0
$\begingroup$

If $f(0)=−10$ and

$f(x)=(6x+4)^2−f(x+2)$ determine $f(3)$ I must be missing something. Thanks.

  • 0
    Indeed, the simple recursion formula determines $f(x)$ for $x\in\mathbb 2Z$. One can choose $f(3)$ arbitrarily (and define $f(x)$ for odd integers $x$ via the recursion formula).2012-12-09
  • 0
    So it could be anything?2012-12-09
  • 0
    If you did not leave out anything important from the problem statement, then yes.2012-12-09
  • 0
    I dont recall leaving anything out only that f(x) is a polynomial2012-12-09
  • 0
    "only"!?!?!?!?!?!?!2012-12-09
  • 0
    haha sorry about that2012-12-09
  • 0
    I thought you knew! :)2012-12-09
  • 0
    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6671/discussion-between-fosho-and-hagen-von-eitzen)2012-12-09
  • 0
    anything? please help me out.2012-12-09

1 Answers 1

0

Under the additional constraint that $f$ is a polynomial, we can calculate $f(3)$.

Note that if $$f(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots +a_1x+a_0$$ (with $a_n\ne 0$) is a polynomial of degree $n$, then $f(x+2)$ is also a polynomial of degree $n$ and has the same leading term, i.e. is of ther form $$f(x+2)=a_nx^n+b_{n-1}x^{n-1}+\ldots +b_1 x+b_0.$$ Therefore $$36x^2+48x+16 = (6x+4)^2=f(x)+f(x+2)= 2a_nx^n+\ldots$$ implies that $n=2$ and $a_2=18$. Thus $$f(x)=18x^2+bx+c.$$ This leads to
$$\begin{align}36x^2+48x+16&=f(x)+f(x+2)\\ &=(18x^2+bx+c)+(18(x+2)^2+b(x+2)+c)\\ &=36x^2+(2b+72)x+(2b+2c+72) \end{align},$$ which implies $2b+72=48$ and $2b+2c+72=16$, hence $b=-12$ and $c=-16$. Thus $$f(3)=36\cdot 9-12\cdot 3-16=110.$$


Remark: We did not use the given fact that $f(0)=-10$. In fact, we could determine $f$ without it and obtain $f(0)=-16$. Thus all in all the problem statement contradicts itself.