Value of an even, periodic function

Question

Let $f$ be an even function from $R$→$R$, with a period of $a=3$ and such that $f(1)=0$. Determine $f(2)$

So in order to meet these criteria, the function must fulfil two conditions.

$$ f(x+a)=f(x), a=3$$

and $$f(-x)=f(x)$$

also $$f(1)=0$$

The function $$f(x)=x-((x-x)+x)$$

meets these criteria. And thus $$f(2)=0$$

The problem I have, is that there seems to be an extensive pool of different functions, which meet these criteria and this solution seems trivial. What am I missing?

Answer 1

Here's a rigorous solution :

Let $f : \mathbb{R} \mapsto \mathbb{R}$ be an even function. Suppose $f(1) = 0$.

Since $f$ is even, we have $f(-1) = f(1) = 0$.
Since $f$ is periodic with period $3$, we have $f(2) = f(-1 + 3) = f(-1) = 0$.

Answer 2

Since the function is even, $f(-1)=f(1)=0$. Now, $f(2)=f(-1+3) = f(-1)=0$. You are not required to determine the function, nor is it unique.