導(dǎo)數(shù)單調(diào)性與極值最值練習(xí)

1、導(dǎo)數(shù)（二）原函數(shù)和其導(dǎo)函數(shù)圖象之間的關(guān)系氐已知辭mm的圖象如.圖所鳳則，訂的圖象可施是D為X上）的導(dǎo)爾數(shù)，則的大敷圖象是（I （2016 +灣化止定申學(xué)）已紂/W滬咖*鈿一中已知?dú)n歡廠口刃在鋌義域今）內(nèi)町電其IO如圖所求記廠乳劃的導(dǎo)哂栽為 廠廠 m 則不等式攻瑋。的解集為（）AJ |,-yU0JJuL23)C. -|ju2,3B t y,0)UtR2ju(|j) 6( -y -yUy,yU-|-,3)6-冊(cè)是贖數(shù)人工)的爭(zhēng)爾將的圖孰廚在同亠令半閒庇幅生標(biāo)系中.則下需中不 叫憶此臨的總( I二.利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性f(x)在點(diǎn)(1,f(1)處的切線與x軸平行。In x k1.已知函數(shù)f (x

2、)x ，曲線ye(1)求實(shí)數(shù)k的值(2)討論f (x)的單調(diào)性1.利用導(dǎo)數(shù)與函數(shù)單調(diào)性的關(guān)系求參1 1匚在(2若函數(shù)f (x) x2ax)上是增函數(shù)，則實(shí)數(shù) a的取值范圍是(2.已知函數(shù)f (x) In2ax 2x(a0)存在單調(diào)遞減區(qū)間，求實(shí)數(shù) a的取值范圍3.若函數(shù)f (x)bx2 2x 5有3個(gè)單調(diào)區(qū)間，則實(shí)數(shù) b的取值范圍為(4.若函數(shù)f(x)2x2In x在定義域內(nèi)的一個(gè)子區(qū)間(k 1,k 1)上不是單調(diào)函數(shù)，則實(shí)數(shù)k的取值范圍是5.已知函數(shù)f (x) ax(a 1)ln(x 1)(a1)，求 f (x)的單調(diào)性6.已知a R，討論函數(shù)f (x)1 3 3a 1 221xx (2aa

3、)x 3的單調(diào)性7.2設(shè)函數(shù)f(x) In x ax (a 2)x(a R),求函數(shù)f (x)的單調(diào)性。8.已知函數(shù) f(x) In x, h(x) x 2 x。(1 )判斷函數(shù)h(x)的單調(diào)性(2 )求證：當(dāng)1x e2時(shí)，不等式x卅恒成立。四含參數(shù)的函數(shù)的極值與最值1.若函數(shù)f (x)小 3八 22x 3x 1(xax .e (x 0)0)在-2,2上的最大值為 2，則實(shí)數(shù)a的取值范圍為2.已知函數(shù)f(x)x3 2ax a在(0, 1)內(nèi)有極小值沒(méi)有極大值，則實(shí)數(shù)a的取值范圍是( )33.已知函數(shù)f(x) x 3x c的圖象與x軸恰有兩個(gè)交點(diǎn)，則 c的值為(1 In x24. 已知函數(shù)f(x

4、)在區(qū)間(a,a )(a 0)上存在極值，則實(shí)數(shù) a的取值范圍是x3( )a5. 已知函數(shù)f(x) x -(a R),求f(x)在1,2上的最大值與最小值x6.已知函數(shù) f (x) ln x a(1 x), a R(1)討論f (x)的單調(diào)性(2)當(dāng)f (x)有最大值，且最大值大于 2a 2時(shí)，求實(shí)數(shù)a的取值范圍能力提升11. 已知(a 1)x 1 Inx 0對(duì)任意x ,2恒成立，則實(shí)數(shù)a的最大值是()2. 若函數(shù)f(x) x3 6x2 9x 10 a有三個(gè)零點(diǎn)，則實(shí)數(shù) a的取值范圍為()3. 設(shè)函數(shù)f (x)在R上可導(dǎo)，其導(dǎo)函數(shù)為f(x)，且函數(shù)y (1 x)f(x)的圖象如圖所示，則下列結(jié)論中一定成立的是()A函數(shù)f (x)有極大值f(2)和極小值f (1) B函數(shù)f (x)有極大值f( 2)和極小值f(1)C函數(shù)f (x)有極大值f和極小值f ( 2)D函數(shù)f (x)有極大值f ( 2)和極小值f (2