Question

已知函数y=3sin(2x+π/6)—1的对称中心和单调区间

Answer 1

3-1=2，-3-1=-4
(2-4)/2
=
=-1
把y
=
-1
代入y
=
3sin(2x
+
π/6)
-
1得：
sin(2x
+
π/6)
=
0
2x+
π/6
=
2kπ
2x
=
2kπ
-
π/6
x
=旦罚测核爻姑诧太超咖
kπ
-
π/12
(k为整数)
所以(kπ
-
π/12
,
-1)是对称中心
由y
=
3sin(2x
+
π/6)
-
1知：
y
'
=
6cos(2x
+
π/6)
令y
'
>
0
得：
cos(2x
+
π/6)>0
2kπ
-
π/2
<
2x
+
π/6
<
2kπ
+
π/2
2kπ
-
2π/3
<
2x
<
2kπ
+
π/3
即：
kπ
-
π/3
<
x
<
kπ
+
π/6
所以当kπ
-
π/3
<
x
<
kπ
+
π/6
时
y
=
3sin(2x
+
π/6)
-
1
单调递增。

Answer 2

2kπ+π/2≤2x+π/6≤2kπ+3π/2
∴kπ+π/6≤x≤kπ+2π/3
即：函数y=3sin(2x+π/6)的单调递减区间是[kπ+π/6,kπ+2π/3],k∈z