Results 1 to 3 of 3

Math Help - help

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    2

    help

    I am totally clueless abt these problems..any help would be appreciated.Thank you. I have the two problems attached.
    Attached Thumbnails Attached Thumbnails help-mathhelp.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,911
    Thanks
    774
    Hello, tamilmassive!

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

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


    We have: . 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 (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 f(x), we have: . \frac{(a^2+bc)x + b(a+d)}{c(a+d)x + (bc+d^2)} \;=\;\frac{-x+2}{-2x+3}


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

    Divide (3) by (2): . \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): . a^2-b^2 \:=\:-1

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

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


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

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

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2008
    Posts
    2
    thank you for your help
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum