y=1/x平方+根号x 的值域怎么求?

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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

求导,取极小值点的值为最小值。或者用不等式定理1/x*x+4*(1/4根x)>=5*(1/4的5/4次方)