Question

已知函数f(x)=xsinx+cosx,x属于〔0，派〕。求f(x)的单调区间和值域！！！速求！

Answer 1

答：
f(x)=xsinx+cosx，0<=x<=π
求导：
f'(x)=sinx+xcosx-sinx=xcosx
令f'(x)=xcosx=0
解得x=0或者x=π/2
当0<=x<=π/2时，f'(x)>=0，f(x)是单调增函数
π/2<=x<=π时，f'(x)<=0，f(x)是单调减函数
所以：x=π/2时f(x)取得最大值f(π/2)=π/2
f(0)=1
f(π)=-1
所以：
f(x)的单调增区间为[0，π/2]
单调减区间为[π/2，π]
值域为[-1，π/2]

Answer 2

Answer 3

答:
f(x)=xsinx+cosx，0<=x<=π
求导:
f'(x)=sinx+xcosx-sinx=xcosx
令f'(x)=xcosx=0
解得x=0或者x=π/2
当0<=x<=π/2时，f'(x)>=0，f(x)是单调增函数
π/2<=x<=π时，f'(x)<=0，f(x)是单调减函数
所以:x=π/2时f(x)取得最大值f(π/2)=π/2
f(0)=1
f(π)=-1
所以:
f(x)的单调增区间为[0，π/2]
单调减区间为[π/2，π]
值域为[-1，π/2]