A real-valued function $f$ satisfies the relation $f(x)f(y)=f(2xy+3)+3f(x+y)-3f(y)+6y$, for all real numbers $x$ and $y$, then the value of $f(8)$ is?

Question

My approach for this problem :-
Let $x=0$ and $y=1$ then we get $f(0)f(1)=f(3)+6$.
Let $x=1$ and $y=0$ then we get $f(1)f(0)=f(3)+3f(1)-3f(0)$
Using the above two extracted equations we can say that :-
$f(1)f(0)=f(3)+3f(1)-3f(0)$
$\Rightarrow f(3)+6 = f(3)+3f(1)-3f(0)$
$\Rightarrow f(3)+6 = f(3)+3f(1)-3f(0)$
$\Rightarrow f(1)=2+f(0)$

Like this I tried putting the values as $x=0$ and $y=-1$ & $x=-1$ and $y=0$ but I couldn't get to any useful equation that will help me to solve this.

OK ! So I have been practicing questions on functions for a while and I came across these kinds of problems where they give you an equation of function(s) and they ask you to find the value of the function at a given certain value and I have been getting stuck at these questions every now and then. It seems that for these questions there are no hard and fast rule to solve these and one has to manipulate the given problem in one way or the another to get to the solution.
But is it like that? Is there a methodology, an approach or a way to solve these kinds of problem?

Need help with this problem!

Thanks in advance !

Answer 1

We can generalise your argument.

Set $x := y$ and $y := x$.

We get $f(y)f(x) = f(2xy+3) + 3f(x+y) - 3f(x) + 6x$

Which means that

$$\require{cancel}\bcancel{f(2xy + 3)} + \bcancel{3 f(x+y)} - 3f(y) + 6y = \bcancel{f(2xy + 3)} + \bcancel{3 f(x+y)} - 3f(x) + 6x$$

$$ \frac{f(y) - f(x)}{y - x} = 2 $$ for all reals $x$ and $y$

This implies that our function has a constant slope $2$.

$f(x) = 2x + b \ \ ;b \in \mathbb R$

We just need to find which values of $b$ work.

Substituting $x = 0 = y$ as Severin Schraven suggested we get,

$$ f(0)^2 = f(3) \\ (2\times 0 + b)^2 = 2 \times 3 + b \\ b^2 - b - 6 = 0 \\ (b + 2) (b - 3) = 0 $$

As such, $b = -2$ or $b = 3$.

Now let's substitute $x = 1 = y$: $$ f(1)^2 = f(5) + 3 f(2) - 3f(1) + 6 \\ (2 \times 1 + b)^2 = (2 \times 5 + b) + 3(2 \times 2 + b) - 3(2 \times 1 + b) + 6 \\ 4 + 4 b + b^2 = 10 + b + 12 \xcancel{+ 3b} \bcancel{- 6} \xcancel{- 3b} \bcancel{+ 6} \\ b^2 + 3b -18 = 0 \\ (b+6)(b-3) = 0 $$ As such, $b = -6$ or $b = 3$. This means that $b \in \{ -2, 3 \} \cap \{ -6, 3\} \implies b = 3$.

We have to verify that the function $f(x) = 2x + 3$ satisfies the functional equation. That shouldn't be a problem.

$f(8) = 2 \times 8 + 3 = 19$

Now, as for the motivation for this approach:

We can see that some of the terms in our function are symmetric but other terms are not. In particular, the left hand side is entirely symmetric.

Incidentally enough, the symmetric terms are the hardest ones to manipulate so we can exploit their symmetry to cancel them out.