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?
I notice that the function $f(x)$ involves a square root in the numerator, so I need to make sure that the radicand (the expression under the square root) is non-negative. This means that $2x-6 \geq 0$, or equivalently, $x \geq 3$. However, I also notice that the function has a fraction, and I need to avoid dividing by zero. This means that $x-3 \neq 0$, or equivalently, $x \neq 3$. So, combining these two conditions, I get that $x > 3$. The question asks for the smallest possible integer value for $x$, so I need to find the first integer that is larger than 3. I can easily see that this is 4, so the smallest possible integer value for $x$ is 4. 