已知a>0,函数f(x)=-2asin(2x+ π 6 )+2a+b,当x∈[0,π 2 ]时,-5≤f(x)≤1(1)求常数a,b的值;(2)设g(x)=f(x+π/2 )且lgg(x)>0,求g(x)的单调区间就是第二问开头答案是f(x)=f(x+π2)=-4
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![已知a>0,函数f(x)=-2asin(2x+ π 6 )+2a+b,当x∈[0,π 2 ]时,-5≤f(x)≤1(1)求常数a,b的值;(2)设g(x)=f(x+π/2 )且lgg(x)>0,求g(x)的单调区间就是第二问开头答案是f(x)=f(x+π2)=-4](/uploads/image/z/3938103-63-3.jpg?t=%E5%B7%B2%E7%9F%A5a%EF%BC%9E0%2C%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3D-2asin%EF%BC%882x%2B+%CF%80+6+%EF%BC%89%2B2a%2Bb%2C%E5%BD%93x%E2%88%88%5B0%2C%CF%80+2+%5D%E6%97%B6%2C-5%E2%89%A4f%EF%BC%88x%EF%BC%89%E2%89%A41%EF%BC%881%EF%BC%89%E6%B1%82%E5%B8%B8%E6%95%B0a%2Cb%E7%9A%84%E5%80%BC%EF%BC%9B%EF%BC%882%EF%BC%89%E8%AE%BEg%EF%BC%88x%EF%BC%89%3Df%EF%BC%88x%2B%CF%80%2F2+%EF%BC%89%E4%B8%94lgg%EF%BC%88x%EF%BC%89%EF%BC%9E0%2C%E6%B1%82g%EF%BC%88x%EF%BC%89%E7%9A%84%E5%8D%95%E8%B0%83%E5%8C%BA%E9%97%B4%E5%B0%B1%E6%98%AF%E7%AC%AC%E4%BA%8C%E9%97%AE%E5%BC%80%E5%A4%B4%E7%AD%94%E6%A1%88%E6%98%AFf%EF%BC%88x%EF%BC%89%3Df%EF%BC%88x%2B%CF%802%EF%BC%89%3D-4)
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已知a>0,函数f(x)=-2asin(2x+ π 6 )+2a+b,当x∈[0,π 2 ]时,-5≤f(x)≤1(1)求常数a,b的值;(2)设g(x)=f(x+π/2 )且lgg(x)>0,求g(x)的单调区间就是第二问开头答案是f(x)=f(x+π2)=-4
已知a>0,函数f(x)=-2asin(2x+ π 6 )+2a+b,当x∈[0,π 2 ]时,-5≤f(x)≤1
(1)求常数a,b的值;
(2)设g(x)=f(x+π/2 )且lgg(x)>0,求g(x)的单调区间
就是第二问开头答案是f(x)=f(x+π2)=-4sin(2x+π/6)-1 f(x+π/2)怎么会连等后面的式子呢 网上的答案都是一样的 就是第二问开头 他两连等我不明白 题中也没给啊
已知a>0,函数f(x)=-2asin(2x+ π 6 )+2a+b,当x∈[0,π 2 ]时,-5≤f(x)≤1(1)求常数a,b的值;(2)设g(x)=f(x+π/2 )且lgg(x)>0,求g(x)的单调区间就是第二问开头答案是f(x)=f(x+π2)=-4
sin肯定属于【-1/2,1】,a大于零,那么f(x)min=-2a+2a+b=-5,f(x)max=a+2a+b=1,第二题.抱歉,我最近在学导数,已经混乱了
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