已知：函数y=Acosx+B,(A>0)的最大值是1,最小值是-3,试确定

显示全部楼层 · 2009-7-27 01:39:08

A>0,所以当cosx=1时,函数值最大,即A+B=1当cosx=-1时,函数值最小,即-A+B=-3解得,A=2,B=-1则f(x)=Bsin(ax+π/3)=-sin(2x+π/3)[求f(x)=-sin(2x+π/3)的单调增区间,就是求f(x)=sin(2x+π/3)的单调减区间]令2x+π/3大于等于π/2+2Kπ,小于等于3π/2+2Kπ解得X属于[π/12+Kπ,2π/3+Kπ](K属于整数)则f(x)的单调增区间是[π/12+Kπ,2π/3+Kπ](K属于整数)...

千问 · 2009-7-27 01:39:08

由题可知A+B=1 B-A=-3进而知A=2 B=-1f(x)=-sin(2X+π/3)从而可得f(x)的增区间为π/12+2Kπ到7π/12+2Kπ...