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Function composition help

  1. Dec 15, 2008 #1
    1. The problem statement, all variables and given/known data

    F o G(x)
    F(x)=1/x+1
    g(x)=2x-x/x+3
    Simplify

    2. Relevant equations

    F o G=f(g(x))

    3. The attempt at a solution
    Here is the order of things I did.

    F(2x-1/x+3)
    to
    1/(2x-1/x+3)+1
    Ok, this sounds stupid but, I can solve these, but when I get stuff like this that has fractions in fraction I don't know how to do it, I am sad to admit that I don't know how to solve it when there is a fraction in a fraction. Please, don't skip any steps, if someone could show me how to do this step by step not skiping anything or leaving anything out, I will forever be gratful, and yes, I mispelled gratful, I am bad at spelling.
     
  2. jcsd
  3. Dec 15, 2008 #2

    cepheid

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    Staff Emeritus
    Science Advisor
    Gold Member

    THE RULE (in words): Let's say you have a fraction in the denominator. If something is in the denominator of the fraction in the denominator (i.e. on the bottom of the bottom), it ends up on top. If something is in the numerator of the fraction in the denominator (i.e. top of the bottom), it ends up on the bottom.

    THE REASON FOR THE RULE (in words): Because dividing by something is the same thing as multiplying by its reciprocal! Dividing by 2 is the same thing as multiplying by 1/2. Dividing by 1/2 is the same thing as multiplying by 2.

    Hence: a / (1/2) = a*(2) / 1

    Notice that the thing on the bottom of the bottom ended up on the top (with a).

    THE RULE (algebraically):

    [tex] \frac{\frac{a}{b}}{\frac{c}{d}} = \frac {ad}{bc} [/tex]

    THE REASON (algebraically):

    [tex] \frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b} \left(\frac{c}{d}\right)^{-1} = \frac{a}{b} \left(\frac{d}{c}\right) [/tex]

    [tex] = \frac{ad}{bc} [/tex]
     
    Last edited: Dec 15, 2008
  4. Dec 16, 2008 #3
    Do the rules change if its just.....

    x+1/(x-1/x+2)?

    Would it be (x+1)(x+2)/x-1?

    And what if its (x+1/x+2)/x-1 to being with?
     
  5. Dec 16, 2008 #4
    You need to use parantheses, x+1/(x-1/x+2) means x + 1/(x - (1/x) + 2) which I doubt you meant, I would rewrite it as (x+1)/[(x-1)/(x+2)]. And yes that would be (x+1)(x+2)/(x-1) once again use parantheses.

    If you have (x+1)/(x+2) / (x-1) just note that that's the same as (x+1)/(x+2) * 1/(x-1) and you know that (a/b)*(c/d) = (a*c)/(b*d).
     
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