已知f(x)=ax^2+1/bx+c（a，b，c属于Z）是奇函数，又f(1)=2, f(2)<3.(1)求a,b,c的值;(2)当x属于(-∞,-1]时,试判断f(x)的单调性并加以证明.
热心网友
f(x)=[ax^2+1]/[bx+c]f(1)=[a+1]/[b+c]=2f(-1)=[a^2+1]/[c-b]=-2a+1=2(b+c)=2(b-c)===c=0a+1=2bf(x)=[ax^2+1]/bxf(2)=[4a+1]/2b=[4a+4-3]/2b=[8b-3]/2b=4-3/2b0,then b0b=1,a=1if b3/2 无解so f(x)=(x^2+1)/x=x+1/xif x10so f(x)在(-∞,-1]是增函数