Question

已知函数f(x)=xsinx＋cosx当x∈四分之π，，π，求函数fx单调区间

Answer 1

f'(x)=sinx+xcosx-sinx=xcosx
当x属于（π/4,π/2）f'(x)>0,函数f(x)递增
当x属于（π/2,π）f'(x)<0,函数f(x)递减

Answer 2

f(x) = xsinx + cosx
f'(x) = sinx + xcosx -sinx
=xcosx
单调
增加=[π/4, π/2]
减小=[π/2, π]