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求二次函数在闭区间上的最值的方法很多,但无论哪一种方法都没有图象法直观明了。下面通过三道高考题来说明这个问题。例1 (85年高考题)求函数u=-x~2+4x-2在区间[0,3]上的最大值和最小值。解先作出y=-x~2++4x-2在闭区间[0,3]上的图象(如图一),然后观察图象可得:y=-x~2+4x-2在闭区间[0,3]上的最大值是顶点的纵坐标2,最小值是当x=0时的函数值-2。例2 (89年高考题)如果|x|≤π/4,
There are many ways to find the best value of the quadratic function in the closed interval, but no one method is intuitive and clear without the image method. The following questions are illustrated by three college entrance examination questions. Example 1 (85-year college entrance examination question) Find the maximum and minimum values of the function u=-x~2+4x-2 in the interval [0,3]. The solution is to first make an image of y=-x~2++4x-2 in the closed interval [0,3] (see Figure 1), and then observe the image to get: y=-x~2+4x-2 The maximum value in the closed interval [0,3] is the ordinate 2 of the vertex, and the minimum value is the function value -2 when x=0. Example 2 (89 college entrance examination question) If |x| ≤ π/4,