導(dǎo)數(shù)的應(yīng)用時(shí)PPT學(xué)習(xí)教案

1、會(huì)計(jì)學(xué)1導(dǎo)數(shù)的應(yīng)用時(shí)導(dǎo)數(shù)的應(yīng)用時(shí)f xxg xkx kRxf xxkxxxf xg x001 1( )ln(1), ( )()(1)0( );(2)10,(0,)( )( ).、已已知知函函數(shù)數(shù)證證明明：當(dāng)當(dāng)時(shí)時(shí)，證證明明：當(dāng)當(dāng)時(shí)時(shí)，存存在在使使得得對(duì)對(duì)任任意意的的恒恒有有例題：考點(diǎn)一、利用導(dǎo)數(shù)解決不等式問(wèn)題例題：考點(diǎn)一、利用導(dǎo)數(shù)解決不等式問(wèn)題F xf xxxx x(1)( )( )ln(1),0,)證證明明： 令令xF xxx1( )111 則則xF x(0,)( )0當(dāng)當(dāng)時(shí)時(shí)，F(xiàn) x( )0,)所所以以在在上上單單調(diào)調(diào)遞遞減減xF xFxf xx0( )(0)0,0( )所所以以當(dāng)當(dāng)時(shí)時(shí)，

2、即即當(dāng)當(dāng)時(shí)時(shí)，第1頁(yè)/共13頁(yè)f xxg xkx kRkxxxf xg x001 1( )ln(1), ( )()(2)10,(0,)( )( ).、已已知知函函數(shù)數(shù)證證明明：當(dāng)當(dāng)時(shí)時(shí)，存存在在使使得得對(duì)對(duì)任任意意的的恒恒有有例題：考點(diǎn)一、利用導(dǎo)數(shù)解決不等式問(wèn)題例題：考點(diǎn)一、利用導(dǎo)數(shù)解決不等式問(wèn)題G xf xg xxkx x(2)( )( )( )ln(1),0,)令令kxkG xkxx1(1)( )11則則kG xG x0( )0,( )0,)當(dāng)當(dāng)時(shí)時(shí)，所所以以在在上上單單調(diào)調(diào)遞遞增增G xGx0( )(0)0,所所以以故故任任意意正正實(shí)實(shí)數(shù)數(shù)均均滿滿足足題題意意第2頁(yè)/共13頁(yè)f xxg

3、xkx kRkxxxf xg x001 1( )ln(1), ( )()(2)10,(0,)( )( ).、已已知知函函數(shù)數(shù)證證明明：當(dāng)當(dāng)時(shí)時(shí)，存存在在使使得得對(duì)對(duì)任任意意的的恒恒有有例題：考點(diǎn)一、利用導(dǎo)數(shù)解決不等式問(wèn)題例題：考點(diǎn)一、利用導(dǎo)數(shù)解決不等式問(wèn)題kkG xxkk1101( )0,10 當(dāng)當(dāng)時(shí)時(shí)，令令得得xxxG xk0011,(0,),( )0取取對(duì)對(duì)任任意意有有G xx0( )0,)從從而而在在上上單單調(diào)調(diào)遞遞增增G xGf xg x( )(0)0,( )( )所所以以即即kxxf xg x0010,(0,)( )( )綜綜上上所所述述：當(dāng)當(dāng)時(shí)時(shí)，總總存存在在使使得得對(duì)對(duì)任任意意恒

4、恒有有第3頁(yè)/共13頁(yè)xf xexaxaf xxf xa212( )1.(1)0( )0;(2)0( )0. 、已已知知函函數(shù)數(shù)當(dāng)當(dāng)時(shí)時(shí)，求求證證：當(dāng)當(dāng)時(shí)時(shí)，若若不不等等式式恒恒成成立立，求求實(shí)實(shí)數(shù)數(shù) 的的取取值值范范圍圍例題：考點(diǎn)一、利用導(dǎo)數(shù)解決不等式問(wèn)題例題：考點(diǎn)一、利用導(dǎo)數(shù)解決不等式問(wèn)題xaf xex(1)0( )1 解解： 當(dāng)當(dāng)時(shí)時(shí)，xfxe( )1xfx(,0)( )0 當(dāng)當(dāng)時(shí)時(shí)，xfx(0,)( )0當(dāng)當(dāng)時(shí)時(shí)，f x( )(,0)(0,)所所以以在在上上單單調(diào)調(diào)遞遞減減，在在上上單單調(diào)調(diào)遞遞增增f xfmin( )(0)0所所以以f x( )0所所以以第4頁(yè)/共13頁(yè)xf xexa

5、xxf xa212( )1.(2)0( )0. 、已已知知函函數(shù)數(shù)當(dāng)當(dāng)時(shí)時(shí)，若若不不等等式式恒恒成成立立，求求實(shí)實(shí)數(shù)數(shù) 的的取取值值范范圍圍例題：考點(diǎn)一、利用導(dǎo)數(shù)解決不等式問(wèn)題例題：考點(diǎn)一、利用導(dǎo)數(shù)解決不等式問(wèn)題xfxeax(2)( )12 xxh xfxeaxh xea( )( )12,( )2 令令則則aah x121( )00,)2當(dāng)當(dāng)，即即時(shí)時(shí)，在在上上恒恒成成立立，h x( )0,)所所以以在在上上單單調(diào)調(diào)遞遞增增h xhfxf( )(0),( )(0)0所所以以即即f x( )0,)所所以以在在上上單單調(diào)調(diào)遞遞增增f xf( )(0)0所所以以a12所所以以當(dāng)當(dāng)時(shí)時(shí)滿滿足足條條件件

6、第5頁(yè)/共13頁(yè)xf xexaxxf xa212( )1.(2)0( )0. 、已已知知函函數(shù)數(shù)當(dāng)當(dāng)時(shí)時(shí)，若若不不等等式式恒恒成成立立，求求實(shí)實(shí)數(shù)數(shù) 的的取取值值范范圍圍例題：考點(diǎn)一、利用導(dǎo)數(shù)解決不等式問(wèn)題例題：考點(diǎn)一、利用導(dǎo)數(shù)解決不等式問(wèn)題aah xxa121( )0ln22當(dāng)當(dāng)，即即時(shí)時(shí)，令令得得xln ah xh x0,2 )( )0,( )當(dāng)當(dāng)時(shí)時(shí)，所所以以單單調(diào)調(diào)遞遞減減f xf( )(0)0,所所以以不不符符合合題題意意a1(,2綜綜上上所所述述：實(shí)實(shí)數(shù)數(shù) 的的取取值值范范圍圍是是 xah xhfxf0,ln2 )( )(0)0,( )(0)0所所以以當(dāng)當(dāng)時(shí)時(shí)，即即f xa( )0

7、,ln2 )所所以以在在上上單單調(diào)調(diào)遞遞減減第6頁(yè)/共13頁(yè)1.利用導(dǎo)數(shù)證明不等式的方法利用導(dǎo)數(shù)證明不等式的方法證明f(x)g(x),x(a,b),可以構(gòu)造函數(shù)F(x)=f(x)-g(x),如果F(x)0,則F(x)在(a,b)上是減(增)函數(shù),同時(shí)若F(a)()0,由減(增)函數(shù)的定義可知,x(a,b)時(shí),有F(x)0,即證明了f(x)g(x).其一般步驟是:構(gòu)造可導(dǎo)函數(shù)研究單調(diào)性或最值得出不等關(guān)系整理得出結(jié)論.考法指導(dǎo)考法指導(dǎo)1、第7頁(yè)/共13頁(yè)2.不等式成立不等式成立(恒成立恒成立)問(wèn)題問(wèn)題(1)f(x)a恒成立f(x)mina,(x)a成立f(x)maxa.(2)f(x)b恒成立f(x

8、)maxb,f(x)b成立f(x)minb.(3)f(x)g(x)恒成立 F(x)min0.(4)x1M,x2N,f(x1)g(x2)f(x1)ming(x2)min.x1M,x2N,f(x1)g(x2)f(x1)ming(x2)max.x1M,x2N,f(x1)g(x2)f(x1)maxg(x2)min.x1M,x2N,f(x1)g(x2)f(x1)maxg(x2)max.考法指導(dǎo)考法指導(dǎo)1、第8頁(yè)/共13頁(yè)xef xxeyf xPf xxx213( ).(1)( )(2,)2(2)( )2(ln )、已已知知函函數(shù)數(shù)求求曲曲線線在在點(diǎn)點(diǎn)處處的的切切線線方方程程；證證明明：例題：考點(diǎn)一、利用導(dǎo)數(shù)解決不等式問(wèn)題例題：考點(diǎn)一、利用導(dǎo)數(shù)解決不等式問(wèn)題第9頁(yè)/共13頁(yè)xef xf xxxx13( ).(2)( )2(ln )、已已知知函函數(shù)數(shù)證證明明：例題：考點(diǎn)一、利用導(dǎo)數(shù)解決不等式問(wèn)題例題：考點(diǎn)一