第8講冪函數(shù),冪、指對函數(shù)綜合問題

7.冪函數(shù),“冪、指、對”函數(shù)綜合問題[基礎(chǔ)秘訣](問中學(xué)問1寫出冪函數(shù)的定義,并總結(jié)冪函數(shù)的圖象和性質(zhì)問2g(xxa(a0)f(xax(a0,a1的單調(diào)性有何不同?問3總結(jié)“冪、指、對”問題的解題方法問4“基本函數(shù)法”?“輔助函數(shù)法[范例評注](例中學(xué) 已知冪函數(shù)y(m23m3)xm2m2的圖象不過原點(diǎn),則m的取值范圍 例 若(a1)3(32a)3,則a的取值范圍 例 將下列四數(shù)從小到大排列起來 0.40. lg0.4 log0.例 當(dāng)0<a<b<1時(shí),下列不等式中正確的1A.(1a)b(1bC.(1a)b(1

B.(1a)a(1D.(1a)a(1例 對于函數(shù)f(x)定義域中任意的x1,x2(x1x2),有如下結(jié)論①f(x1x2)f(x1)f(x2) ②f(x1x2)f(x1)f(x2)③(xx)[f(x)f(x)]0 ④f(x1x2)f(x1)f(x2) 當(dāng)f(x)2x時(shí),上述結(jié)論中正確結(jié)論的序號 例 設(shè)f(x)lg(mx2mx1)4f(x)R,m的取值范圍f(x)R,m的取值范圍例 設(shè)f(x)

12x4x3

x(,1]時(shí)有意義,a的取值范圍例 已知函數(shù)f(x)lg1x1求f(x)的定義域和值域求函數(shù)的奇偶性和單調(diào)性4x例 已知f(x) 2x1,求f(19)f(19)的值例10f(x)定義在(0)上,xy(0)f(xy)=f(xf(y),x>1f(xf(1)的值f(x)是增函數(shù)[7](做中學(xué),用中學(xué)y B.yx

C.y D.y

yx3

yx2 (3)y

yx3點(diǎn)(2,2)f(x)的圖象上,點(diǎn)(2,1)g(x)的圖象上4問當(dāng)x為何值時(shí)有:(1)f(x)=g (2)f(x)>g (3)f(x)<ga(2005.)設(shè)0a1,函數(shù)f(x)log(a2x2ax2),則使f(x)0的x的aA.(, B.(0, C.(,loga D.(loga3,f(xlg(x22xm的值域R,mA. B.m C.m D.x設(shè)0x2，求函數(shù)y 232x5的最大值、最小值fx)|lgx|,0abf(a)f(babf(xlg(mx2mx1有最小值,m的取值范圍4,a的取值范圍

4(a (a 2xlog2a1 0恒成立f(xaf(x)的值域

2x

R上的奇函數(shù)f(x)f1x),f1(15的值幻燈片幻燈片77冪函數(shù)冪、指、對”函數(shù)綜合問題問寫出冪函數(shù)的定義并總結(jié)冪函數(shù)的圖象和性質(zhì)定義：yxa(aR的函數(shù)叫做冪函數(shù)圖象冪函數(shù)基本圖(6圖①y②y⑤③②②y③y1④y⑤y⑥y⑥1⑤O④⑥⑤x ③幻燈片問問寫出冪函數(shù)的定義并總結(jié)冪函數(shù)的圖象和性質(zhì)冪函數(shù)的通性所有冪函數(shù)在(0,+)上都有定義 圖象都過點(diǎn)y=xa(a>0的圖象都過y=xa(a0的圖象都不過原y=xa(a>0)在區(qū)間[0,+)上是增函y=xa(a<0)在區(qū)間(0,+)上是減函冪函數(shù)y=xa(a<0)有兩條漸進(jìn)線x0y奇偶性和凹向:類比基本圖幻燈片幻燈片問問yxa(a0)yax(a0,a的單調(diào)性易錯(cuò)易混,怎樣區(qū)別“真假猴王y=xyaa?單調(diào)性誰是“指”,“冪幻燈片問問解總結(jié)“冪指對函數(shù)注意三個(gè)方法的解題方法?幻燈片幻燈片問問 “基本函數(shù)法”?“輔助函數(shù)法”解基本函數(shù)法利用基本函數(shù)的圖象和性質(zhì)來解題的方法,叫做基本函數(shù)法.確定基本函數(shù)類總結(jié)基本函數(shù)的圖象和性質(zhì)把其它問題化歸為基本函數(shù)來解決輔助函數(shù)法幻燈片[[范例評注](例中學(xué)例已知冪函數(shù)y(m23m3)xmm2的圖象不過原點(diǎn),則m的取值范圍是m1m22解 3m32m2m2m1m幻燈片幻燈片例例探解若(a1)3(32a3則ayx1分類 3 (,1)∪(2,3(2)32aa10a(1)a132a02a32(3)a132aaa(,1)∪(2,3幻燈片例例 0.40.lg0.4 log0.以0為界：lg以1為界：log0.30.2>log0.以0.20.3為界∴l(xiāng)g0.4<50.5<0.40.3<log0.答幻燈片幻燈片例例當(dāng)0<a<b<1時(shí),下列不等式中正確的是(D1A.(1a)b(1bC.(1a)b(1B.(1a)a(1D.(1a)a(1解法解法(1a)a(1a)b(1檢驗(yàn)特值a1,b42 (1a)a(3)4()4(11144幻燈片例例對于函數(shù)f(x)定義域中任意的xx(xx有如下結(jié)論①f(x1x2)f(x1)f(x2②f(x1x2)f(x1)f(x2 ③(xx)[f(x)f(x)] 12④f(x1x2)f(x1)f(x2當(dāng)f(x)=2x時(shí),上述結(jié)論中正確結(jié)論的序號是①③④22解①正確,②不正確f(x)=2x的圖象如圖由f(x)是減函數(shù)知③正確由f(x)的圖象下凸知④正確y(12例例1設(shè)f )4f(x)R求實(shí)數(shù)mf(x)的值域是R求實(shí)數(shù)m的取值范圍(1)m取值m0,或m 0m1答(2)m取值m0m1答 評注基本函數(shù)法對數(shù)函數(shù)和二次函數(shù)的圖象和性質(zhì)幻燈片例例設(shè)fx12x4x在x(,1]上有意義3解求實(shí)a的取值范圍問題等價(jià)于1+2x+4xa0x(,1]上恒成立12x4xa0a(1)x(1)x··42引進(jìn)輔助函數(shù)g(x)(1)x(1)x x(,因?yàn)間(x)是減函數(shù)，所以gxg(1)113 x(1]上恒成立ag(1)3424∴實(shí)數(shù)a的取值a43評注輔助函數(shù)法:把不等式恒成立化為輔助函數(shù)的最值問題.幻燈片幻燈片例例f(x)11求的定義域和值域求函數(shù)的奇偶性和單調(diào)性解11x01x∴定義域?yàn)橛蓇 (0,)f(x)lgu的值域?yàn)?(2)f(x是奇函數(shù)1fxlg21是減函數(shù)1幻燈片例例解fx412x1求f(19)+f(19)的值x令g(x) 12x,則fx由gx)2x4x1 2x 22x222知g(xg(x∴f(19)+f=[g(19)+1]+[g(19)+1]=幻燈片幻燈片例例f(xy)=f(xf(y且當(dāng)x>1f(x求f(1)的值證明f(x)是增函數(shù)(1)f(xyf(xf(y)xy=1，得f(1)=f(1)+f(1)f(1)=0．(2)證明設(shè)x>x>0則x21f(x2 xxf(x)f(x)f x)f(xx11221f(x)f(x)f(x)f(x)x1112x111x1f(x2f(x1),f(x)是增函數(shù)評注把抽象函數(shù)f(x)類比于對數(shù)函數(shù)lgx便能掌握問題的思路和結(jié)論幻燈片[[檢測7]解1.21提:(1)yx是偶函數(shù),在33yx2定義在[0,)上,其上的圖象類比yyx2是偶函數(shù)(4)yx3是奇函數(shù),類比yx幻燈片幻燈片2,2)在冪函數(shù)f(x)的圖象上,點(diǎn) 14圖象上,問當(dāng)x為何值時(shí)有(1)f(x)=g (2)f(x)>g (3)f(x)<g解f(xxa由(2)a2a2,f(x設(shè)g(xxb由(2)b1b2g(x4y當(dāng)1x1且x0時(shí)f(x)11 x幻燈片2x設(shè)0a1,函數(shù)f(x)log 2a2),則使f(x)0x的取值范圍是a (C令tax>0f(x)0t22t2>1t>3t1(舍ax>3(0a1)x(,loga3),故選解函數(shù)f(x)=lg(x2+2x+m)的值域?yàn)槿≈捣秶莿t實(shí)數(shù)m A.B.mC.mD.=44m0m1故選幻燈片幻燈片解設(shè)0x2,求函數(shù)y 232x5的最大值、最小值t2x[1, yf(t)1t23t2t0=3[1xf(3)12f(1)5 f(4)2 52 12幻燈片7.7.fx|lgx|,0abf(af(b求證ab證由b>a>0 f(x)=|lgx|f(a)>f(b)(lga)2>(lgb)2(lga+lgb)(lgalgb)>結(jié)合①得lga+lgb0lg(ab)0ab幻燈片幻燈片4的取值范圍．解“m取值”m0,或m00m1.(答幻燈片設(shè)對所有實(shí)數(shù)x,不等式x224(a1)2xa2a2(a1)20—恒成立,a的取值范圍解設(shè)t a1,則①化為2(3t)x22tx2t—3t0時(shí),6x60不恒成立3t0,恒