1、求导f'(x)＝x^2＋2ax＋b,f(x)在x=1处取得极值,则f'(1)=0,即1+2a+b=0,所以b=-1-2a.2、f'(x)=x^2+2ax+b=x^2+2ax-1-2a＝(x-1)(x+1+2a),令f'(x)=0得x=1或=-1-2a.当a=-1时,f'(x)＝(x-1)^2≥0,所以f(x)在(-∞,+∞)内单...