已知凼数f(x)=x2 1bx c是奇凼数,且f(1)=2,(1)求f(x)的解...
解:(1)∵f(x)是奇函数,∴f(-x)=-f(x),即x2 1-bx c=-x2 1bx c,即-bx c=-bx-c,则c=-c,解得c=0,∵f(1)=2,∴f(1)=1 1b=2b=2,解得b=1,故f(x)=x2 1x.(2)∵f(x)=x2 1x=x 1x,函数f(x)在(0,1)上单调递减,证明:设0<x1<x2<1,则f(x1)-f(x2)=x1 1x1-x2-1x2=(x1-x2) x2-x1x1⋅x2=(x1-x2)⋅x1x2-1x1⋅x2,∵0<x1<x2<1,∴x1-x2<0,0<x1x2<1,x1x2-1<0,∴f(x1)-f(x2)=(x1-x2)⋅x1x2-...全部
解:(1)∵f(x)是奇函数,∴f(-x)=-f(x),即x2 1-bx c=-x2 1bx c,即-bx c=-bx-c,则c=-c,解得c=0,∵f(1)=2,∴f(1)=1 1b=2b=2,解得b=1,故f(x)=x2 1x.(2)∵f(x)=x2 1x=x 1x,函数f(x)在(0,1)上单调递减,证明:设0<x1<x2<1,则f(x1)-f(x2)=x1 1x1-x2-1x2=(x1-x2) x2-x1x1⋅x2=(x1-x2)⋅x1x2-1x1⋅x2,∵0<x1<x2<1,∴x1-x2<0,0<x1x2<1,x1x2-1<0,∴f(x1)-f(x2)=(x1-x2)⋅x1x2-1x1x2>0,即f(x1)>f(x2),∴f(x)在(0,1)上的单调递减.。
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