# A Hard Function 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

Note by Jason Chrysoprase
5 years, 1 month ago

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I'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 $\mathbb{R} - \{58\}$

- 5 years, 1 month ago

So every $x$ we put, the answer is always real ?

So how about $f(0)$?

- 5 years, 1 month ago

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.

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

Domain

${x\in \mathbb{R} : x \neq 58}$

Range

${f \in \mathbb{R} : f \neq 58}$

- 5 years, 1 month ago

Staff - 5 years, 1 month ago

- 5 years, 1 month ago

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

Staff - 5 years, 1 month ago

Thank you sir :-)

- 5 years, 1 month ago

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

- 5 years, 1 month ago

Anyone ?

- 5 years, 1 month ago

why not me???

- 5 years, 1 month ago

Sorry, forgot u :(

- 5 years, 1 month ago