一道高一数学已知函数f(x)=Asin(ωx+φ),x属于R(其中A大于0,ω大于0,0小于φ小于二分之π)的周期为π,且图像上一个最低点为M(3分之2π,-2)(1)求f(x)的解析式(2)当x属于[0,12分之π],求f(x)的最值(3)求f(x)的单调区间写下解答过程,谢谢
2019-05-30
一道高一数学
已知函数f(x)=Asin(ωx+φ),x属于R(其中A大于0,ω大于0,0小于φ小于二分之π)的周期为π,且图像上一个最低点为M(3分之2π,-2)
(1)求f(x)的解析式
(2)当x属于[0,12分之π],求f(x)的最值
(3)求f(x)的单调区间
写下解答过程,谢谢
优质解答
(1)由图像上一个最低点为M(2π/3,-2)可得A=2,
T=2π/ω=π,ω=2,
f(2π/3)=2sin(2x+φ)=-2,
2*2π/3+φ=3π/2,φ=π/6,
f(x)=2sin(2x+π/6).
(2)、x∈[0,π/12],函数单调递增,
最小值是f(0)=2sin(π/6)=1.
最大值=f(π/12)=2sin(π/3)=√3.
(3).对于f(x)=sinx单调递增区间,2kπ-π/2≤x≤2kπ+π/2,
2kπ-π/2≤2x+π/6≤2kπ+π/2,
kπ-π/3≤x≤kπ+π/6,
单调递减区间,2kπ+π/2≤x≤2kπ+3π/2,
2kπ+π/2≤2x+π/6≤2kπ+3π/2,
kπ+π/6≤x≤kπ+2π/3.
(1)由图像上一个最低点为M(2π/3,-2)可得A=2,
T=2π/ω=π,ω=2,
f(2π/3)=2sin(2x+φ)=-2,
2*2π/3+φ=3π/2,φ=π/6,
f(x)=2sin(2x+π/6).
(2)、x∈[0,π/12],函数单调递增,
最小值是f(0)=2sin(π/6)=1.
最大值=f(π/12)=2sin(π/3)=√3.
(3).对于f(x)=sinx单调递增区间,2kπ-π/2≤x≤2kπ+π/2,
2kπ-π/2≤2x+π/6≤2kπ+π/2,
kπ-π/3≤x≤kπ+π/6,
单调递减区间,2kπ+π/2≤x≤2kπ+3π/2,
2kπ+π/2≤2x+π/6≤2kπ+3π/2,
kπ+π/6≤x≤kπ+2π/3.