高中常见分段函数题型归纳

提高兴趣增强自信对接高考分层教学总结规律规范答题.2x2x[1,0];123,f(x).例.x.22|x1|2,(|x1)f(x)1,(|x1)1x2f[f()]12例求.14f[f(f(32)12f()|1|2321().2121232,例.:,.,0.exx1g(x)g(g())2练设练设lnx,x0.则e2(xx1f(x)x((x2f[f(2)]则31提高兴趣增强自信对接高考分层教学总结规律规范答题4x3(x0)f(x)x3(0x1)x5(x1)例.f(x)f(0)3maxf(x)f(1)4maxx0,,当0x1,,当x1,x5154,f(x)4.max例设a求.2.,2若2；当；若.若；当当a；2..yf(x)yg(x)yx,例,和2提高兴趣增强自信对接高考分层教学总结规律规范答题yg(x)yx2,1,,f(x)）2x2(1x0).f(x)2(0x2)2x2(1x2)C.f(x)1(2x4)x22x6(1x2).f(x)3(2x4)x2x[2,0],yx1,2,1个12xyy(x2)11x1,f(x)2x2(x[1,0]),当x[0,1],1212,y2x1yx,2,1,f(x)x2(x[0,2])y2(x2)112x4,,122x2(1x0)f(x)2(0x2)x2,例2月112l2式；?l23提高兴趣增强自信对接高考分层教学总结规律规范答题0设tt2t。2例ye|lnx||x1|）y1xO1Byy11xxO1O1DC例3a.22.例3a.4提高兴趣增强自信对接高考分层教学总结规律规范答题∴.例.在R∴在R.∵∴注.yf(x)x0f(x)31,设f(x)Rx例,,yg(x),求g(x).x0,则x0,f(x)31,f(x),xRf(x)f(x),且f(0)0,f(x)13x,31(0)log(1)(0)xxxxf(x)0(x0)13(x0)3g(x)0(x0)xxlog)(0)x,.31af(),1(x)+()f(x)ff1x9例为则f(a4)((x2xxf(x)x(x(x2例.x0,x0,f(x)(x)(x1)x(x1)f(x),当x0,225提高兴趣增强自信对接高考分层教学总结规律规范答题f(0)f(0)0,当0x0,,f(x)(x)(x1)x(x1)f(x),x22意xRf(x)f(x),f(x).x与.(xxx3f(x)x(x2例.当x；12121212当；12当x12x。1212..f(x).当x0,例f(x)|12x||2x|.x0,(x)3x11,f(x),当f'2f'(x)2x0,f(x),f(x)f(x)R在;R在.f(x)3x(x2)3x1(x2)(,]12,.2x(,1]xf(x)1f(x)logxx)4x例81,的2xxlogx2(,1]x214,则2x22,得14,,若81,则1,x3),x3.x42x(,1]xf(x)1f(x)logxx)4x例81,的2xxlogx2(,1]x214,则2x22,得14,,若81,则6提高兴趣增强自信对接高考分层教学总结规律规范答题1,x3),x3.x41x2x||xx练algx1,x1,f(x)0,x0.()()0fxbfxcx练R有7）b0c0且b0c0且b0c0且b0c0且f(x)(,)f(2x)f(2x)f(7x)f(7x)练在，f(1)f0.yf(x)f(x)0[.21(xxf(x)12(xf(x)1xx例,若0,则0）.(1,1)B.(1,)C.(,2)(0,)D.(,1))和知f(x)1x00,(,1)).12f(x)1x00x10x001,x00,当,2011,,当,x10x,0(,1)).((x4x1(xx2f(x)()1fxx例,）A．(,2][0,10](,2][0,1][2,0][1,10](,2][1,10]7提高兴趣增强自信对接高考分层教学总结规律规范答题x1,f(x)1(x1)1x或x0,01,当x1,2xxf(x)14x11x13x10,1x,,x2或0x,A.((x2xf(x)4x1(x()1fxx）例,A．(,2][0,10](,2][0,1](,2][1,10][2,0][1,10]x1,f(x)1(x1)1x或x0,x01,当x1,2xf(x)14x11x13