文 [1]、[2 ]谈到了下面一道不等式题的“分析”与“简捷证明” ,但对题目本质结构的揭示还不够深刻 .我们想从新的角度作出新的分析 ,并立即加以推广 .题目　已知二次函数 ｆ(ｘ) =ａｘ2 +ｂｘ+ｃ ,当 -1≤ｘ≤ 1时 ,有 -1≤ｆ(ｘ) ≤ 1.求证 :当 -2≤ｘ≤ 2时 ,有 -7≤ ｆ(ｘ Articles [1], [2] talk about the following analysis of “inequalities” and “concise proof”, but the disclosure of the essential structure of the topic is not yet profound enough. We want to make new analysis from a new perspective and immediately promote it. The problem is known as the quadratic function f(x) = ax2 + bx + c. When -1 ≤ x ≤ 1, there is -1 ≤ f(x) ≤ 1. Verify that when -2 ≤ x ≤ 2, there is - 7≤ f(x