The function $f(x)$ satisfies
\[f(x + y) = f(x) f(y)\]for all real numbers $x$ and $y.$  Find all possible values of $f(0).$

Enter all the possible values, separated by commas.
Setting $x = y = 0,$ we get
\[f(0) = f(0)^2,\]so $f(0) = 0$ or $f(0) = 1.$  The constant functions $f(x) = 0$ and $f(x) = 1$ show that both $\boxed{0,1}$ are possible values of $f(x).$
01
Level 3
Intermediate Algebra
Problem: The function $f(x)$ satisfies
\[f(x + y) = f(x) f(y)\]for all real numbers $x$ and $y.$  Find all possible values of $f(0).$

Enter all the possible values, separated by commas.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the possible values of \( f(0) \), we start by substituting \( x = 0 \) and \( y = 0 \) into the functional equation \( f(x + y) = f(x) f(y) \). This gives us:

\[ f(0 + 0) = f(0) f(0) \]
\[ f(0) = f(0)^2 \]

This equation implies that \( f(0) \) is a fixed point of the function \( g(t) = t^2 \). The solutions to the equation \( f(0) = f(0)^2 \) can be derived by rearranging it:

\[ f(0)^2 - f(0) = 0 \]
\[ f(0)(f(0) - 1) = 0 \]

From this factorization, we see that \( f(0) \) can either be \( 0 \) or \( 1 \). 

Next, we need to verify if both values are consistent with the original functional equation. 

1. If \( f(0) = 0 \):
   Substituting \( y = 0 \) into the original equation gives:
   \[ f(x + 0) = f(x) f(0) \]
   \[ f(x) = f(x) \cdot 0 \]
   This implies \( f(x) = 0 \) for all \( x \). Thus, \( f(x) = 0 \) is a valid solution.

2. If \( f(0) = 1 \):
   Again substituting \( y = 0 \) gives:
   \[ f(x + 0) = f(x) f(0) \]
   \[ f(x) = f(x) \cdot 1 \]
   This does not impose any restrictions on \( f(x) \). 

To explore further, we can consider the case where \( f(x) \) is not identically zero. A common solution to the functional equation is \( f(x) = e^{kx} \) for some constant \( k \). In this case, we find:

\[ f(0) = e^{k \cdot 0} = e^0 = 1 \]

Thus, both values \( 0 \) and \( 1 \) are consistent with the functional equation. 

In conclusion, the possible values of \( f(0) \) are \( 0 \) and \( 1 \). Therefore, the final answer is \( 01 \).<|END_OF_LATENT|> The final answer is $01$.