f(x)=a·b=(1+sin(2x),sinx-cosx)·(1,sinx+cosx)
=1+sin(2x)+(sinx-cosx)(sinx+cosx)
=1+sin(2x)+sinx^2-cosx^2
=1+sin(2x)-cos(2x)
=1+√2sin(2x-π/4)
sin(2x-π/4)∈[-1,1]
√2sin(2x-π/4)∈[-√2,√2]
故:f(x)∈[1-√2,1+√2]
f(x)=a·b=(1+sin(2x),sinx-cosx)·(1,sinx+cosx)
=1+sin(2x)+(sinx-cosx)(sinx+cosx)
=1+sin(2x)+sinx^2-cosx^2
=1+sin(2x)-cos(2x)
=1+√2sin(2x-π/4)
sin(2x-π/4)∈[-1,1]
√2sin(2x-π/4)∈[-√2,√2]
故:f(x)∈[1-√2,1+√2]