求函数y=3sin(2x+π/4),x∈[0,π]的单调递减区间我的解法:令z=2x+π/4,则y=3sin z其单调递减区间为[π/2+2kπ,3π/2+2kπ]∴π/2+2kπ≤2x+π/4≤3π/2+2kπ∴-3π/8+kπ≤x≤5π/8+kπ设A=[0,π],B=[-3π/8+kπ,5π/8+kπ]∵A∩

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求函数y=3sin(2x+π/4),x∈[0,π]的单调递减区间我的解法:令z=2x+π/4,则y=3sin z其单调递减区间为[π/2+2kπ,3π/2+2kπ]∴π/2+2kπ≤2x+π/4≤3π/2+2kπ∴-3π/8+kπ≤x≤5π/8+kπ设A=[0,π],B=[-3π/8+kπ,5π/8+kπ]∵A∩
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求函数y=3sin(2x+π/4),x∈[0,π]的单调递减区间我的解法:令z=2x+π/4,则y=3sin z其单调递减区间为[π/2+2kπ,3π/2+2kπ]∴π/2+2kπ≤2x+π/4≤3π/2+2kπ∴-3π/8+kπ≤x≤5π/8+kπ设A=[0,π],B=[-3π/8+kπ,5π/8+kπ]∵A∩
求函数y=3sin(2x+π/4),x∈[0,π]的单调递减区间
我的解法:
令z=2x+π/4,则y=3sin z
其单调递减区间为
[π/2+2kπ,3π/2+2kπ]
∴π/2+2kπ≤2x+π/4≤3π/2+2kπ
∴-3π/8+kπ≤x≤5π/8+kπ
设A=[0,π],B=[-3π/8+kπ,5π/8+kπ]
∵A∩B=[0,5π/8]
∴其单调递减区间为[0,5π/8]
但是我用计算器算了一下,0开始还是单调递增的,中间过了某个数才开始递减,这是怎么回事?

求函数y=3sin(2x+π/4),x∈[0,π]的单调递减区间我的解法:令z=2x+π/4,则y=3sin z其单调递减区间为[π/2+2kπ,3π/2+2kπ]∴π/2+2kπ≤2x+π/4≤3π/2+2kπ∴-3π/8+kπ≤x≤5π/8+kπ设A=[0,π],B=[-3π/8+kπ,5π/8+kπ]∵A∩
x∈[π/8+kπ,5π/8+kπ],k∈Z
递减区间为 [π/8,5π/8].