已知二次函数y=x^2-2ax+1在区间(2,3)内是单调函数,则实数a的取值范围a≤2或a≥3、 求过程!!!!

2024-12-15 16:40:55
推荐回答(2个)
回答1:

画图,抛物线,,,,开口向上
单调函数
如果是单调递增函数,则对称轴在2的左边
x=a<=2
如果是单调递减函数,则对称轴在3的右边
x=a>=3
所以a≤2或a≥3

如果对称轴在2,3之间,,,则其单调区间为
[2,a] 单调递减
[a,3] 单调递增

回答2:

函数图象是抛物线,开口向上。对称轴是x=a,所以当a≤2时,(2,3)落在对称轴右边,单调增,
当a≥3时,(2,3)落在对称轴左边,单调减,若a在2,3之间,则不具单调性。这画图很容易理解

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