# Thread: Simplifying a composite function.

1. ## Simplifying a composite function.

I was hoping someone could help me out with the following problem:

f(g(x))

f=(1+x)/2x
g=x/(3x+2)

I need to simplify that and find out the domain of the composite.

For the life of me I can't remember how exactly to work these two fractions together. I haven't done something like that in years, and the only thing that makes sense to me is to replace all the x's in g with f's equation, but then I get a mess of fractions. Still I went with it and got (x-1)/x, so the domain is all real numbers but 0, but I can't help but feel that something is wrong.

2. ## Re: Help with a problem

First simplify this,

$\displaystyle f(g(x))= \frac{1+\frac{x}{3x+2}}{2\left(\frac{x}{3x+2}\righ t)}$

3. ## Re: Help with a problem

Okay... I first put the 1 to 3x+2/3x+2 and added it to x/3x+x to get 3x+x+2/3x+2

For the bottom, I just multiplied by 2 and got 2x/3x+2. I'm not sure what the t is doing there so I just ignored it, otherwise I would have just put it in the numerator.

Then I multiplied the top fraction by the bottom one (with the numerator/denominator reversed)... and at this point I'm guessing I've messed up as I got

9x^2 + 3x^2 + 6x + 6x + 2x + 4 all over 5x^2 + 4x

Then I ended up with 2(6x^2 + 7x + 2)/x(5x + 4)

I actually understand what I'm going in my calculus class right now, so I guess I'm just rusty a few algebra rules, which I haven't really used for a while now. Like how to replace fractions in variables inside fractions with variables, apparently.

4. ## Re: Help with a problem

Oh wait, I think I got it! This time I redid it (I didn't add 1 and multiply 2 to both numerator and denominator last time) and got

(4x + 2)/(6x + 4) / (2x)/(6x + 4)

Then I multiplied both top and bottom fractions by (6x + 4) to get rid of those denominators, and then simplifying what I had left I got (2x + 1)/x.

This would make the domain all real numbers except 0, right?

And I just had a question now... why is it we can't multiply the first fraction by the reverse of the second, like when normally dividing fractions? That seemed the most obvious solution to me but I keep getting weird answers, so that must not be right.

5. ## Re: Help with a problem

$\displaystyle f(g(x))= \frac{1+\frac{x}{3x+2}}{\frac{2x}{3x+2}}$

$\displaystyle = \frac{\frac{3x+2}{3x+2}+\frac{x}{3x+2}}{\frac{2x}{ 3x+2}}$

$\displaystyle = \frac{\frac{4x+2}{3x+2}}{\frac{2x}{3x+2}}$

$\displaystyle = \frac{4x+2}{3x+2}\times \frac{3x+2}{2x}$

$\displaystyle = \frac{4x+2}{2x}$

$\displaystyle = \frac{4x}{2x}+\frac{2}{2x}$

$\displaystyle = 2+\frac{1}{x}$

Same as you have...