If $f$ is a polynomial, and $f(-2)=3$, $f(-1)=-3=f(1)$, $f(2)=6$, and $f(3)=5$, then what is the minimum possible degree of $f$?
1. **Given Points and Polynomial Degree**:
   We are given the points $f(-2)=3$, $f(-1)=-3$, $f(1)=-3$, $f(2)=6$, and $f(3)=5$. To determine the minimum possible degree of the polynomial $f$, we start by noting that a polynomial of degree $n$ is uniquely determined by $n+1$ points. Since we have 5 points, the polynomial could be of degree at most 4.

2. **Checking for Degree 3**:
   To check if a polynomial of degree 3 could fit these points, we assume $f(x)$ is a cubic polynomial:
   \[
   f(x) = ax^3 + bx^2 + cx + d
   \]
   We need to determine if there exist coefficients $a$, $b$, $c$, and $d$ such that the polynomial passes through the given points.

3. **System of Equations**:
   Substituting the given points into the polynomial, we get the following system of equations:
   \[
   \begin{cases}
   a(-2)^3 + b(-2)^2 + c(-2) + d = 3 \\
   a(-1)^3 + b(-1)^2 + c(-1) + d = -3 \\
   a(1)^3 + b(1)^2 + c(1) + d = -3 \\
   a(2)^3 + b(2)^2 + c(2) + d = 6 \\
   a(3)^3 + b(3)^2 + c(3) + d = 5
   \end{cases}
   \]
   Simplifying these equations, we get:
   \[
   \begin{cases}
   -8a + 4b - 2c + d = 3 \\
   -a + b - c + d = -3 \\
   a + b + c + d = -3 \\
   8a + 4b + 2c + d = 6 \\
   27a + 9b + 3c + d = 5
   \end{cases}
   \]

4. **Solving the System**:
   We solve this system of linear equations to find the coefficients $a$, $b$, $c$, and $d$. However, solving this system shows that there is no consistent solution for $a$, $b$, $c$, and $d$ that satisfies all five equations simultaneously. This indicates that a cubic polynomial cannot fit all the given points.

5. **Conclusion**:
   Since a cubic polynomial (degree 3) cannot fit the given points, the minimum possible degree of the polynomial $f$ must be 4.

The final answer is $\boxed{4}$.