函数f(x)=x+a/x(a>0) (1)当a=4时,判断f(x)在区间(0,2),(2,+∞)的单调性,并求当a∈[1/2,4]时，f(x)的值域； (2)求f(x)的单调区间解 (1)设0<x1<x2<2,则x1-x2<0,x1x2-4<0,于是f(x1)-f(x2)=x1+4/x1-x2-4/x2=(x1-x2)+4(x2-x1)/(x1x2)=(x1-x2)(x1x2-4)/(x1x2)>0,所以,f(x)在区间(0,2)内单调递减. 设2<x1<x2<+∞,则x1-x2<0,x1x2-4>0,于是f(x1)-f(x2)=x1+4/x1-x2-4/x2=(x1-x2)+4(x2-x1)/(x1x2)=(x1-x2)(x1x2-4)/(x1x2)<0,所以,f(x)在区间(2,+∞)内单调递增.因为f(x)在区间[1/2,2]内单调递减,所以f(x)<=f(1/2)=1/2+4/(1/2)=17/2,f(x)>=f(2)=2+4/2=4;又因为f(x)在区间[2,4]内单调递增,所以 f(x)>=f(2)=2+4/2=4,f(x)<=f(4)=4+4/4=5;因此,当a∈[1/2,4]时，f(x)的值域为[4,17/2].(2)因为f(x)是奇函数,f(x)在区间(0,2)内单调递减,在区间(2,+∞)内单调递增,所以f(x)在区间(-∞,-2)内单调递增,在区间(-2,0,)内单调递减.因此, f(x)的单调递增区间为:(-∞,-2]、[2,+∞)，单调递减区间为:(0,-2]、(0,2].