作业帮设f(x)＝–2x+2,记f1(x)＝f(x),fn(x)＝f[fn-1(x)],n≥2,n∈N,则函数y＝fn(x)的图像恒过定点 .
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/29 10:27:11
![设f(x)＝–2x+2,记f1(x)＝f(x),fn(x)＝f[fn-1(x)],n≥2,n∈N,则函数y＝fn(x)的图像恒过定点 .](/uploads/image/z/5223408-24-8.jpg?t=%E8%AE%BEf%28x%29%EF%BC%9D%E2%80%932x%2B2%2C%E8%AE%B0f1%28x%29%EF%BC%9Df%28x%29%2Cfn%28x%29%EF%BC%9Df%5Bfn-1%28x%29%5D%2Cn%E2%89%A52%2Cn%E2%88%88N%2C%E5%88%99%E5%87%BD%E6%95%B0y%EF%BC%9Dfn%28x%29%E7%9A%84%E5%9B%BE%E5%83%8F%E6%81%92%E8%BF%87%E5%AE%9A%E7%82%B9+.)
设f(x)＝–2x+2,记f1(x)＝f(x),fn(x)＝f[fn-1(x)],n≥2,n∈N,则函数y＝fn(x)的图像恒过定点 .
设f(x)＝–2x+2,记f1(x)＝f(x),fn(x)＝f[fn-1(x)],n≥2,n∈N,则函数y＝fn(x)的图像恒过定点 .
设f(x)＝–2x+2,记f1(x)＝f(x),fn(x)＝f[fn-1(x)],n≥2,n∈N,则函数y＝fn(x)的图像恒过定点 .
解不动点方程：f(x)=-2x+2=x得：x=2/3
因此函数恒过定点（2/3,2/3)