Given $f(x) = \frac{\sqrt{x-1}}{x-2}$, what is the smallest possible integer value for $x$ such that $f(x)$ has a real number value?
To find the smallest possible integer value for $x$, I need to consider two conditions: $x-1 \geq 0$ and $x-2 \neq 0$. The first condition is necessary because the square root of a negative number is not a real number, and the second condition is necessary because the denominator of a fraction cannot be zero. To satisfy the first condition, I can rewrite it as $x \geq 1$. This means that any integer value of $x$ that is greater than or equal to 1 will make the square root term real. To satisfy the second condition, I can rewrite it as $x \neq 2$. This means that any integer value of $x$ that is not equal to 2 will make the denominator non-zero. Combining the two conditions, I can say that any integer value of $x$ that is greater than or equal to 1 and not equal to 2 will make $f(x)$ a real number. The smallest possible integer value that meets these criteria is 1, because it is the smallest integer that is greater than or equal to 1 and not equal to 2. To check that this is a valid solution, I can plug in $x=1$ into $f(x)$ and see if I get a real number value. $f(1) = \frac{\sqrt{1-1}}{1-2} = \frac{\sqrt{0}}{-1} = \frac{0}{-1} = 0$ 0 is a real number, so $f(1)$ has a real number value, and therefore $x=1$ is the smallest possible integer value for $x$ such that $f(x)$ has a real number value. 