设函数f(x)=2sin(π/2x+π/5),若对于任意实数x都有f(x1)小于等于f(x)小于等于f(x2)则绝对直x1-x2最小值
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![设函数f(x)=2sin(π/2x+π/5),若对于任意实数x都有f(x1)小于等于f(x)小于等于f(x2)则绝对直x1-x2最小值](/uploads/image/z/3928994-26-4.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3D2sin%28%CF%80%2F2x%2B%CF%80%2F5%29%2C%E8%8B%A5%E5%AF%B9%E4%BA%8E%E4%BB%BB%E6%84%8F%E5%AE%9E%E6%95%B0x%E9%83%BD%E6%9C%89f%EF%BC%88x1%29%E5%B0%8F%E4%BA%8E%E7%AD%89%E4%BA%8Ef%EF%BC%88x%29%E5%B0%8F%E4%BA%8E%E7%AD%89%E4%BA%8Ef%28x2%EF%BC%89%E5%88%99%E7%BB%9D%E5%AF%B9%E7%9B%B4x1-x2%E6%9C%80%E5%B0%8F%E5%80%BC)
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设函数f(x)=2sin(π/2x+π/5),若对于任意实数x都有f(x1)小于等于f(x)小于等于f(x2)则绝对直x1-x2最小值
设函数f(x)=2sin(π/2x+π/5),若对于任意实数x都有f(x1)小于等于f(x)小于等于f(x2)则绝对直x1-x2最小值
设函数f(x)=2sin(π/2x+π/5),若对于任意实数x都有f(x1)小于等于f(x)小于等于f(x2)则绝对直x1-x2最小值
即f(x1)是最小值,f(x2)是最大值
所以x=x1,sin(π/2x+π/5)=-1
xx=2,sin(π/2x+π/5)=1
最大最小至少相差半个周期
T=2π/(π/2)=4
所以 |x1-x2|最小=T/2=2
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