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Question Number 199781 by cortano12 last updated on 09/Nov/23

Answered by qaz last updated on 09/Nov/23

$$f\left(x\right)=\frac{2(1-2x)}{x(1-x)}-f\left(\frac{1}{1-x}\right)$$$$=\frac{2(1-2x)}{x(1-x)}-(\frac{2(1-\frac{2}{1-x})}{\frac{1}{1-x}(1-\frac{1}{1-x})}-f\left(\frac{1}{1-\frac{1}{1-x}}\right))$$$$=\frac{2(1-2x)}{x(1-x)}-\frac{2(1+x)(1-x)}{x}+f(1-\frac{1}{x})$$$$=\frac{2(1-2x)}{x(1-x)}-\frac{2(1+x)(1-x)}{x}+(\frac{2(1-2(1-\frac{1}{x}))}{(1-\frac{1}{x})(1-(1-\frac{1}{x}))}-f\left(\frac{1}{1-(1-\frac{1}{x})}\right))$$$$=\frac{2(1-2x)}{x(1-x)}-\frac{2(1+x)(1-x)}{x}+\frac{2x(2-x)}{x-1}-f\left(x\right)$$$$\Rightarrow f\left(x\right)=\frac{(1-2x)}{x(1-x)}-\frac{(1+x)(1-x)}{x}+\frac{x(2-x)}{x-1}$$

Commented by Frix last updated on 09/Nov/23

$$\mathrm{Nice}.\mathrm{Factorize}\mathrm{it}\mathrm{to}\mathrm{get}$$$$f\left(x\right)=\frac{x+1}{x-1}$$

Commented by cortano12 last updated on 10/Nov/23

$$\mathrm{yes}....\mathrm{nice}$$

Commented by York12 last updated on 10/Nov/23

$$$$$$\mathrm{Bruh}\mathrm{where}\mathrm{to}\mathrm{learn}\mathrm{functional}\mathrm{equations}$$

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