Question

讨论函数f(x)=ax+1/x+2(a≠1/2)在(x>-2)上的单调性

Answer 1

令,x2>x1,则有X2-X1>0,X1*X2>0,

f(x2)-f(x1)=(ax2+1)/(x2+2)-(ax1+1)/(x1+2)
=[2a(x2-x1)+(x1-x2)]/[x1*x2+2(x1+x2)+4]
=[(x2-x1)(2a-1)]/[x1*x2+2(x1+x2)+4].

因为:X2-X1>0,X1*X2>0,(X>-2)则有
[X1*X2+2(X1+X2)+4]>0,(a≠1/2)
讨论:
1)当(2a-1)>0时,a>1/2,有,f(x2)-f(x1)>0,
f(x2)>f(x1),x2>x1,
则,f(x)在X>-2上是单调递增函数.
2)当(2a-1)<0时,a<1/2,有,f(x2)-f(x1)<0,
f(x2)<f(x1),x2>x1,
则,f(x)在X>-2上是单调递减函数.