Functions Question 608
If $ f(x)=\frac{x-3}{x+1} $ , then $ f(x) $
[f{f(x)}] $ equals [RPET 1996]
Options:
x
?x
C) $ \frac{x}{2} $
D) $ -\frac{1}{x} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ f,[f(x)]=\frac{f(x)-3}{f(x)+1} $ $ =\frac{( \frac{x-3}{x+1} )-3}{( \frac{x-3}{x+1} )+1}=\frac{x-3-3x-3}{x-3+x+1}=\frac{-2x-6}{2x-2}=\frac{-x-3}{x-1} $ Now $ f,[f(f(x))]=f,( \frac{-x-3}{x-1} ) $ $ f(x)=(2,,4]-{3} $ .
BETA
