If f(3)=M+1 for some rational/reciprocal function f(x) with the denominator of (x+2). If g(x) is the inverse function of f(x) and g(m)=2, find a possible function f(x) in the form f(x)= (ax+b)/(cx+d). There are many possible answers.

Originally Posted by FrancisG
If f(3)=M+1 for some rational/reciprocal function f(x) with the denominator of (x+2). If g(x) is the inverse function of f(x) and g(m)=2, find a possible function f(x) in the form f(x)= (ax+b)/(cx+d). There are many possible answers.
You have posted the same question twice.
That is strictly forbidden on this helpsite.

Sorry I posted this to the wrong forum so I re-posted it and forgot to delete this one.

are $M$ and $m$ intended to be distinct numbers, or does $M=m$ ?

Originally Posted by romsek
are $M$ and $m$ intended to be distinct numbers, or does $M=m$ ?
suppose $M$ and $m$ are abitrary

$f(3)=M+1$

$f(x) = \dfrac{a x + b}{x+2}$

$g(x) = f^{-1}(x)$

$g(m)=2$

ok, let's digest all of this

$f(3) = M+1 = \dfrac{3a+b}{5}$

$5M+5 = 3a+b$

$g(m)=2 \Rightarrow f(2)=m$

$f(2)=m \Rightarrow \dfrac{2a+b}{4}=m$

$2a+b = 4m$

so now we have two equations in $a,~b$

$\begin{pmatrix}3 &1\\2 &1\end{pmatrix}\begin{pmatrix}a \\ b\end{pmatrix} = \begin{pmatrix}5(M+1)\\4m\end{pmatrix}$

and this can be solved to produce

$a=5-4m+5M,~b=-10+12m-10M$

this results in

$f(x) = \dfrac{(5-4m+5M)x + (-10+12m-10M)}{x+2}$