函数f(x)=xcosx-sinx在【0,2π】上的最大值是?最小值是?
问题描述:
函数f(x)=xcosx-sinx在【0,2π】上的最大值是?最小值是?
答
f(x)=xcosx-sinxf'(x) = -xsinx + cosx - cosx =0xsinx =0x = 0 or 2πf''(x) = -(xcosx + sinx)f''(0) = 0f''(2π) = 2π >0 (min)minf(x) = f(2π) = -2π x在(0,2π)f'(x)