> 数学 >
幂函数y=f(x)图象过点(2,√2),则f(9)=?(答案为3) 已知x^(2/3)>x^(3/5),则x的取值范围 (答案x1)
人气:347 ℃ 时间:2020-06-18 18:55:54
解答
(1)∵幂函数y=f(x)图象过点(2,√2),∴设y=f(x)=x^a,∴√2=2^a∴a=1/2∴y=f(x)=x^1/2,∴f(9)=9^1/2=3(2)∵x^(2/3)>x^(3/5)∴(x^(2/3))^15>(x^(3/5))^15∴x^10>x^9∴x^9(x-1)>0∴x∈(-∞,0)∪(1,+∞)...
推荐
猜你喜欢
© 2026 79432.Com All Rights Reserved.
电脑版|手机版