设函数f(x)=2sin(π/2x+π/5),若对任意x∈R,都有f(x1)≤f(x)≤f(x2)对于这个问题,为什么当f(x1)=-2,f(x2)=2 原式才满足我还想问f(x1)≤f(x)≤f(x2)我有点不懂
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![设函数f(x)=2sin(π/2x+π/5),若对任意x∈R,都有f(x1)≤f(x)≤f(x2)对于这个问题,为什么当f(x1)=-2,f(x2)=2 原式才满足我还想问f(x1)≤f(x)≤f(x2)我有点不懂](/uploads/image/z/5410690-34-0.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3D2sin%28%CF%80%2F2x%2B%CF%80%2F5%EF%BC%89%2C%E8%8B%A5%E5%AF%B9%E4%BB%BB%E6%84%8Fx%E2%88%88R%2C%E9%83%BD%E6%9C%89f%28x1%29%E2%89%A4f%28x%29%E2%89%A4f%28x2%29%E5%AF%B9%E4%BA%8E%E8%BF%99%E4%B8%AA%E9%97%AE%E9%A2%98%2C%E4%B8%BA%E4%BB%80%E4%B9%88%E5%BD%93f%28x1%29%3D-2%2Cf%28x2%29%3D2+%E5%8E%9F%E5%BC%8F%E6%89%8D%E6%BB%A1%E8%B6%B3%E6%88%91%E8%BF%98%E6%83%B3%E9%97%AEf%28x1%29%E2%89%A4f%28x%29%E2%89%A4f%28x2%29%E6%88%91%E6%9C%89%E7%82%B9%E4%B8%8D%E6%87%82)
设函数f(x)=2sin(π/2x+π/5),若对任意x∈R,都有f(x1)≤f(x)≤f(x2)对于这个问题,为什么当f(x1)=-2,f(x2)=2 原式才满足我还想问f(x1)≤f(x)≤f(x2)我有点不懂
设函数f(x)=2sin(π/2x+π/5),若对任意x∈R,都有f(x1)≤f(x)≤f(x2)
对于这个问题,为什么当f(x1)=-2,f(x2)=2 原式才满足
我还想问f(x1)≤f(x)≤f(x2)
我有点不懂
设函数f(x)=2sin(π/2x+π/5),若对任意x∈R,都有f(x1)≤f(x)≤f(x2)对于这个问题,为什么当f(x1)=-2,f(x2)=2 原式才满足我还想问f(x1)≤f(x)≤f(x2)我有点不懂
对任意x∈R,都有f(x1)≤f(x)≤f(x2)
所以f(x1)是最小值,f(x2)是最大值
2sin(π/2x+π/5)∈[-2,2]
所以f(x1)=-2,f(x2)=2
问题补充:,则|x1-x2|的最小值为? 2
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