设函数f(x=ax+b) 其中a+b为实数,f1(x)=f(x),f(n+1)(x)=f(fn(x)),n=1,2,3...若f5(x)=32x+93 则ab=?
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![设函数f(x=ax+b) 其中a+b为实数,f1(x)=f(x),f(n+1)(x)=f(fn(x)),n=1,2,3...若f5(x)=32x+93 则ab=?](/uploads/image/z/8600361-33-1.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%3Dax%2Bb%29+%E5%85%B6%E4%B8%ADa%2Bb%E4%B8%BA%E5%AE%9E%E6%95%B0%2Cf1%28x%29%3Df%28x%29%2Cf%28n%2B1%29%28x%29%3Df%28fn%28x%29%29%2Cn%3D1%2C2%2C3...%E8%8B%A5f5%28x%29%3D32x%2B93+%E5%88%99ab%3D%3F)
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设函数f(x=ax+b) 其中a+b为实数,f1(x)=f(x),f(n+1)(x)=f(fn(x)),n=1,2,3...若f5(x)=32x+93 则ab=?
设函数f(x=ax+b) 其中a+b为实数,f1(x)=f(x),f(n+1)(x)=f(fn(x)),n=1,2,3...若f5(x)=32x+93 则ab=?
设函数f(x=ax+b) 其中a+b为实数,f1(x)=f(x),f(n+1)(x)=f(fn(x)),n=1,2,3...若f5(x)=32x+93 则ab=?
由f1(x)=f(x)=ax+b,得到f2(x)=f(f1(x))=a(ax+b)+b=a2x+ab+b,
f3(x)=f(f2(x))=a[a(ax+b)+b]+b=a3x+a2b+ab+b,
同理f4(x)=f(f3(x))=a4x+a3b+a2b+ab+b,
则f5(x)=f(f4(x))=a5x+a4b+a3b+a2b+ab+b=32x+93,
即a5=32①,a4b+a3b+a2b+ab+b=93②,
由①解得:a=2,把a=2代入②解得:b=3,
则ab=6.
故答案为:6
题的括号有问题,在整理一下