<Question>
What is the domain of the real-valued function $f(x)=\frac{2x-7}{\sqrt{x^2-5x+6}}$?
</Question>

<think>
<Plan_1>
Okay, so I need to find the domain of the function f(x) = (2x - 7)/sqrt(x² - 5x + 6). Hmm, let's start by recalling that the domain of a function includes all the real numbers x for which the function is defined. Since this is a rational function with a square root in the denominator, there are a couple of things to consider here.

<prompt_1.1> Identify the conditions for the domain of the given function, considering the square root in the denominator. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, the denominator can't be zero because division by zero is undefined. Also, since the square root is in the denominator, the expression inside the square root must be positive (because square root of a negative number isn't a real number), so the expression under the square root, which is x² - 5x + 6, must be greater than zero. Additionally, the numerator, 2x - 7, doesn't affect the domain as long as it's defined, which it is for all real numbers. So the main focus is on the denominator's square root expression.

So, to summarize, I need to find all x such that x² - 5x + 6 > 0.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Let me solve the inequality x² - 5x + 6 > 0. To do this, I should first factor the quadratic expression.

<prompt_2.1> Solve the inequality x² - 5x + 6 > 0 by factoring the quadratic expression. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
Let me try factoring x² - 5x + 6. Looking for two numbers that multiply to 6 and add up to -5. Hmm, the factors of 6 are 1 and 6, 2 and 3. Since the middle term is -5x, and the constant term is +6, the factors should both be negative because they add up to -5. So, -2 and -3 multiply to 6 and add to -5. Therefore, x² - 5x + 6 factors as (x - 2)(x - 3).

Wait, let me check that: (x - 2)(x - 3) = x² - 3x - 2x + 6 = x² -5x +6. Yep, that's correct.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
So, the inequality becomes (x - 2)(x - 3) > 0. Now, to solve this inequality, I need to find the intervals where the product of (x - 2) and (x - 3) is positive. The critical points are x = 2 and x = 3, which divide the real number line into three intervals: (-∞, 2), (2, 3), and (3, ∞).

I can test each interval to see the sign of the product in each interval.

<prompt_3.1> Test the intervals divided by the critical points to determine where the product (x - 2)(x - 3) is positive. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
First interval: x < 2. Let's pick x = 1. Plugging into (1 - 2)(1 - 3) = (-1)(-2) = 2, which is positive. So the product is positive here.

Second interval: 2 < x < 3. Let's take x = 2.5. Then (2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25, which is negative.

Third interval: x > 3. Let's choose x = 4. (4 - 2)(4 - 3) = (2)(1) = 2, which is positive.

So the inequality (x - 2)(x - 3) > 0 holds when x is in (-∞, 2) or (3, ∞).
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
But wait, I need to check if the inequality is strict or not. Since the original inequality is x² -5x +6 > 0, not ≥, the points where the expression equals zero, which are x = 2 and x = 3, are excluded from the domain. So, those points are vertical asymptotes or points where the function is undefined.

Therefore, putting it all together, the domain of f(x) is all real numbers x such that x < 2 or x > 3.

But let me just double-check to make sure I didn't make any mistakes. Let's verify with a value in each interval.

<prompt_4.1> Verify the solution by checking the values in each interval to ensure the inequality (x - 2)(x - 3) > 0 holds. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
For x = 0 (which is in (-∞, 2)): sqrt(0² -5*0 +6) = sqrt(6) which is defined, so denominator is sqrt(6), numerator is -7, so f(0) = -7/sqrt(6), which is a real number. So x=0 is okay.

For x = 2.5 (which is in (2,3)): sqrt(2.5² -5*2.5 +6). Let's compute inside the sqrt: 6.25 -12.5 +6 = -0.25. So sqrt of negative number is undefined, which is why x=2.5 isn't in the domain. Correct.

For x = 4 (in (3, ∞)): sqrt(16 -20 +6) = sqrt(2), which is defined. So f(4) = (8 -7)/sqrt(2) = 1/sqrt(2), which is fine.

Therefore, the domain is indeed x in (-∞, 2) union (3, ∞).
</execution_4.1>
</Execution_4>

<Final_answer>
Therefore, the domain of the function $f(x)$ is $\boxed{(-\infty, 2) \cup (3, \infty)}$.
</Final_answer>
</think>