已知函数f(x)=3sin^2x+2sinxcosx+cos^2x，x属于R。(1)求函数的最大值及此时x的集合（2）求单调增区间

显示全部楼层 · 2011-12-25 19:30:06

f(x)=3sin^2x+2sinxcosx+cos^2x=2sin^2x+2sinxcosx+1=2sin^2x-1+2sinxcosx+2=-cos(2x)+sin(2x)+2=sin(2x-π/4)+2所以最大值3，最小值-1最大值是sin(2x-π/4)=12x-π/4=2kπ+π/22x=2kπ+3π/4x=kπ+3π/8单调区间y=sin(2x-π/4)的单增区间为2kπ-π/2≤2x-π/4≤2kπ+π/22kπ-π/4≤2x≤2kπ+3π/4kπ-π/8≤x≤kπ+3π/8所以单减区间为kπ+3π/8≤x≤kπ+7π/8...

千问 · 2011-12-25 19:30:06

最大值 2+根号2 x=kπ+(3/8)π 增区间（ -π/8+kπ/2, 3π/8+kπ/2）...