5

$f: \mathbb R \to \mathbb R, f(x^2+x+3)+2f(x^2-3x+5)= 6x^2 -10 x + 17 \forall x \in \mathbb R $, then find the function $f(x)$

I have:

$f(15/4)= 16/3$ (both quadratics intersect at $x=1/2$)

$f(3)= 3$ (by substituting $x=0$ and $x=1$ and then solving the simultaneous equations obtained)

$f(5)= 7$

I am not getting anything fruitful from these.

Could anyone provide me a hint on how to proceed?

Edit:

Someone had commented (now it's deleted) that we can assume $f(x) = ax +b$, how can we do that? I got the right answer using that. Is it always okay to assume that way? When is it a reliable assumption?

Archer
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    "Someone has commented that we can assume f(x)=ax+b, how can we do that?" Don't think you can. If the answer were $g(x) = x^2 + 3$ and you were told $g(2x-6) + g(x^2 -7) = x^4 -24x^2 -24x + 85$, I don't think that strategy would help at all. – fleablood Apr 11 '18 at 07:37
  • In that one we can take g to be quadratic otherwise taking cubic or higher makes no sense to solve since it would lead to coefficients of higher terms than x^4 to be non zero . @fleablood so techinique would work there too. – Orion_Pax Jun 04 '22 at 17:05
  • But yeah we would need to be sure of a proof that answer needs to be a polynomial – Orion_Pax Jun 04 '22 at 17:05

2 Answers2

8

Replace $x$ by $1-x$,

\begin{align*} f((1-x)^2+(1-x)+3)+2f((1-x)^2−3(1-x)+5)&=6(1-x)^2−10(1-x)+17\\ f(x^2-3x+5)+2f(x^2+x+3)&=6x^2-2x+13\\ \end{align*}

Solving with $f(x^2+x+3)+2f(x^2-3x+5)= 6x^2 -10 x + 17$, we have $f(x^2+x+3)=2x^2+2x+3$.

$f(x)=2x-3$.

CY Aries
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  • Same method as for https://math.stackexchange.com/questions/2731250/functional-equation-fx-2f-left-dfrac2002x-right-3x – CY Aries Apr 11 '18 at 07:20
  • Thanks,please see the edit too. – Archer Apr 11 '18 at 07:21
  • As in the last question, we can solve the problem if we can have $f(x^2-3x+5)+2f(x^2+x+3)$. So I just let $y^2-3y+5=x^2+x+3$. From this, I have $(y-2)(y-1)=x(x+1)$. If I replace $x$ by $x-2$, then $x^2+x+3$ will become $x^2-3x+5$ but $x^2-3x+5$ will not go to $x^2+x+3$. The correct way should be writing $(1-y)(2-y)=x(x+1)$. That's why I replace $x$ by $1-x$ as it is a self-inverse function. – CY Aries Apr 11 '18 at 07:29
  • But this holds only for $x\geq\frac{11}{4}$, the smallest possible value of $y^2+y+1$. For values smaller than this $f$ may be arbitrary. – Jens Schwaiger Apr 11 '18 at 09:47
  • Yes. The given condition tells nothing about the function for smaller $x$ – CY Aries Apr 11 '18 at 09:50
  • Is there some sort of theory based on why self inverse substitutions interchanges the functions ? @CYAries – Orion_Pax Jun 04 '22 at 16:46
  • @JensSchwaiger yeah indeed its not anything making not sense , since f(x^2+x+3) will only take values in the domain which is in the range of (x^2+x+3) – Orion_Pax Jun 04 '22 at 17:01
  • @Orion_Pax I just noticed that $x^2+x+3=(x+1/2)^2+11/4$ and $x^2-3x+5=(x-3/2)^2+11/4$. – CY Aries Jun 08 '22 at 00:51
  • Thanks understood – Orion_Pax Jun 09 '22 at 06:57
-1

If $x^2-3x+5=0$, $$f(4x-2)+2f(0)=8x-13 $$

If $x^2+x+3=0$ $$f(0)+2f(-4x+2)=-16x-1$$

$$\rightarrow f(4x-2)+f(-4x+2)+\frac52f(0)=\frac{-27}2$$

put $x=\frac12$ $$\frac92f(0)=-27/2$$ $\rightarrow f(0)=-3$

Then $$f(4x-2)=8x-7$$ Put $4x-2=y$ $$\rightarrow f(y)=8(\frac12+\frac y4)-7=2y-3$$

Takahiro Waki
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