2

I have a quation regarding composite functions.

if $f\left( \dfrac{x^2+1} x \right)=\dfrac{x^4+1}{x^2}$, then what is $f(x)$?

Michael Hardy
  • 1
Masan
  • 161

2 Answers2

5

Hint: Solve $u=(x^2+1)/x$ for $x$ and then substitute $x=x(u)$ into $f((x^2+1)/x)=f(u)=\frac{x(u)^4+1}{x(u)^2}$.

It is important to note that $x\neq 0$ must hold. A useful hint is also that $x(u)^2=ux-1$ and similarly $x(u)^4=(ux-1)^2=u^2x^2-2ux+1=u^2(ux-1)-2ux+1$.

MrYouMath
  • 15,833
  • Caveat: how do you guarantee that it's the only solution (if it's only, and it's doubtful). – xsnl Apr 03 '17 at 15:42
  • @xsnl: As far as I see there should be two solutions. – MrYouMath Apr 03 '17 at 15:53
  • I mean, there's obviously only two polynomial solutions, and there're a lot of other functions besides polynomials. Why there's no other analytic solutions, for example? – xsnl Apr 03 '17 at 15:57
  • @xsnl: Hmm, that is really an excellent question. Do you have any non-polynomial candidate? They should be all related to the polynomial in some way. – MrYouMath Apr 03 '17 at 16:01
1

HINT: $$\left(\frac{x^2+1} x\right)^2=\frac{x^4+2x^2+1}{x^2}=\frac{x^4+1+2x^2}{x^2}$$ Alternatively you have $$\frac{x^2+1} x = x + \frac 1 x \\\frac{x^4+1}{x^2} = x^2 + \frac 1 {x^2}$$

Michael Hardy
  • 1
kingW3
  • 13,496