### Instruction: 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.### Response: To find the possible values of $f(0),$ I need to think about what this equation implies about the function $f.$
One thing I notice is that if I plug in $x = 0,$ I get
\[f(0 + y) = f(0) f(y)\]which simplifies to
\[f(y) = f(0) f(y)\]for any $y.$
This means that either $f(0) = 1$ or $f(y) = 0$ for all $y.$
If $f(0) = 1,$ then the equation becomes
\[f(x + y) = f(x) f(y)\]for any $x$ and $y.$
This is a familiar property of the exponential function, so I suspect that $f(x) = a^x$ for some constant $a.$
To check this, I can plug in $x = y = 1$ and get
\[f(2) = f(1)^2\]which means that $a^2 = a^2,$ which is true for any $a.$
Similarly, I can plug in $x = 1$ and $y = -1$ and get
\[f(0) = f(1) f(-1)\]which means that $1 = a^{-1} a,$ which is also true for any $a.$
So, it seems that any function of the form $f(x) = a^x$ satisfies the equation if $f(0) = 1.$
On the other hand, if $f(y) = 0$ for all $y,$ then the equation becomes
\[f(x + y) = 0\]for any $x$ and $y.$
This is also true for the constant function $f(x) = 0.$
So, another possible function that satisfies the equation is $f(x) = 0.$
In summary, the possible values of $f(0)$ are either $1$ or $0.$