怎样解决导数零点不可求问题

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628400、..、12013fx=2x|log0.5x|-1A1B2C3D41fx=-2xlog2x-10<x≤12xlog2x-1x>1{.∵01fx=-2xlog2x-1x0fxf1=-1<0∴fx01∵1+∞fx=2xlog2x-1f2=3>0∴fx1+∞.fxB.2fx=0()12x=|log0.5x|.y=()12xy=|log0.5x|B.1.2..、1..22012fx=ex-ax-2.1fx2a=1kx>0x-kf'x+x+1>0k.1fx-∞+∞f'x=ex-a.a≤0f'x>0fx-∞+∞a>0x∈-∞lnaf'x<0x∈lna+∞f'x>0.fx-∞lnalna+∞.2a=1x-kf'x+x+1=x-kex-1+x+1.x>0x-kf'x+x+1>0k<x+1ex-1+xx>0.①gx=x+1ex-1+xg'x=-xex-1ex-12+1=exex-x-2ex-12.1hx=ex-x-20+∞.h1<0h2>0hx0+∞g'x0+∞bb∈12.x∈0bg'x<0x∈b+∞g'x>0gx0+∞gb.g'b=0eb=b+2gb=b+1eb-1+b=b+1∈23.①k<gbk·22·20152.2g'x=exex-x-2ex-12g'xhx=ex-x-2hx0+∞.2.、.32014fx=cosx-xπ+2x-83sinx+1gx=3x-πcosx-41+sinxln3-2x()π.1x0∈0π()2fx0=02x1∈π2()πgx1=01x0+x1<π.1x∈0π()2f'x=-1+sinxπ+2x-2x-23cosx<0fx0π()2.f0=π-83>0fπ()2=-π2-163<0x0∈0π()2fx0=0.2hx=3x-πcosx1+sinx-4ln3-2π()xx∈π2[]πt=π-xx∈π2[]πt∈0π[]2.ut=hπ-t=3tcost1+sint-4ln1+2π()tu't=3ftπ+2t1+sint.1t∈0x0u't>0t∈x0π()2u't<0。ut0x0.u0=0t∈0x0ut>0ut0x0.utx0π()2ux0>0uπ()2=-4ln2<0t1∈x0π()2ut1=0.x1=π-t1∈π2()πhx1=hπ-t1=ut1=0.x∈π2()π1+sinx>0gx=1+sinxhxhxx1∈π2()πgx1=0.x1=π-t1t1>x0x0+x1<π.2u't=3ftπ+2t1+sintu't=00π[]2.1“x0∈0π()2fx0=0”x00π()20x0x0π()20x[)0x0π()2.“+++”“”、.、42014fx=exx2-k2x+ln()xke=2.71828….1k≤0fx2fx02k.1fx0·32·5+∞f'x=x-2ex-kxx3.k≤0ex-kx>0x∈02f'x<0fxx∈2+∞f'x>0fx.fx022+∞.21k≤0fx02fx02.k>0gx=ex-kxx∈0+∞.g'x=ex-k=ex-elnk0<k≤1x∈02g'x=ex-k>0gxfx02k>1x∈0lnkg'x<0gx.x∈lnk+∞g'x>0gxgxglnk=k1-lnk.fx02g0>0glnk<0g2>00<lnk<2{e<k<e22.fx02kee2()2.42gx=ex-kxg″x=ex>0gx..42.、.52012fx=lnx+kexke=2.71828…y=fx1f1x.1k2fx3gx=xf'xf'xfxx>0gx<1+e-2.1f'x=1x-lnx-kex.f'1=1-ke=0∴k=1.2f'x=1x-lnx-1ex.kx=1x-lnx-1k'x=-1x2-1x<0kx0+∞.k1=00<x<1kx>0f'x>0x>1kx<0f'x<0.fx011+∞.32x≥1gx=xf'x≤0<1+e-2gx<1+e-20<x<1.0<x<1ex>1gx>0∴gx=1-xlnx-xex<1-xlnx-x.Fx=1-xlnx-xx∈01F'x=-lnx+2.x∈0e-2F'x>0x∈e-21F'x<0x=e-2FxFe-2=1+e-2.gx<Fx≤1+e-2.x>0gx<1+e-2.30<x<1gx=1-xlnx-xex.ex>1gxFx=1-xlnx-x.·42·2015

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