求函数y=3sin(2x+π/4)求单调递减区间.这是高中必修4的题,请按照标准答案答!
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求函数y=3sin(2x+π/4)求单调递减区间.这是高中必修4的题,请按照标准答案答!
求函数y=3sin(2x+π/4)求单调递减区间.这是高中必修4的题,请按照标准答案答!
求函数y=3sin(2x+π/4)求单调递减区间.这是高中必修4的题,请按照标准答案答!
设t=2x+π/4,则y=3sin(2x+π/4)为y=3sin t
由正弦函数的单调性质,可知
正弦函数的单调递增区间为[2kπ-π/2,2kπ+π/2]
所以 y=3sin t的单调递增区间为[2kπ-π/2,2kπ+π/2]
即2kπ-π/2
由正弦函数的单调性质,可知
正弦函数的单调递增区间为[2kπ-π/2,2kπ+π/2]
所以 y=3sin t的单调递增区间为[2kπ-π/2,2kπ+π/2]
即2kπ-π/2<=t<=2kπ+π/2
又因为t=2x+π/4
所以2kπ-π/2<=2x+π/4<=2kπ+π/2 解得 kπ-3π/8<=x<=kπ...
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由正弦函数的单调性质,可知
正弦函数的单调递增区间为[2kπ-π/2,2kπ+π/2]
所以 y=3sin t的单调递增区间为[2kπ-π/2,2kπ+π/2]
即2kπ-π/2<=t<=2kπ+π/2
又因为t=2x+π/4
所以2kπ-π/2<=2x+π/4<=2kπ+π/2 解得 kπ-3π/8<=x<=kπ+π/8 即函数y=3sin(2x+π/4)的单调递增区间为[kπ-3π/8,kπ+π/8]
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太简单了吧 y=3sin(2x+π/4)=3sin[2(x+π/8)],因为函数y=sinx的单调递减区间为[π/2+2kπ,3π/2+2kπ],所以y=3sin(2x+π/4)=3sin[2(x+π/8)]的单调递减区间为2(x+π/8)∈[π/2+2kπ,3π/2+2kπ],即x∈[π/8+kπ,5π/8+kπ],k为(-∞,+∞)之间任意整数。书上有许多这种例子,基本上没什么差别,搞懂了原理...
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太简单了吧 y=3sin(2x+π/4)=3sin[2(x+π/8)],因为函数y=sinx的单调递减区间为[π/2+2kπ,3π/2+2kπ],所以y=3sin(2x+π/4)=3sin[2(x+π/8)]的单调递减区间为2(x+π/8)∈[π/2+2kπ,3π/2+2kπ],即x∈[π/8+kπ,5π/8+kπ],k为(-∞,+∞)之间任意整数。书上有许多这种例子,基本上没什么差别,搞懂了原理 遇到这类题就容易解决了
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T=2π/2=π
初相为π/4单调递减区间为(0,π/2)