【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

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)