已知w>0函数f(x)=sin(wx+π/4)在(π/2,π)上单调递减,则w取值范围是
答案是当x∈(π/2,π)时,wx+π/4∈(πw/2+π/4,πw+π/4)而函数y=sinx的单调递减区间为[π/2,3π/2]那么πw/2+π/4≥π/2,πw+π...
答案是当x∈(π/2,π)时,wx+π/4∈(πw/2+π/4,πw+π/4) 而函数y=sinx的单调递减区间为[π/2,3π/2] 那么πw/2+π/4≥π/2,πw+π/4≤3π/2 所以1/2≤w≤5/4,即w的取值范围是[1/2,5/4] 但我想问这步 πw/2+π/4≥π/2,πw+π/4≤3π/2 为何不用加周期2kπ呢?
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∵函数f(x)=sin(wx+π/4)在(π/2,π)上单调递减
(1)求f(x)单调减区间:
2kπ+π/2<=wx+π/4<=2kπ+3π/2==>2kπ/w+π/(4w)<=x<=2kπ/w+5π/(4w)
f(x)在(π/2,π)上单调递减,并不等于f(x)单调减区间是(π/2,π)
就是说f(x)单调减区间一定是大于等于(π/2,π)
∴π/(4w)<=π/2==>w>=1/2
5π/(4w)>=π==>w<=5/4
所以1/2≤w≤5/4,即w的取值范围是[1/2,5/4]
问这步
πw/2+π/4≥π/2,πw+π/4≤3π/2
为何不用加周期2kπ
实际上应该加:
2kπ+π/(4w)<=2kπ+π/2==>w>=1/2
2kπ+5π/(4w)>=2kπ+π==>w<=5/4
在计算过程中消去了,所以写的时候可以不写
(1)求f(x)单调减区间:
2kπ+π/2<=wx+π/4<=2kπ+3π/2==>2kπ/w+π/(4w)<=x<=2kπ/w+5π/(4w)
f(x)在(π/2,π)上单调递减,并不等于f(x)单调减区间是(π/2,π)
就是说f(x)单调减区间一定是大于等于(π/2,π)
∴π/(4w)<=π/2==>w>=1/2
5π/(4w)>=π==>w<=5/4
所以1/2≤w≤5/4,即w的取值范围是[1/2,5/4]
问这步
πw/2+π/4≥π/2,πw+π/4≤3π/2
为何不用加周期2kπ
实际上应该加:
2kπ+π/(4w)<=2kπ+π/2==>w>=1/2
2kπ+5π/(4w)>=2kπ+π==>w<=5/4
在计算过程中消去了,所以写的时候可以不写
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