【y=1/x平方+根号x的值域怎么求?】
<p>问题：【y=1/x平方+根号x的值域怎么求?】<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">程显蒙的回答：<div class="content-b">网友采纳 y=1/(x^2)+x^(1/2) 由于存在根号,所以x大于零 y=1/(x^2)+(1/4)*x^(1/2))+(1/4)*x^(1/2))+(1/4)*x^(1/2))+(1/4)*x^(1/2) >=5*五次根号下(1/(x^2)*(1/4)*x^(1/2))*(1/4)*x^(1/2))*(1/4)*x^(1/2))*(1/4)*x^(1/2)) =5*（2的负五分之八次方） 当1/(x^2)＝(1/4)*x^(1/2)时取等号,即x＝2的负五分之四次方时有最小值5*（2的负五分之八次方）,最大是正无穷大
页:
[1]