f-R-R-continuous-such-that-x-R-f-x-f-f-x-1-f-2004-2003-f-1999-

Question Number 950 by 123456 last updated on 06/May/15

f : R \to R, continuous such that \forall x \in R, f (x) f [f (x)] = 1, f (2004) = 2003, f (1999) = ?

Commented by 123456 last updated on 06/May/15

f (2004) f [f (2004)] = 2003 f (2003) = 1 \Leftrightarrow f (2003) = \frac{1}{2003}

f (2003) f [f (2003)] = \frac{1}{2003} f (\frac{1}{2003}) = 1 \Leftrightarrow f (\frac{1}{2003}) = 2003 = f (2004)

f (\frac{1}{2003}) f [f (\frac{1}{2003})] = 2003 f (2003) = 1

Commented by 123456 last updated on 06/May/15

f (x) f [f (x)] = 1

f [f (x)] f {f [f (x)]} = 1

f (x) = f {f [f (x)]} \lor f [f (x)] = 0

\exists x, f (x) = 0 \Rightarrow f (x) f [f (x)] = 1 \Rightarrow 0 f (0) = 1

f (x) = f {f [f (x)]}

f (2004) = f {f [f (2004)]} = 2003

f (x) = y

f (x) f [f (x)] = 1

y f (y) = 1

f (y) = \frac{1}{y}

Answered by prakash jain last updated on 06/May/15

f (2003) = \frac{1}{2003} and f (2004) = 2003

Given that f (x) is continous :

\exists x such that f (x) = 1999

Since f (f (x)) = \frac{1}{f (x)}

p u t f (x) = 1999

f (1999) = \frac{1}{1999}