已知函数y=loga(3-ax),(a>0,a≠1)在[0,1]上单调递减,则实数a的取值范围为_.
已知函数y=loga(3-ax),(a>0,a≠1)在[0,1]上单调递减,则实数a的取值范围为______.
数学人气:160 ℃时间:2019-10-23 07:27:43
优质解答
设t=g(x)=3-ax,则∵a>0,a≠1,∴t=3-ax在定义域上单调递减,要使函数y=loga(3-ax),(a>0,a≠1)在[0,1]上单调递减,则有y=logat在定义域上为单调递增,则须有a>1g(1)>0,即a>1g(1)=3−a>0,解得1<a...
我来回答
类似推荐
猜你喜欢
- 1如何验证 物体无论悬浮或漂浮 浮力=重力
- 223和69的最小公倍数和最大公因数各是几?
- 3We _________ should help old people.
- 4已知H3PO4的PKa1=2.16 PKa2=7.21 PKa3=12.32.0.01mol.Na2HPO4溶液的pH约为多少
- 5The factories p_____ waste water into the river.
- 6I love sitcoms very much. (提问) ________ do you _______ sitcoms?
- 7英语的词语类型分类
- 8I'm sorry I( )(not) know you( )(are)here.
- 9there are so many different t-----of bread that i do not know which to buy.
- 10lucy is going to go shopping ,很多写going to shoping