函數的單調性與周一、函數的

4講函數的單調性與我們在初中研究了一次函數、二次函數的圖象，研究了yaxb(a0)yax2bxc(a0[1]

yyx22x

的圖象，并觀察說出在定義域(,g(xx22x1的圖象在對稱軸左方的區間(,1)是下降的，在對稱軸右方的區間)g(x)隨自變量x的增大而增大。1：yf(x)(xA，對于區間(ab1x1x2(a,b)，x1x2f(x1f(x2yf(x(ab內是增函數（increasingfunction2x1x2(a,b)，x1x2f(x1f(x2yf(x間(ab內是減函數(decreasingfunction)2、yf(xf(xf(x1f(x是增（減）函數圖象自左到右上升（下降2、圖象的峰（谷）函數增（減）變減（增）點函數的極大（小）xIf(xM（f(xMx0If(x0 value[3]如圖是定義在閉區間[-5，5]f(xf(x1f(xxx[說 1]函數的單調性與定義的區間有關，它是函數的局部性質[說明3]初等函數均可分段單調x1f(xx21(x(1,1的單調性2f(xx22x3f(xx22x3的單調區間3．已知函數fxR上的減函數，則滿足f

1x1xA． B． C．1,0∪ D．,1∪（1）f(0)1x0f(x1f(xR設集A(x,y)f(x2fy2f(1)B(xy)f(axy2)1,aR，ABa的取值范圍。5yf(u)(uR是增函數，u(x)(xR是減函數，求復合函f((xR的單二、函數的定義：yf(x)(xAxAf(xTf(xT0是常數)f(x為周期函數，T是f(x)的周期說明：（1）周期函數的定義域至少一Tf(xkTf(x的周期f(xxf(xTf(xf(x2Tf(xxf(axf(axa0)f(x是周期為4a的周期函數。f(x課后練習f(x)x22xyx(1y612x參考答問題3（1）5,5,1,25,1,2,5是增 2 （2）x1x20,1)x1f(x)f(x)

1

1(xx)11 x x x 1 2 2(xx)x2x1(x

1 x x2) x1 (xx)(x1x2

12由x1x2得，x1x2 又x1,x2(0,x1x2(0, x1x21 則(xx)(x1x21) f(x)f f(x在(0,1)問題 例 x1,x2(1,1)且x1ax(x21)x(x2a(xx2xx2)(xxf(x)f(x) 1x22x21 2 (x21)(x21 2 (x21)(x212 ax1x2(x2x1)(x2 a(xx)(xx (x21)(x2 (x21)(x2由x1 x2x1x1,x2(1,x20,x20,x21x211212x1x21,x1x21（1）a0(x2x1)(x1x21) (x21)(x2 （2）a0a(x2x1)(x1x21) (x21)(x2 （3）a0a(x2x1)(x1x21) (x21)(x2

f(x1f(x2f(x1)f(x2f(x1)f(x2

f(x在1,1f(x在區間1,1f(x在1,1x22xf(x)

x g(x)利用圖象求單調區間（ xx2x(,1),(0,1)減(1,0),(1,f

x22x2xx

x3或x1<x

(,1),(13)

(1,1),(3,例 而f(5)(0, f(50)f(5)f f(0) x f(x)(0, f(0)f(x)f(于是f(x) f(（2）x1x2Rx1f(x1f(x2f(x1x2x2f(x2f(x1x2f(x2f(x2f(x2)f(x1x2)∵x2R由（1）及題設f(x2) 又x1 x1x2由（1）f(x1x2f(x2f(x1x21f(xR

f(x1x2)1f(x1)f(x2Ax,yfx2y2f(1)x,yx2y2Bx,yf(axy2)f(0)x,yaxy2∵A∩B

x2y2 axy2x2ax2)2 (1a2x24ax30316a243(1a2) a23 a a3x1x2Rx1∵u(x)在R上的單調遞減(x1)(x2) 即u1u2又∵yf(u)在R上是增函數f(u1)f(u2) y1y2則x1x2時y1y2f((x))在R上單調遞減6x 0x解：f(x)x 0x

x2k,2k2kx2k0, f(x)f(x2k)x2k 課后練習22x3x22x

x