若函数f(x)=asin(x-π/3)+b满足f(π/3)+f(π/2)=7且f(π)-f(0)=2√3求
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若函数f(x)=asin(x-π/3)+b满足f(π/3)+f(π/2)=7且f(π)-f(0)=2√3求
(1)f(x)的解析式(2)f(x)的单调区间(3)f(x)的最小值(4)使f(x)=1/2的x的集合
(1)f(x)的解析式(2)f(x)的单调区间(3)f(x)的最小值(4)使f(x)=1/2的x的集合
1.f(π/3) + f(π/2)=[asin(π/3 - π/3) + b] + [asin(π/2 - π/3) + b] =(asin0 + b) + [asin(π/6) + b]
=b + (1/2)a + b =7
f(π) - f(0) =[asin(π - π/3) + b] - [asin(0 - π/3) + b] =[asin(2π/3) + b] - [asin(-π/3) + b]
=[(√3/2)a + b] - [(-√3/2)a + b] = (√3)a =2√3
由此解得:a=2 ,b=3
∴f(x)=2sin(x - π/3) + 3
2.增区间:2kπ-π/2
=b + (1/2)a + b =7
f(π) - f(0) =[asin(π - π/3) + b] - [asin(0 - π/3) + b] =[asin(2π/3) + b] - [asin(-π/3) + b]
=[(√3/2)a + b] - [(-√3/2)a + b] = (√3)a =2√3
由此解得:a=2 ,b=3
∴f(x)=2sin(x - π/3) + 3
2.增区间:2kπ-π/2
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