×

# 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
1 year, 1 month ago

Sort by:

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 $$\mathbb{R} - \{58\}$$ · 1 year, 1 month ago

Domain

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

Range

$${f \in \mathbb{R} : f \neq 58}$$ · 1 year, 1 month ago

The answer is $$\mathbb{R} - {58}$$ since there is no $$x$$ value for $$58$$ · 1 year, 1 month ago

I don't know functions well , can you please explain a bit more on how you took the range? · 1 year, 1 month ago

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

So how about $$f(0)$$? · 1 year, 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. · 1 year, 1 month ago

Check this out Staff · 1 year, 1 month ago

Wow, that guy really wrote the same problem. We both came from Indonesia :) · 1 year, 1 month ago

Practice the challenge quizzes in the chapter and read the wikis. Staff · 1 year, 1 month ago

Thank you sir :-) · 1 year, 1 month ago

Anyone ? · 1 year, 1 month ago

why not me??? · 1 year, 1 month ago