Problem: Given the linear fractional transformation of x into f1​(x)=x+12x−1​. Define fn+1​(x)= f1​(fn​(x)) for n=1,2,3,⋯. Assuming that f35​(x)=f5​(x), it follows that f28​(x) is equal to
Answer Choices:
A. x
B. x1​
C. xx−1​
D. 1−x1​
E. None of these
Solution:
If g(x) is the inverse of the transformation f1​(x), then g(fn+1​(x))=fn​(x). Since f35​(x)=f5​(x), successive application of this formula yields f31​(x)=f1​(x)∴f30​(x) =g(f1​(x))=x∴f29​(x)=g(x)∴f28​(x)=g(g(x)). But g(x)=2−xx+1​∴f28​(x)= g(g(x))=1−x1​.