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if $f(3x-1)=9x^2+6x-7$ determine all the $f(x)$ functions. I tried in this way : $t=3x-1 \Rightarrow x=(t+1)/3$ $f(t)=9(t+1)^2/9-6((t+1)/3)-7((t+1)/3)\ldots$ but unfortunately I get the original function. Thanks in advance.

user1729
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Liviu Solcovenco
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  • You do not get the original function after substituting. However you did make a mistake in substitution, in the last (constant) term, which just remains -7 after substitution. – hardmath Oct 14 '13 at 14:26

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Using your method, substituting $t = 3x - 1 \iff x = \dfrac {t+1}{3} $, we have $$\begin{align} f(t) & = \dfrac{9(t+1)^2}{9} + \dfrac{6(t+1)}{3} - 7 \\ \\ & = (t+1)^2 +2(t+1) - 7 \\ \\ &= t^2 + 4t - 4 \end{align}$$

Hence, $$f(x) = x^2 + 4x - 4$$

amWhy
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  • Thanks for the reply. I doubt wether it is correct in the sense that t≠x (t=3x-1) . – Liviu Solcovenco Oct 14 '13 at 14:34
  • Liviu - This is indeed correct. You use the substitution to determine what $f(x)$ is by using a "dummy variable" t. Use the final equation, and evaluate $f(3x - 1)$ by substituting $3x - 1$ everywhere you see $x$ in $f(x) = x^2 + 4x - 4$, and you'll get the original equation, $f(3x-1)$. – amWhy Oct 14 '13 at 14:41
  • amWhy, thank you very much ! – Liviu Solcovenco Oct 14 '13 at 14:45
  • I learned something too. This method is nice. My method was over complicated and thus not worth publishing. My question is though after some experimenting, how would you approach such a problem if t cannot be expressed in terms of x? Or does that fall outside the boundaries of "doable"? – imranfat Oct 14 '13 at 15:39