课件参考教案

函数的单调性与周一、函数的

4讲函数的单调性与我们在初中研究了一次函数、二次函数的图象，研究了yaxb(a0)yax2bxc(a0[1]

yyx22x

的图象，并观察说出在定义域(,g(xx22x1的图象在对称轴左方的区间(,1)是下降的，在对称轴右方的区间)g(x)随自变量x的增大而增大。1：yf(x)(xA，对于区间(ab1x1x2(a,b)，x1x2f(x1f(x2yf(x(ab内是增函数（increasingfunction2x1x2(a,b)，x1x2f(x1f(x2yf(x间(ab内是减函数(decreasingfunction)2、yf(xf(xf(x1f(x是增（减）函数图象自左到右上升（下降2、图象的峰（谷）函数增（减）变减（增）点函数的极大（小）xIf(xM（f(xMx0If(x0 value[3]如图是定义在闭区间[-5，5]f(xf(x1f(xxx[说 1]函数的单调性与定义的区间有关，它是函数的局部性质[说明3]初等函数均可分段单调x1f(xx21(x(1,1的单调性2f(xx22x3f(xx22x3的单调区间3．已知函数fxR上的减函数，则满足f

1x1xA． B． C．1,0∪ D．,1∪（1）f(0)1x0f(x1f(xR设集A(x,y)f(x2fy2f(1)B(xy)f(axy2)1,aR，ABa的取值范围。5yf(u)(uR是增函数，u(x)(xR是减函数，求复合函f((xR的单二、函数的定义：yf(x)(xAxAf(xTf(xT0是常数)f(x为周期函数，T是f(x)的周期说明：（1）周期函数的定义域至少一Tf(xkTf(x的周期f(xxf(xTf(xf(x2Tf(xxf(axf(axa0)f(x是周期为4a的周期函数。f(x课后练习f(x)x22xyx(1y612x参考答问题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 则(xx)(x1x21) f(x)f f(x在(0,1)问题 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在区间1,1f(x在1,1x22xf(x)

x g(x)利用图象求单调区间（ xx2x(,1),(0,1)减(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）及题设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上的单调递减(x1)(x2) 即u1u2又∵yf(u)在R上是增函数f(u1)f(u2) y1y2则x1x2时y1y2f((x))在R上单调递减6x 0x解：f(x)x 0x

x2k,2k2kx2k0, f(x)f(x2k)x2k 课后练习22x3x22x

x