作业帮已知函数f(x)=sinxcosx+(cosx)^2(1)求f(x)的最小正周期（2）求f(x)在区间[-兀/6,兀/3]上的最大值和最小值
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![已知函数f(x)=sinxcosx+(cosx)^2(1)求f(x)的最小正周期（2）求f(x)在区间[-兀/6,兀/3]上的最大值和最小值](/uploads/image/z/15096568-40-8.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsinxcosx%2B%28cosx%29%5E2%281%29%E6%B1%82f%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%EF%BC%882%EF%BC%89%E6%B1%82f%28x%29%E5%9C%A8%E5%8C%BA%E9%97%B4%5B-%E5%85%80%2F6%2C%E5%85%80%2F3%5D%E4%B8%8A%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BC)
已知函数f(x)=sinxcosx+(cosx)^2(1)求f(x)的最小正周期（2）求f(x)在区间[-兀/6,兀/3]上的最大值和最小值
已知函数f(x)=sinxcosx+(cosx)^2(1)求f(x)的最小正周期（2）求f(x)在区间[-兀/6,兀/3]上的最大值和最小值
已知函数f(x)=sinxcosx+(cosx)^2(1)求f(x)的最小正周期（2）求f(x)在区间[-兀/6,兀/3]上的最大值和最小值
f(x)=sinxcosx+cos^2 x
=sin2x/2+(1+cos2x)/2
=1/2*(sin2x+cos2x)+1/2
=√2/2*sin(2x+π/4)+1/2
∴f(x)的最小正周期T=2π/2=π
∵f(x)的单调增区间为[-π/4+2kπ,3π/4+2kπ],k∈Z
当k=1时,单调增区间为[-π/4,3π/4],且[-π/6,π/3]在此区间内
∴此时f(x)的最大值为
f(π/3)=√2/2*sin(2*π/3+π/4)+1/2
=√2/2*sin(11π/12)+1/2
=√2/2*(√6-√2)/4+1/2
=(√3-1)/4+1/2
最小值为
f(-π/6)=√2/2*sin(-2*π/6+π/4)+1/2
=√2/2*sin(-π/12)+1/2
=√2/2*[-(√6-√2)/4]+1/2
= -(√3-1)/4+1/2
解:
f(x)=sinxcosx+(cosx)^2
=1/2sin2x+1/2cos2x+1/2
=根号2/2sin(2x+π/4)+1/2
(1)
T=2π/2=π
(2)
-π/6<=x<=π/3
-π/3<=2x<=2π/3
-π/12<=2x+π/4<=11π/12
所以sin(2x+π/4)∈[(√2 - √6)/ 4,1]
所以f(x)=根号2/2sin(2x+π/4)+1/2∈[(3-根号12)/2,(根号2 + 1)/2]