I am to simplify $f(g(x))$ where $f(x)=\frac{x+1}{x+4}$, $g(x)=\frac{1}{x}$ then find the domain.

The solution provided is $\frac{1+x}{1+4x}$ with domain $x≠0; x≠-\frac{1}{4}$

I was unable to arrive at the simplified solution let alone the domain. I arrived at $\frac{1}{x^2}+\frac{5}{x}+4$ via the following steps:


= $\frac{\frac{1}{x}+1}{\frac{1}{x}+4}$

= $(\frac{1}{x}+4)(\frac{1}{x}+1)$ # multiply out denominator

= $\frac{1}{x^2}+\frac{5}{x}+4$ # used FOIL method to multiply out binomial

How can I arrive at $\frac{1+x}{1+4x}$? Baby, more granular steps appreciated.

  • 1
    $\begingroup$ What? Then $\frac45=20$, multiplying out the denominator? So I give you $\frac12\$$ and you give me $2\$$? $\endgroup$
    – ajotatxe
    Nov 20 '19 at 16:55
  • $\begingroup$ Please explain (or reexamine) your "multiplying out denominator". That is where the error is and if you look at it more closely you might see your error. As it is, I cant actually see what you were trying to do so I can't actually tell you where you made your error (other than "that's just wrong"). $\endgroup$
    – fleablood
    Nov 20 '19 at 17:48
  • $\begingroup$ I suspect you were trying to do something along the lines of $\frac MN = k \implies M = k\cdot N$ but some combined them in a weird and inconsistent way that just didn't work. $\endgroup$
    – fleablood
    Nov 20 '19 at 17:51

It is $$f(g(x))=\frac{\frac{1}{x}+1}{\frac{1}{x}+4}\times \frac{x}{x}=\frac{1+x}{1+4x}$$


I don't understand your "multiplying out the denominator"

You seem to be claiming that $\frac {NUMERATOR}{DENOMINATOR} = (DENOMINATOR)(NUMERATOR)$. Why?

You are correct with $\frac {\frac 1x + 1}{\frac 1x + 4}$ but you have "fractions within fractions" and you want to simplify that. The standard trick is to reduce the fractions by multiplying "top and bottom" by something that will reduce the fractions, usually the demoninator of the fraction in fractions. (I suspect this is what you were trying to do but somehow got it wrong.)

To get rid of the $x$ in the denominators of the fractions we multiply top and bottom by $x$

$\frac {\frac 1x + 1}{\frac 1x + 4}\cdot \frac xx=$

$\frac {x(\frac 1x + 1)}{x(\frac 1x + 4)} =$

$\frac {\frac xx + 1\cdot x}{\frac xx + 4\cdot x}=$

$\frac {1 + x}{1 + 4x}$

And that's it.

  • $\begingroup$ Thank you for the details here, this has helped my understanding. $\endgroup$
    – Doug Fir
    Nov 20 '19 at 18:00

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