求使函数取得最大值,最小值的自变量x的集合,并写出最大值最小值。(1)y=3sin(2x+π/4) x∈R (2)y=3/2cos(1/2x-π/6) x∈R (3)y=1/2sin(1/2x+π/3) x∈R 麻烦写下具体过程
(1)y=3sin(2x+π/4) x∈R
最大值3,最小值-3
当2x+(π/4)=2kπ+(π/2)(k∈Z)时取得最大值
===> 2x=2kπ+(π/4)
===> x=kπ+(π/8)(k∈Z)
当2x+(π/4)=2kπ-(π/2)(k∈Z)时取得最小值
===> 2x=2kπ-(3π/4)
===> x=kπ-(3π/8)(k∈Z)
(2)y=3/2cos(1/2x-π/6) x∈R
最大值3/2,最小值-3/2
当(1/2)x-(π/6)=2kπ(k∈Z)时取得最大值
===> (1/2)x=2kπ+(π/6)
===> x=4kπ+(π/3)(k∈Z)
当(1/2)x-(π/6)=2kπ+π(k∈Z)时取得最小值
===> (1/2)x=2kπ+(7π/6)
===> x=4kπ+(7π/3)(k∈Z)
(3)y=1/2sin(1/2x+π/3) x∈R
最大值1/2,最小值-1/2
当(1/2)x+(π/3)=2kπ+(π/2)(k∈Z)时取得最大值
===> (1/2)x=2kπ+(π/6)
===> x=4kπ+(π/3)(k∈Z)
当(1/2)x+(π/3)=2kπ+(3π/2)(k∈Z)时取得最小值
===> (1/2)x=2kπ+(7π/6)
===> x=4kπ+(7π/6)(k∈Z)。
。
1)y=3sin(2x+π/4) x∈R 最大值3,最小值-3 取得最大值时sin(2x+π/4)=1 2x+π/4 =2kπ+π/2 x =kπ+π/8 ,k∈Z 当2x+(π/4)=2kπ-(π/2)(k∈Z)时取得最小值 -3 ===> 2x=2kπ-(3π/4) ===> x=kπ-(3π/8)(k∈Z) 下面的都类似