Question

判断函数y=fx=x^2-x/1在区间(1,+无穷)的单调性，并用定义证明你的结论

Answer 1

判断函数y=f(x)=(x^2-1)/x在区间(0,正无穷大)上的单调性,并用定义证明你的结论
设x2>x1>0,
那么
f(x2)-f(x1)=
(x2^2-1)/x2
-(x1^2-1)/x1
=[x1(x2^2-1)-x2(x1^2-1)]/x1x2
=[x1x2^2-x1-x2x1^2+x2]
/x1x2
=[
x1x2(x2-x1)+(x2-x1)]
/x1x2
x2-x1>0
所以
f(x2)-f(x1)>0
即f(x)在（0，
+00）为单调递增

Answer 2

设x1>x2>1
f(x1)-f(x2)=(x1)^2-(1/x1)-(x2)^2+(1/x2)=(x1+x2)(x1-x2)-(x2-x1)/x1x2
=(x1-x2)(x1+x2+1/x1x2)>0
在该区间单调递增