Question

【高一数学】已知f(x)在R上满足f(-x)+f(x)=0,且在[0,+∞]上为增函数,若f(1/2)=1,则-1

Answer 1

f(x)+f(-x)=0 -> f(x)=-f(-x) -> f(x)为R上奇函数,f(0)=0

因为f(x)在(0,+oo)单调递增

所以f(x)在R上递增

f(-1/2)=-f(1/2)=-1

-1=f(-1/2)