函数f(x)=xcosx-sinx在【0,2π】上的最大值是？最小值是？
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f(x)=xcosx-sinx
f'(x) = -xsinx + cosx - cosx =0
xsinx =0
x = 0 or 2π
f''(x) = -(xcosx + sinx)
f''(0) = 0
f''(2π) = 2π >0 (min)
minf(x) = f(2π) = -2π
x在(0,2π)
f'(x) <0 (decreasing function)
max f(x) = f(0) = 0
f'(x) = -xsinx + cosx - cosx =0
xsinx =0
x = 0 or 2π
f''(x) = -(xcosx + sinx)
f''(0) = 0
f''(2π) = 2π >0 (min)
minf(x) = f(2π) = -2π
x在(0,2π)
f'(x) <0 (decreasing function)
max f(x) = f(0) = 0
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