函数y=f (x) (x≠0)是奇函数,且当x∈(0,+ ∝)时是增函数,若f (1)=0,求不等式f [x (x-1/2)]
问题描述:
函数y=f (x) (x≠0)是奇函数,且当x∈(0,+ ∝)时是增函数,若f (1)=0,求不等式f [x (x-1/2)]
答
函数y=f (x) (x≠0)是奇函数,且当x∈(0,+ ∝)时是增函数,若f (1)=0,求不等式f [x (x-1/2)]