1. ## help

I am totally clueless abt these problems..any help would be appreciated.Thank you. I have the two problems attached.

2. Hello, tamilmassive!

3. Let $\displaystyle f(x) \:=\:\frac{2-x}{3-2x}\quad\left(x \neq \frac{3}{2}\right)$

Find a function $\displaystyle g(x)$ of the form .$\displaystyle \frac{ax +b}{cx+d}$ .so that: $\displaystyle g(g(x)) \:=\:f(x).$
Specify at what point $\displaystyle g$ is defined.
I will assume that $\displaystyle a,b,c,d$ are integers.

We have: . $\displaystyle g(g(x)) \;=\;\frac{a\left(\dfrac{ax+b}{cx+d}\right) + b}{c\left(\dfrac{ax+b}{cx+d}\right) + d}$

Multiply top and bottom by $\displaystyle (cx+d)\!:\;\;\frac{a(ax+b) + b(cx+d)}{c(ax+b) + d(cx+d)} \;=\;\frac{(a^2+bc)x + (ab+bd)}{(ac + cd)x + (bc+d^2)}$

Since this is to be equal to $\displaystyle f(x)$, we have: . $\displaystyle \frac{(a^2+bc)x + b(a+d)}{c(a+d)x + (bc+d^2)} \;=\;\frac{-x+2}{-2x+3}$

Equating coefficients: . $\displaystyle \begin{array}{ccccccccc} {\color{blue}(1)} & a^2+bc &=& -1 & & {\color{blue}(2)} & b(a+d) &=& 2 \\ {\color{blue}(3)} & c(a+d) &=&-2 & & {\color{blue}(4)} & bc+d^2 &=& 3 \end{array}$

Divide (3) by (2): . $\displaystyle \frac{c(a+d)}{b(a+d)} \:=\:\frac{\text{-}2}{2}\quad\Rightarrow\quad \frac{c}{b} \:=\:-1\quad\Rightarrow\quad c \:=\:-b$

Substitute into (1): . $\displaystyle a^2-b^2 \:=\:-1$

. . Let $\displaystyle a \,= \,0,\;b \,= \,1$ . . . then $\displaystyle c \,= \,-1$

Substitute into (4): . $\displaystyle -1 + d^2\:=\:3\quad\Rightarrow\quad d \,=\,2$

Therefore: . $\displaystyle \boxed{g(x) \;=\;\frac{1}{2-x}}$

$\displaystyle g(x)$ is defined for all real $\displaystyle x \neq 2.$

3. thank you for your help