Question

求函数f(x)=2+2sinxcosx+sinx+cosx的最大值和最小值

Answer 1

f(x)=2+2sinxcosx+sinx+cosx=sin^2 x +cos^2 x+2sinxcosx+sinx+cosx+1=(sinx+cosx)^2+(sinx+cosx)+1令u=sinx+cosx=√2·[(√2/2)·sinx+(√2/2)·cosx]=√2·[cos(π/4)·sinx+sin(π/4)·cosx]=√2·sin(x+π/4).∵...