I am little confused as to how I can solve rational polynomials such as $\ f\circ f \ $ if $f (x) = x + \frac{1}{x} $.

$$f(f(x)) = x + \frac{1}{x}+ \frac {1}{x+1/x}$$

Am I only allowed to multiply the numerator and denominator by the LCD or can I multiply by the reciprocal of the second term and then begin determining the function? It seems that I can but I don't know if that is just a fluke for this problem.

I then multiply each side by x to have equal denominators and numerators for both terms before adding them.

Finally I add the two terms and get $\dfrac {x^4+2x}{x} \ $

Am I correct?

  • $\begingroup$ By "solve", do you mean "simplify"? $\endgroup$ Commented Oct 4, 2014 at 16:05
  • $\begingroup$ Yes, I guess. My math teachers have told me to avoid saying simplify since it can mean many things to many different people. $\endgroup$
    – Cetshwayo
    Commented Oct 4, 2014 at 16:09

1 Answer 1


First, work on the third term to clear the fraction in the denominator, and we can do this by multiplying by $1 = \frac xx$:

$$\frac{1}{x+\frac 1x} = \left(\frac xx\right)\cdot \frac 1{x + \frac 1x} = \frac{x}{x(x + \frac 1x)} = \frac x{x^2 + 1}$$

Now, we can find the common denominator of all terms (lcm) $= x(x^2+1)$.

$$x + \frac 1x + \frac x{x^2+1} = \frac{x^2(x^2 + 1)}{x(x^2+1)} + \frac {x^2 + 1}{x(x^2 + 1)} + \frac{x^2}{x(x^2 + 1)} = \frac{x^2(x^2 + 1) + (x^2 + 1) + x^2}{x(x^2 + 1)}$$

Now, it's simply a matter of simplifying the numerator, to get $$\frac{x^4 + 3x^2 +1}{x(x^2 + 1)}$$

  • $\begingroup$ Okay, since you are multiplying one part of the polynomial in the numerator by x, why don't you do the same to the right side? Isn't that a rule. You must distribute the x. $\endgroup$
    – Cetshwayo
    Commented Oct 4, 2014 at 16:11
  • $\begingroup$ I multiplied the term by $\frac xx = 1$: I multiplied both numerator AND denominator by $x$. $\endgroup$
    – amWhy
    Commented Oct 4, 2014 at 16:15
  • $\begingroup$ So the answer is $\frac {x^4+3x^2+1}{x^2+1} \ $ $\endgroup$
    – Cetshwayo
    Commented Oct 4, 2014 at 16:20
  • $\begingroup$ Almost: we need for the denominator to be $x(x^2+1)$. $\endgroup$
    – amWhy
    Commented Oct 4, 2014 at 16:21
  • $\begingroup$ So I can multiply the x through the numerator and denominator as the first step? $\endgroup$
    – Cetshwayo
    Commented Oct 4, 2014 at 16:22

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