设函数f(x)在[0,2π]上连续,且f(0)=f(2π),证明:至少存在一点ξ∈[0,π],使得f(ξ)=f(ξ+π)
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![设函数f(x)在[0,2π]上连续,且f(0)=f(2π),证明:至少存在一点ξ∈[0,π],使得f(ξ)=f(ξ+π)](/uploads/image/z/6889962-66-2.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8%5B0%2C2%CF%80%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E4%B8%94f%280%29%3Df%282%CF%80%29%2C%E8%AF%81%E6%98%8E%EF%BC%9A%E8%87%B3%E5%B0%91%E5%AD%98%E5%9C%A8%E4%B8%80%E7%82%B9%CE%BE%E2%88%88%5B0%2C%CF%80%5D%2C%E4%BD%BF%E5%BE%97f%28%CE%BE%29%3Df%28%CE%BE%2B%CF%80%EF%BC%89)
设函数f(x)在[0,2π]上连续,且f(0)=f(2π),证明:至少存在一点ξ∈[0,π],使得f(ξ)=f(ξ+π)
设函数f(x)在[0,2π]上连续,且f(0)=f(2π),证明:至少存在一点ξ∈[0,π],使得f(ξ)=f(ξ+π)
设函数f(x)在[0,2π]上连续,且f(0)=f(2π),证明:至少存在一点ξ∈[0,π],使得f(ξ)=f(ξ+π)
令g(x)=f(x+π)-f(x),则g(x)在[0,π]上连续;由题意有g(0)=f(π)-f(0),g(π)=f(2π)-f(π),则g(0)=-g(π);
若g(0)=g(π)=0,则取ξ=0或π;若g(0)=g(π)不等于0,则有g(0)*g(π)
令g(x)=f(x+π)-f(x),则g(x)在[0,π]上连续;由题意有g(0)=f(π)-f(0),g(π)=f(2π)-f(π),则g(0)=-g(π);
若g(0)=g(π)=0,则取ξ=0或π;若g(0)=g(π)不等于0,则有g(0)*g(π)<0,根据连续函数的介值性定理得:必存在ξ∈[(0,π),使得g(ξ)=0,即f(ξ)=f(ξ+π)
总上可得:至少存在一点ξ∈...
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令g(x)=f(x+π)-f(x),则g(x)在[0,π]上连续;由题意有g(0)=f(π)-f(0),g(π)=f(2π)-f(π),则g(0)=-g(π);
若g(0)=g(π)=0,则取ξ=0或π;若g(0)=g(π)不等于0,则有g(0)*g(π)<0,根据连续函数的介值性定理得:必存在ξ∈[(0,π),使得g(ξ)=0,即f(ξ)=f(ξ+π)
总上可得:至少存在一点ξ∈[0,π],使得f(ξ)=f(ξ+π)
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