作业帮已知a> 0,函数f(x)=-2asin(2x+ π/6)+2a+b,当x∈[0,π/2]时,-5≤ f(x)≤ 1.求f(x)的单调区间?
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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.求f(x)的单调区间?](/uploads/image/z/2691665-17-5.jpg?t=%E5%B7%B2%E7%9F%A5a%3E+0%2C%E5%87%BD%E6%95%B0f%28x%29%3D-2asin%282x%2B+%CF%80%2F6%29%2B2a%2Bb%2C%E5%BD%93x%E2%88%88%5B0%2C%CF%80%2F2%5D%E6%97%B6%2C-5%E2%89%A4+f%28x%29%E2%89%A4+1.%E6%B1%82f%28x%29%E7%9A%84%E5%8D%95%E8%B0%83%E5%8C%BA%E9%97%B4%3F)
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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.求f(x)的单调区间?
已知a> 0,函数f(x)=-2asin(2x+ π/6)+2a+b,当x∈[0,π/2]时,-5≤ f(x)≤ 1.求f(x)的单调区间?
已知a> 0,函数f(x)=-2asin(2x+ π/6)+2a+b,当x∈[0,π/2]时,-5≤ f(x)≤ 1.求f(x)的单调区间?
sin(2x+ π/6)的单调区间是在x∈[0, π/6]为增,在x∈[π/6, π/2]为减,
因为系数-2a小于零,所以f(x)在x∈[0, π/6]为减,在x∈[π/6, π/2]为增
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