Question

I need answer quickly

Answer 1

$f(x)f( \frac{1}{x} ) = f(x) + f( \frac{1}{x} ) \\ = > f(x){ \large(}f( \frac{1}{x}) - 1{ \large)}= f( \frac{1}{x}) \\ = > f(x) = \frac{f( \frac{1}{x}) }{f( \frac{1}{x}) - 1} \: \: \: \: \: \: (1) \: \\ = > replace \: x \: by \: \frac{1}{x} \: we \: get \\ = > f( \frac{1}{x}) = \frac{f(x)}{f(x) - 1} \: \: \: \: \: \: \: (2)\\ = > multiply \: 1 \: and \: 2 \: now \\ = > f(x)f( \frac{1}{x} ) = \frac{f(x)}{(f(x) - 1)} \frac{f( \frac{1}{x}) }{(f( \frac{1}{x}) - 1)} \\ = > (f(x) - 1)(f( \frac{1}{x})) - 1) = 1 \\ = > take \: f(x) - 1 = g(x) \\ = > we \: get \: \\ = > g(x)g( \frac{1}{x} ) = 1 \\ = > this \: is \: a \: defined \: funtion \: which \: gives \: \\ g(x) = {x}^{n} \\ try \: with \: some \: example \: to \: cheak \: its \: correctness$

hence we got

$g(x)= {x}^{n} \\ which\: is \:equal \:to\: \\f(x)-1= {x}^{n}\\ f(x)= {x}^{n}+1\\given \: that \: f(-2)=33\\=>{x}^{n}+1=33\\=>{-2}^{n}=32\\ therefore \:n=5\\now \:we \:got \:\\f(x)={x}^{5}+1 \\f(1)={1}^{5}+1\\f(1)=2$

alternative :

$f(x)f( \frac{1}{x} ) = f(x) + f( \frac{1}{x} ) \\substitute\: x=1\\ we \:get \:f(1)f( \frac{1}{1} ) = f(1) + f( \frac{1}{1} ) \\{f(1)}^{2}=2f(1)\\f(1)=2$

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