f（x+2）＝f（x）是周期函数吗 f(x)=f(-x+2）是否是周期函数
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\u89e3\u9898\u8fc7\u7a0b\u5982\u4e0b\uff1a
f(x+2)=-f(x)
\u89e3\uff1a
=f(x+4)
=f(x+2+2)
=-f(x+2)
=-[-f(x)]
=f(x)
\u2234 f(x)\u7684\u5468\u671f\u662f4
\u6269\u5c55\u8d44\u6599\u6c42\u51fd\u6570\u5468\u671f\u7684\u65b9\u6cd5\uff1a
\u5bf9\u4e8e\u51fd\u6570y=f\uff08x\uff09\uff0c\u5982\u679c\u5b58\u5728\u4e00\u4e2a\u4e0d\u4e3a\u96f6\u7684\u5e38\u6570T\uff0c\u4f7f\u5f97\u5f53x\u53d6\u5b9a\u4e49\u57df\u5185\u7684\u6bcf\u4e00\u4e2a\u503c\u65f6\uff0cf\uff08x+T\uff09=f\uff08x\uff09\u90fd\u6210\u7acb\uff0c\u90a3\u4e48\u5c31\u628a\u51fd\u6570y=f\uff08x\uff09\u53eb\u505a\u5468\u671f\u51fd\u6570\uff0c\u4e0d\u4e3a\u96f6\u7684\u5e38\u6570T\u53eb\u505a\u8fd9\u4e2a\u51fd\u6570\u7684\u5468\u671f\u3002\u4e8b\u5b9e\u4e0a\uff0c\u4efb\u4f55\u4e00\u4e2a\u5e38\u6570kT\uff08k\u2208Z\uff0c\u4e14k\u22600\uff09\u90fd\u662f\u5b83\u7684\u5468\u671f\u3002
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\u5468\u671f\u51fd\u6570\u7684\u6027\u8d28\uff1a
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f(x)=-f(x+2),\u5c06x\u52a02\u5f97f(x+2)=-f(x+4)\uff0c\u53c8\u56e0\u4e3af(x)=-f(x+2),\u6240\u4ee5-f(x)=-f(x+4)\uff0c\u6240\u4ee5f(x)=f(x+4)\uff0c\u5373\u51fd\u6570\u662f\u4ee54\u4e3a\u5468\u671f\u7684\u5468\u671f\u51fd\u6570
是的以2为周期的函数
因为f（x+T）＝f（x）则为以 T为周期的函数
是周期函数，x每增加，y相同，你可以画图来帮助理解下，从（0,2）随便画一条线，根据
f（x+2）＝f（x），那么（2,4）还是这条线，原样平移过来了，（4,6）区间同样，这样周期性的重复，所以是周期函数
周期函数的定义是如果存在正数T，是对于任意x属于D都有x+T属于D，并且使f（x）=f（x+T）恒成立
假设f(x)是周期函数有f（x+T）=f（2—x）得x+T=2—x
T=2—2x是以x为自变量的函数不能确定周期T所以不是周期函数
如果是在函数定义域内的话，那就是周期为2的周期函数，这个你可以看下书上的定义
(x)=f(2+x)对定义域内的任意x都成立，则f(x)以x=1为对称轴的函数。
f(x)=f(2-x)对定义域内的任意x都成立，则f(x)是周期函数
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