设函数f(x)=ax+b,其中a,b为实数,f1(x)=f(x),fn+1(x)=f(fn(x)),n=1,2,3…若f7(x)=128x+381,求a+b?
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![设函数f(x)=ax+b,其中a,b为实数,f1(x)=f(x),fn+1(x)=f(fn(x)),n=1,2,3…若f7(x)=128x+381,求a+b?](/uploads/image/z/10173023-71-3.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dax%2Bb%2C%E5%85%B6%E4%B8%ADa%2Cb%E4%B8%BA%E5%AE%9E%E6%95%B0%2Cf1%28x%29%3Df%28x%29%2Cfn%2B1%28x%29%3Df%28fn%28x%29%29%2Cn%3D1%2C2%2C3%E2%80%A6%E8%8B%A5f7%28x%29%3D128x%2B381%2C%E6%B1%82a%2Bb%3F)
设函数f(x)=ax+b,其中a,b为实数,f1(x)=f(x),fn+1(x)=f(fn(x)),n=1,2,3…若f7(x)=128x+381,求a+b?
设函数f(x)=ax+b,其中a,b为实数,f1(x)=f(x),fn+1(x)=f(fn(x)),n=1,2,3…若f7(x)=128x+381,求a+b?
设函数f(x)=ax+b,其中a,b为实数,f1(x)=f(x),fn+1(x)=f(fn(x)),n=1,2,3…若f7(x)=128x+381,求a+b?
f2(x)=f[f(x)]=a(ax+b)+b=a²x +b(a+1)
f3(x)=f[f2(x)]=a[a²x+b(a+1)] +b=a³x+ab(a+1)+b=a³x+b(a²+a+1)
f4(x)=f[f3(x)]=a[a³x+b(a²+a+1)]+b=a^4·x+b(a³+a²+a+1)
.
从而 a^7=128,(1+a+a²+...+a^6)b=381
解得 a=2,b=3