Question

求函数y=sinx-cosx+2sinxcosx的最值及相应x的值.

Answer 1

求导的

y'=cosx+sinx+2(cosx)^2-2(sinx)^2

=cosx+2(cosx)^2+1/8+sinx-2(sinx)^2-1/8

=2(cosx+1/4)^2-2(sinx-1/4)^2=0

cosx+1/4=sinx-1/4 或cosx=sinx

极值点：x=arccos四分之根2+ (1/4+2k)pi 或x=(K+1/4)pi

最大值 1 根2对应x为(K+1/4)pi k为整数