>
数学
>
f﹙x﹚=2sin﹙2x-π/3﹚+1的值域是什么?
人气:213 ℃ 时间:2020-06-05 08:33:10
解答
最小值 = 2(- 1) + 1 = - 1
最大值 = 2(1) + 1 = 3
值域为[- 1,3]谢谢,不过我题目掉了一个条件,x∈﹙0,π/4﹚.对不起啊。1 + 2sin(2x - π/3)在0≤x≤π/4上单调递增最小值为f(0) = 2sin(- π/3) + 1 = - √3 + 1最大值为f(π/4) = 2sin(2 * π/4 - π/3) + 1 = 2值域为[1 - √3,2]
推荐
求函数f(x)=2sin(2x-π/3)的值域(6/π
f(x)=2sin(2x+π/6)+1,求函数f(x)在【-π/6,π/3】上的值域
f(x)=2-cos(2x-π/3)-2sin²x,x∈{0,π/2},求f(x)值域
y=-2sin(2x-π/3) x∈[0,5/3]的值域为多少
若f(x)=2sin(1\2x-π\3)其中x属于0-2π,则f(x)的值域
换一种英语句型表达,意思不变.
She has a big head.改为否定句
6x-7=2(x-y) x+y/2-1=0解二元一次方程
猜你喜欢
I'm sorry,but what you said is of the least importance to us
在画地理剖面图时,什么是垂直比例尺
写一篇初一英语作文:《My vacation》
黄铜 酸洗 着色
方程x³+2x²y=2009的整数解为
明天的天气会不会有雨?
Jim is one of the boys ___ from England A who is B who are C that coems
(a+1)(a-1)(a^2+1)(a^4+1)(a^8+1)(a^16+1)
© 2026 79432.Com All Rights Reserved.
电脑版
|
手机版