Question

已知函数f(x)=sinx/2cosx/2+cos的平方x/2. 1.求函数的单调区间。 2.求函数在【-π/4,π】上的最大值和最小

Answer 1

解：
函数f(x)=sin(x/2)cos(x/2)+cos²(x/2)
=(sin(x/2)+cos(x/2))cos(x/2)
=√2(√2/2sin (x/2)+√2/2cos(x/2))cos(x/2)
=√2(sin (x/2) cosπ/4+cos (x/2) sinπ/4)cos(x/2)
=√2sin(x/2+π/4)cos(x/2)
=√2/2sin(x+π/4)+ 1/2

1、
f(x) = √2/2sin(x+π/4)+ 1/2
sin(x),cos(x)的定义域为R,值域为〔-1,1〕
即： -π/2+2kπ≤x+π/4≤π/2+2kπ
函数f(x)的单增区间为-3π/4+2kπ≤x≤π/4+2kπ.(k∈z).
即：【-3π/4+2kπ，π/4+2kπ】
2、
-π/4≤x≤π
0≤x+π/4≤5π/4
所以x+π/4=5π/4，sin(x+π/4)最小=-√2/2
函数f(x)的最小值为
f(x) =√2/2sin(x+π/4)+ 1/2=-√2/2×√2/2+ 1/2=-1/2+1/2=0
x+π/4=π/2，sin(x+π/4)最大=1
函数f(x)的最大值为
f(x) =√2/2sin(x+π/4)+ 1/2=√2/2+ 1/2

Answer 2

楼上相当详细了。