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Here we have $$f(x)+f\left(\frac{x-1}{x}\right)=2 x+4$$ Find the function $f(x)$.

Firstly, I let $t=\frac{x}{x-1}$ then I got the equation $$f\left(\frac{1}{t-1}\right)+f(t)=\frac{4t-2}{t-1}$$ After that I don’t find any notices to do more.

Please kindly give me a hint . Thank beforehand!

Blue
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Petter Green
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    Hint: Substitute $(x-1)/x$ in place of $x$ to obtain a second equation, then do this substitution again to obtain a third one. Solve the resulting system. – Gary Jun 15 '21 at 08:03
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    Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – Martin R Jun 15 '21 at 08:04

1 Answers1

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I'll give you some hints to can work with.

  1. Replace $x$ by $1 - \frac{1}{x}$ throughout to get another functional equation.

  2. In the new equation you obtain, replace $x$ by $1 - \frac{1}{x}$ again!

Why is this a good idea? Note that $$1 - \frac{1}{\left(1 - \frac{1}{\left(1 - \frac{1}{x} \right)} \right)} = x$$ and $$1 - \frac{1}{\left(1 - \frac{1}{x} \right)} = \frac{-1}{x-1}$$ You'll be left with three equations, where you can eliminate $f(1-\frac{1}{x})$ and $f(\frac{-1}{x-1})$ (the usual way of solving a system of linear equations) and obtain $f(x)$.

stoic-santiago
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