已知f(x)是定义在R 上的偶函数,且f(x+2)=-1/f(x)在R上恒成立,当x在[2,3]时,f(x)=x.求f(2009)为什么不是f(x+2)=-1/f(x)所以f(x+2)*f(x)=-1令x=x+2有f(x+2)*f(x+4)=-1所以f(x)=f(x+4)所以函数是周期为4的周期函数f(2009)=f(
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![已知f(x)是定义在R 上的偶函数,且f(x+2)=-1/f(x)在R上恒成立,当x在[2,3]时,f(x)=x.求f(2009)为什么不是f(x+2)=-1/f(x)所以f(x+2)*f(x)=-1令x=x+2有f(x+2)*f(x+4)=-1所以f(x)=f(x+4)所以函数是周期为4的周期函数f(2009)=f(](/uploads/image/z/2691025-25-5.jpg?t=%E5%B7%B2%E7%9F%A5f%28x%29%E6%98%AF%E5%AE%9A%E4%B9%89%E5%9C%A8R+%E4%B8%8A%E7%9A%84%E5%81%B6%E5%87%BD%E6%95%B0%2C%E4%B8%94f%28x%2B2%29%3D-1%2Ff%28x%29%E5%9C%A8R%E4%B8%8A%E6%81%92%E6%88%90%E7%AB%8B%2C%E5%BD%93x%E5%9C%A8%5B2%2C3%5D%E6%97%B6%2Cf%28x%29%3Dx.%E6%B1%82f%282009%29%E4%B8%BA%E4%BB%80%E4%B9%88%E4%B8%8D%E6%98%AFf%28x%2B2%29%3D-1%2Ff%28x%29%E6%89%80%E4%BB%A5f%28x%2B2%29%2Af%28x%29%3D-1%E4%BB%A4x%3Dx%2B2%E6%9C%89f%28x%2B2%29%2Af%28x%2B4%29%3D-1%E6%89%80%E4%BB%A5f%28x%29%3Df%28x%2B4%29%E6%89%80%E4%BB%A5%E5%87%BD%E6%95%B0%E6%98%AF%E5%91%A8%E6%9C%9F%E4%B8%BA4%E7%9A%84%E5%91%A8%E6%9C%9F%E5%87%BD%E6%95%B0f%282009%29%3Df%28)
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