Question

已知函数fx=lnx-x分之1...判断fx的单调性,用定义法证明

Answer 1

f(x)=lnx-1/x的定义域为x>0
f(x)在定义域内是增函数.
设0<x1<x2
f(x1)-f(x2)=lnx1-1/x1-(lnx2-1/x2)
=lnx1-lnx2+1/x2-1/x1
=ln(x1/x2)+(x1-x2)/(x1x2)
∵0<x1<x2
∴x1/x2<1
ln(x1/x2)<0
x1-x2<0
x1x2>0
∴(x1-x2)/(x1x2)<0
ln(x1/x2)+(x1-x2)/(x1x2)<0
即：f(x1)-f(x2)<0
f(x1)<f(x2)
函数f(x)为增函数.</f(x2)
</x1<x2
</x1<x2