()gx先求的極值點(diǎn)，由得：()gx()0gx0?ln20020x1x0x1????即：③ln20020x1xx1???由基本不等式代入上式得：lnxx1??20020x1x1x1????故：即：20020x1x10x1?????()(

3、)0201x110x1????由于，即，故：，即2011x1??20110x1???0x10??0x1?即：的極值點(diǎn)()gx0x1?在時(shí)，由于有界，而無界0xx1??22x11x1???lnx0?第3頁(yè)當(dāng)時(shí)，用洛必達(dá)法則得：x1?，則ln(ln)(ln)limlimlim()22x1x1x12xx2xx2x112xx1x1?????????()K10?用數(shù)值解如下：x0.30.40.50.60.70.80.91.0()Kx0.20620

4、.12730.07580.04220.02090.00830.00180.0000x1.11.21.31.41.51.61.71.8()Kx0.00150.00550.01140.01860.02690.03590.04540.0553其中，的最小值是，即，所以本題結(jié)果是.()Kx()K10?()()KxK1?k0?2、函數(shù)第2題已知函數(shù)，，，連續(xù)，若存在均屬于()ln2fxxax??a0?x0?()fx區(qū)間的，且，使，證明：[]13?

5、?1????()()ff???lnlnln322a53???解析：⑴求出函數(shù)的導(dǎo)函數(shù)()fx函數(shù)：①()ln2fxxax??其導(dǎo)函數(shù)：②()2112axfx2axxx????()()12ax12axx???⑵給出函數(shù)的單調(diào)區(qū)間()fx由于，由②式知：的符號(hào)由的符號(hào)決定.x0?()fx()12ax?當(dāng)，即：時(shí)，，函數(shù)單調(diào)遞增；12ax0??1x2a?()fx0?()fx當(dāng)，即：時(shí)，，函數(shù)單調(diào)遞減；12ax0??1x2a?()fx0?()f