Based on 2013 Olympiad Problem

A function \(f\) is defined for all \(x\) and has the following property,

\(\large f(x) = \frac{ax+b}{cx+d}\)

If \(a, b, c\) and \(d\) are real number and the function above satisfy \(f(19)=19, f(97) = 97, f(f(x)) = x,\) for every \(x\) value, except \(\large -\frac{d}{c}\), find range of the function

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## Comments

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TopNewestI'll assume that the domain is \(\mathbb{R} - \{-\frac{d}{c} \} \)

If \(\dfrac{b}{a} \neq \dfrac{d}{c} \), then the range is \(\mathbb{R}-\{\frac{a}{c}\}\) if \(c \neq 0\) and \(\mathbb{R}\) otherwise.

If equality holds above, then \(f(x)\) is a constant function, which contradicts the fact that \(f(19) \neq f(97)\).

In conclusion, the range is \(\mathbb{R}-\{\frac{a}{c}\}\) if \(c \neq 0\) otherwise \(\mathbb{R}\).

Edit:After finding \(a,b, c, d\),

Either \(f(x)=x\) or \(f(x)=\dfrac{58x-19 \times 97}{x-58} \)

So, the range is either \(\mathbb{R}\) or \(\mathbb{R} - \{58\}\)

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So every \(x\) we put, the answer is always real ?

So how about \(f(0)\)?

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Oops, sorry for late reply. Whats the problem if we put \(f(0)\).

\(f(0) = \dfrac{a×0 + b}{c×0 + d} = \dfrac{b}{d}\).

Since \(\dfrac{-d}{c}\) is removed from its domain, the answer \(\dfrac{b}{d}\) is real too.

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I don't know functions well , can you please explain a bit more on how you took the range?

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The answer is \(\mathbb{R} - {58}\) since there is no \(x\) value for \(58\)

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Domain

\({x\in \mathbb{R} : x \neq 58}\)

Range

\({f \in \mathbb{R} : f \neq 58}\)

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Check this out

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Sir, Can you please suggest me where can I learn about functions & solving these type of questions? Honestly, I don't know much about this topic , please help me, thank you.!

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Practice the challenge quizzes in the chapter and read the wikis.

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Wow, that guy really wrote the same problem. We both came from Indonesia :)

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@Abhay Kumar @Ashish Siva @Margaret Zheng @Mark C @Rishabh Tiwari @Rishabh Cool @Pi Han Goh

Anyone ?

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why not me???

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Sorry, forgot u :(

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