由f (x)＝ax2＋bx得 f (-1)＝a-b,f (1)＝a＋b,f (-2)＝4a-2b∴a＝ [f (1)＋f(-1)],b＝ [f (1)-f(-1)] 则f(-2)＝2[f (1)＋f (-1)]-[f (1)-f (-1)]＝3f (-1)＋f (1)由条件1≤f(-1)≤2,2≤f (1)≤4可得3×1＋2≤3f(-1)...