Compute the domain of the function  $$f(x)=\frac{1}{\lfloor x^2-7x+13\rfloor}.$$
The discriminant of the quadratic is $7^2-4(13)=-3<0$, so the quadratic has no real roots and is always positive for real inputs.  The function is undefined if $0\leq x^2-7x+13<1$, since $\lfloor x^2-7x+13 \rfloor = 0$ in that case. Since the quadratic is always positive, we consider the inequality $x^2-7x+13<1$.

To find when $x^2-7x+13=1$, subtract 1 from both sides to obtain $x^2-7x+12=0$ and factor as $(x-3)(x-4)=0$, so $x=3$ or $x=4$.  The parabola $x^2-7x+12$ is negative between these points, so we must exclude the interval $(3,4)$ from the domain. So the domain of $f$ is $\boxed{(-\infty,3] \cup [4,\infty)}$.
(-\iny,3]\cup[4,\iny)
Level 5
Algebra
Problem: Compute the domain of the function  $$f(x)=\frac{1}{\lfloor x^2-7x+13\rfloor}.$$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the domain of the function, we need to identify the values of x for which the function is defined. The function f(x) is defined as long as the denominator, which is the floor function of the quadratic expression, is not equal to zero. 

First, we analyze the quadratic expression inside the floor function: 

x^2 - 7x + 13. 

To find when this expression is zero, we can use the quadratic formula:

x = [7 ± sqrt((-7)^2 - 4*1*13)] / (2*1) = [7 ± sqrt(49 - 52)] / 2 = [7 ± sqrt(-3)] / 2.

Since the discriminant is negative, the quadratic has no real roots. This means that the expression x^2 - 7x + 13 is always positive for all real x. 

Next, we evaluate the minimum value of the quadratic to understand the behavior of the floor function. The vertex of the quadratic occurs at x = -b/(2a) = 7/2 = 3.5. 

Calculating the value at the vertex:

f(3.5) = (3.5)^2 - 7(3.5) + 13 = 12.25 - 24.5 + 13 = 0.75.

Since the minimum value of the quadratic is 0.75, the floor function will take values starting from 0. 

The floor function, denoted as ⌊x^2 - 7x + 13⌋, will yield values of 0, 1, 2, etc., but we need to find when it equals zero. The floor function equals zero when:

0 ≤ x^2 - 7x + 13 < 1.

Solving the inequality:

1. For x^2 - 7x + 13 < 1:
   x^2 - 7x + 12 < 0.
   Factoring gives (x - 3)(x - 4) < 0.

   The critical points are x = 3 and x = 4. The quadratic is negative between these points, so:

   3 < x < 4.

2. For the floor function to be zero, we need to exclude the interval (3, 4) from the domain of f(x).

Thus, the domain of f(x) is all real numbers except for the interval (3, 4). In interval notation, this is expressed as:

(-∞, 3] ∪ [4, ∞). 

This reasoning leads us to conclude that the final answer is indeed (-∞, 3] ∪ [4, ∞).<|END_OF_LATENT|> The final answer is $(-\iny,3]\cup[4,\iny)$.