Given $f(x) = \frac{\sqrt{2x-6}}{x-3}$, what is the smallest possible integer value for $x$ such that $f(x)$ has a real number value?
In order for $f(x)$ to have a real number value, the expression inside the square root in the numerator must be non-negative and the denominator must not be 0. So we have the two conditions $2x-6\ge0 \Rightarrow x \ge 3$ and $x-3 \ne 0 \Rightarrow x \ne 3$. We see that $x=\boxed{4}$ is the smallest integer value that satisfies both conditions.