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已知函数f(x)=根号2sin(2x+6π).x∈[π/12,π/3].求f(x)的值域.
人气:390 ℃ 时间:2019-10-19 02:27:03
解答

f(x)=(√2)sin(2x+6π)=(√2)sin(2x)
设y=2x,则:f(x)=(√2)siny
因为x∈[π/12,π/3],所以:y∈[π/6,2π/3]
显然,当y=π/2时,siny有最大值,即f(x)有最大值:f(x)|max=(√2)sin(π/2)=√2
当y=π/6时,siny有最小值,即f(x)有最小值:f(x)|min=(√2)sin(π/6)=√2/6
所以,f(x)的值域是:f(x)∈[√2/6,√2]
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