Question

10.函数+f(x)=3sin(-2x+π/4),+求此函数的最小正周期、单调递增区问？

Answer 1

∵ f(x)=3sin(-2x+π/4)=3sin(-2x-2π+π/4)
=3sin【-2（x+π）+π/4】=f[（x+π)]
-2x+π/4∈[2kπ-π/2,2kπ+π/2]
-2x∈[2kπ-3π/4,2kπ+π/4]
x∈[kπ-π/8,kπ+3π/8]
∴ f(x)=3sin(-2x+π/4)的最小正周期π，单调递增区间：[kπ-π/8,kπ+3π/8]，k为整数 。

Answer 2

94位粉丝
f(x)=3sin(2x+π/4)
单增:
2x+π/4∈[2kπ-π/2,2kπ+π/2]
x∈[kπ-3π/8,kπ+π/8]
k∈z
所以单调增区间为
[kπ-3π/8,kπ+π/8]
k∈z
单调减区间为
单调增区间向右平移π/2个单位
为
[kπ+π/8,kπ+5π/8]
k∈z
最小正周期为
2π/2=π