已知函数fx=1/2x-sinx,x∈[0,2π],求单调区间和最值
人气：185 ℃ 时间：2020-02-06 06:13:23
解答
f'(x)=1/2-cosx
令f'(x)=0 x1=π/3 x2=5π/3
f''(x)=sinx f''(π/3)=√3/2>0 f''(5π/3)=-√3/2
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