Christmas Streak 65/88: Functional Equation #1

Substituting $x=3$ into the equation gives

$f(3)+2f(-2)=27 \dots (1)$

Substituting $x=-2$ into the equation gives

$f(-2)+2f(3)=12 \dots (2)$

Treating (1) and (2) as simultaneous equations allows us to eliminate f(-2) to find that

$\boxed{f(3)=-1}$

Let's try to find a closed form for $f(x)$ :

We know that $f(x)+2f(1-x)=3x^2$ (A)

Substituting $x$ by $1-x$ , we get : $f(1-x)+2f(x)=3(1-x)^2$ (B)

Computing 2(B)-(A), we obtain : $3f(x)=6(x-1)^2-3x^2$ , so $f(x)=x^2-4x+2$

Thus $f(3)=-1$