函数f(x)=x²-2(a-1)x+2在区间(-无穷,4]上递减,则a的取值范围是------- A.[-3,+无穷)B.(-无穷,3]C.(-无穷,-3)D.[3,+无穷)
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![函数f(x)=x²-2(a-1)x+2在区间(-无穷,4]上递减,则a的取值范围是------- A.[-3,+无穷)B.(-无穷,3]C.(-无穷,-3)D.[3,+无穷)](/uploads/image/z/8506447-7-7.jpg?t=%E5%87%BD%E6%95%B0f%28x%29%3Dx%26%23178%3B-2%28a-1%29x%2B2%E5%9C%A8%E5%8C%BA%E9%97%B4%EF%BC%88-%E6%97%A0%E7%A9%B7%2C4%5D%E4%B8%8A%E9%80%92%E5%87%8F%2C%E5%88%99a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E6%98%AF-------+A.%5B-3%EF%BC%8C%2B%E6%97%A0%E7%A9%B7%EF%BC%89B.%EF%BC%88-%E6%97%A0%E7%A9%B7%EF%BC%8C3%5DC.%EF%BC%88-%E6%97%A0%E7%A9%B7%EF%BC%8C-3%EF%BC%89D.%5B3%EF%BC%8C%2B%E6%97%A0%E7%A9%B7%EF%BC%89)
函数f(x)=x²-2(a-1)x+2在区间(-无穷,4]上递减,则a的取值范围是------- A.[-3,+无穷)B.(-无穷,3]C.(-无穷,-3)D.[3,+无穷)
函数f(x)=x²-2(a-1)x+2在区间(-无穷,4]上递减,则a的取值范围是-------
A.[-3,+无穷)
B.(-无穷,3]
C.(-无穷,-3)
D.[3,+无穷)
函数f(x)=x²-2(a-1)x+2在区间(-无穷,4]上递减,则a的取值范围是------- A.[-3,+无穷)B.(-无穷,3]C.(-无穷,-3)D.[3,+无穷)
最简单,最好理解的方法:都知道二次函数对称轴两边一遍递增,一遍递减,这个函数开口向上,则对称轴以左递减,那么要想在区间(-无穷,4]上递减,对称轴x=a-1大于或等于4,也就是说给定范围在对称轴左边就行了.那么得a大于等于5 ,根据你的题目条件这是没问题的,或许是你的题目错了吧,很正常的.
抛物线开口向上,对称轴为x=a-1,递减区间为(-无穷,a-1),确保(-无穷,4]上递减,对称轴在4的右边或者等于4,所以a-1≥4即a≥5
导函数为f(x)=2x-2*(a-1)
是一个一次递增函数
只需令2*4-2*(a-1)《0
解得a》5
你答案错了