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
f(-1)=f(1)=0,f[x(x-1/2)]
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)]
f(-1)=f(1)=0,f[x(x-1/2)]