Question

判断并证明函数f(x)=ax²+1/x(1<a<3)在【1,2】上的单调性

已知函数f(x)=ax/x^2-1(a>0)
判断函数f(x)在（-1,1）上的单调性并用定义证明

Answer 1

x1,x2∈(-1,1) x1>x2
f(x1)-f(x2)
=ax1/(x1^2-1)-ax2/(x2^2-1)
=a[x1(x2^2-1)-x2(x1^2-1)]/[(x1^2-1)(x2^2-1)]
=a[x1x2(x2-x1)-(x1-x2)]/[(x1^2-1)(x2^2-1)]
=a(x2-x1)(x1x2+1)/[(x1^2-1)(x2^2-1)]
因为a>0 x1>x2 x2-x1<0
x1*x2>-1 x1x2+1>0
(x1^2-1)(x2^2-1)>0
所以
f(x1)-f(x2)<0
函数单调递减