函数f:[0,1]--->[0,1],满足f1(x)=f(x),...,fn(x)=f(fn-1(x)),n=2,3,4. 2x ,0<=x<=1/2现设f(x)= { ,则x属于[0,1]且满足f4(x)=x的x的个数是( ) 2-2x,1/2<=x<=1A、4
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![函数f:[0,1]--->[0,1],满足f1(x)=f(x),...,fn(x)=f(fn-1(x)),n=2,3,4. 2x ,0<=x<=1/2现设f(x)= { ,则x属于[0,1]且满足f4(x)=x的x的个数是( ) 2-2x,1/2<=x<=1A、4](/uploads/image/z/15084733-13-3.jpg?t=%E5%87%BD%E6%95%B0f%EF%BC%9A%5B0%2C1%5D---%26gt%3B%5B0%2C1%5D%2C%E6%BB%A1%E8%B6%B3f1%28x%29%3Df%28x%29%2C...%2Cfn%28x%29%3Df%28fn-1%28x%29%29%2Cn%3D2%2C3%2C4.+++++++++++++++++2x+%2C0%26lt%3B%3Dx%26lt%3B%3D1%2F2%E7%8E%B0%E8%AE%BEf%28x%29%3D+%7B+++++++++++++++++++++++++++++%2C%E5%88%99x%E5%B1%9E%E4%BA%8E%5B0%2C1%5D%E4%B8%94%E6%BB%A1%E8%B6%B3f4%28x%29%3Dx%E7%9A%84x%E7%9A%84%E4%B8%AA%E6%95%B0%E6%98%AF%EF%BC%88++%EF%BC%89++++++++++++++++2-2x%2C1%2F2%26lt%3B%3Dx%26lt%3B%3D1A%E3%80%814)
函数f:[0,1]--->[0,1],满足f1(x)=f(x),...,fn(x)=f(fn-1(x)),n=2,3,4. 2x ,0<=x<=1/2现设f(x)= { ,则x属于[0,1]且满足f4(x)=x的x的个数是( ) 2-2x,1/2<=x<=1A、4
函数f:[0,1]--->[0,1],满足f1(x)=f(x),...,fn(x)=f(fn-1(x)),n=2,3,4.
2x ,0<=x<=1/2
现设f(x)= { ,则x属于[0,1]且满足f4(x)=x的x的个数是( )
2-2x,1/2<=x<=1
A、4个 B、8个 C、16个 D、24个 (怎么做,看不懂这是什么图像)
函数f:[0,1]--->[0,1],满足f1(x)=f(x),...,fn(x)=f(fn-1(x)),n=2,3,4. 2x ,0<=x<=1/2现设f(x)= { ,则x属于[0,1]且满足f4(x)=x的x的个数是( ) 2-2x,1/2<=x<=1A、4
不太会画图...所以我大概说一下
你可以试话f1(x)的图,然后你会发现是一个先递增再递减的函数,不妨称之为山= =+;
然后你试画f2的图,就会发现就跟你的图差不多,是两个更窄一点的山,定义域依然在0-1之间,而且山顶同高,都为1;
其实按规律来说,fn就是在0,1区间里等分2^(n-1)个区间,然后每个区间都是一个山
所以f4的时候就是8个山
f4(x)=x的解,就是y=f4和y=x的交点,所以一共有16个