If $f(0)=−10$ and
$f(x)=(6x+4)^2−f(x+2)$ determine $f(3)$ I must be missing something. Thanks.
If $f(0)=−10$ and
$f(x)=(6x+4)^2−f(x+2)$ determine $f(3)$ I must be missing something. Thanks.
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.