Question

数学问题！十万火急！！！！！！！！

函数g(x)与f(x)=√2 sin(π/8x+π/4) 关于x=8对称,求y=f(x)+g(x)的单调增区间

请写出详细步骤，谢谢!!!!!!!!!

Answer 1

解：因为：函数g(x)与f(x)=√2 sin(π/8x+π/4) 关于x=8对称
所以：g(x)=f(8-x)=√2 sin(π/8(8-x)+π/4) =√2 sin(-π/8x+5π/4) =-√2 sin(π/8x-π/4)
y=f(x)+g(x)=√2 sin(π/8x+π/4) -√2 sin(π/8x-π/4)
=sinπ/8x+cosπ/8x-sinπ/8x+cosπ/8x
=2cosπ/8x
所以递增区间为：2kπ-π≤π/8x≤2kπ
即：16k-8≤x≤16k
递增区间是[16k-8,16k] ，其中 k∈Z

Answer 2

函数g(x)与f(x)=√2 sin(π/8x+π/4) 关于x=8对称
g(x)=f(8-x)=√2 sin(π/8(8-x)+π/4) =√2 sin(-π/8x+5π/4) =√2 sin(π/8x-π/4)
y=f(x)+g(x)=√2 sin(π/8x+π/4) +√2 sin(π/8x-π/4)
=sinπ/8x+cosπ/8x+sinπ/8x-cosπ/8x
=2sinπ/8x
所以y递增时：2kπ-π/2<=π/8x<=2kπ+π/2
16k-4<=x<=16k+4
递增区间是[16k-4,16k+4] k∈Z

Answer 3

解：因为：函数g(x)与f(x)=√2 sin(π/8x+π/4) 关于x=8对称
g(x)=f(8-x)=√2 sin(π/8(8-x)+π/4) =√2 sin(-π/8x+5π/4) =√2 sin(π/8x-π/4)
再求导就行了