来自舒振宇的问题
【y=1/x平方+根号x的值域怎么求?】
y=1/x平方+根号x的值域怎么求?
1回答
2020-12-26 22:12
【y=1/x平方+根号x的值域怎么求?】
y=1/x平方+根号x的值域怎么求?
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