（1）x>0有,f(x)=x(x-1)=(x-1/2)-1/4 显然,f(x)在(0,1/2]单调减(1/2,∞)单调增 （2）x≤0有,f(x)=-x(x-1)=-(x-1/2) 1/4 显然,f(x)在(-∞,-1/2]单调增(-1/2,0]单调减 综上,f(x)在(-1/2,0]∪(0,1/2]单调减,(-∞,-1/2]...