设函数f(x)=sinxcosx+cosx^2,求f(x)的最小正周期,当x属于【0,π/2】时,求函数f(x)的最大值和最小值答案是f(x)=sinxcosx+cosx^2=½(sin2x+cos2x+1)=½[√2sin(2x+π/4)+1],即最小正周期为π.当x属于[0,π/2]时,2x∈[
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![设函数f(x)=sinxcosx+cosx^2,求f(x)的最小正周期,当x属于【0,π/2】时,求函数f(x)的最大值和最小值答案是f(x)=sinxcosx+cosx^2=½(sin2x+cos2x+1)=½[√2sin(2x+π/4)+1],即最小正周期为π.当x属于[0,π/2]时,2x∈[](/uploads/image/z/12884861-29-1.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dsinxcosx%2Bcosx%5E2%2C%E6%B1%82f%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%2C%E5%BD%93x%E5%B1%9E%E4%BA%8E%E3%80%900%2C%CF%80%2F2%E3%80%91%E6%97%B6%2C%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BC%E7%AD%94%E6%A1%88%E6%98%AFf%28x%29%3Dsinxcosx%2Bcosx%5E2%3D%26%23189%3B%28sin2x%2Bcos2x%2B1%29%3D%26%23189%3B%5B%E2%88%9A2sin%282x%2B%CF%80%2F4%29%2B1%5D%2C%E5%8D%B3%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E4%B8%BA%CF%80.%E5%BD%93x%E5%B1%9E%E4%BA%8E%5B0%2C%CF%80%2F2%5D%E6%97%B6%2C2x%E2%88%88%5B)
设函数f(x)=sinxcosx+cosx^2,求f(x)的最小正周期,当x属于【0,π/2】时,求函数f(x)的最大值和最小值答案是f(x)=sinxcosx+cosx^2=½(sin2x+cos2x+1)=½[√2sin(2x+π/4)+1],即最小正周期为π.当x属于[0,π/2]时,2x∈[
设函数f(x)=sinxcosx+cosx^2,求f(x)的最小正周期,当x属于【0,π/2】时,求函数f(x)的最大值和最小值
答案是f(x)=sinxcosx+cosx^2=½(sin2x+cos2x+1)=½[√2sin(2x+π/4)+1],即最小正周期为π.
当x属于[0,π/2]时,2x∈[0,π],2x+π/4∈[π/4,5π/4],sin(2x+π/4)∈[-√2/2,1],f(x)∈[0,(√2+1)/2],故f(x)最大值为(√2+1)/2,最小值为0.
我想知道答案里的sinπ/4为什么会等于-√2/2?
设函数f(x)=sinxcosx+cosx^2,求f(x)的最小正周期,当x属于【0,π/2】时,求函数f(x)的最大值和最小值答案是f(x)=sinxcosx+cosx^2=½(sin2x+cos2x+1)=½[√2sin(2x+π/4)+1],即最小正周期为π.当x属于[0,π/2]时,2x∈[
sin(2x+π/4)∈[-√2/2,1]这一步对吗?
sin π/2=1为最大
sin 5π/4=-√2/2为最小
两者构成该[-√2/2,1]
算 f(x)的区间 时并不一定要取x区间的端点
令a=2x+π/4∈[π/4,5π/4],
sina∈[-√2/2,1],
在a=5π/4时取到-√2/2
并非sinπ/4=-√2/2