由f(x+2)=f(x),知函数y=f(x)(x∈R)是周期为2函数,
x∈[-1,1]时,f(x)=x2,作出它的图象则y=f(x)(注意拓展它的区间)
其图象如下:
∴函数f(x)的图象与y=|lgx|的图象交点个数是10个.
故答案为:10.
已知函数y=f(x)(x∈R)满足f(x+2)=f(x),且x∈[-1,1]时,f(x)=x2,那么y=f(x)的图象与y=|lgx|的图象的交点个数有_个.
已知函数y=f(x)(x∈R)满足f(x+2)=f(x),且x∈[-1,1]时,f(x)=x2,那么y=f(x)的图象与y=|lgx|的图象的交点个数有______个.
数学人气:200 ℃时间:2019-10-26 06:37:29
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