Question

①求值域的题若x>1,求y=x+[4/(x-1)]的值域请麻烦给出过程来,

Answer 1

因x>1,故x-1>0.由均值不等式可知,y=x+[4/(x-1)].===>y-1=(x-1)+[4/(x-1)]≥4,等号仅当x=3时取得.故y-1≥4.===>y≥5.故函数值域为[5,+∞).