已知函数f(x)满足f(x+y)+f(x-y)=2f(x)f(y),x和y都属于R,且f(0)≠0,试证明f(x)是偶函数
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![已知函数f(x)满足f(x+y)+f(x-y)=2f(x)f(y),x和y都属于R,且f(0)≠0,试证明f(x)是偶函数](/uploads/image/z/5933105-17-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%E6%BB%A1%E8%B6%B3f%28x%2By%29%2Bf%28x-y%29%3D2f%28x%29f%28y%29%2Cx%E5%92%8Cy%E9%83%BD%E5%B1%9E%E4%BA%8ER%2C%E4%B8%94f%280%29%E2%89%A00%2C%E8%AF%95%E8%AF%81%E6%98%8Ef%28x%29%E6%98%AF%E5%81%B6%E5%87%BD%E6%95%B0)
已知函数f(x)满足f(x+y)+f(x-y)=2f(x)f(y),x和y都属于R,且f(0)≠0,试证明f(x)是偶函数
已知函数f(x)满足f(x+y)+f(x-y)=2f(x)f(y),x和y都属于R,且f(0)≠0,试证明f(x)是偶函数
已知函数f(x)满足f(x+y)+f(x-y)=2f(x)f(y),x和y都属于R,且f(0)≠0,试证明f(x)是偶函数
令x=y=0
故:f(0+0)+f(0-0)=2f(0)f(0)
即:2f(0)=2 f(0)f(0)
因为f(0)≠0
故:f(0)=1
令x=0
故:f(y)+f(-y)=2f(0)f(y)
故:f(y)+f(-y)=2f(y)
故:f(-y)=f(y)
因为y任意
故:f(-x)=f(x)
故:f(x)是偶函数(定义域R关于原点对称)
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