Help please

Since b^2-4ac = -3 for $x^2-7x+13$, that means $x^2-7x+13$ has no zeroes.

So, the domain of $f(x)=\frac{1}{x^2-7x+13}$ is R.

x  can be all real numbers except those that make  $\lfloor x^2 - 7x + 13\rfloor = 0$

...which means...

x  can be all real numbers except those that make  $0\ \leq\ x^2-7x+13\ <\ 1$

To find the  x  values that make the inequality true, it helps to look at a graph of  y = x² - 7x + 13:

We can see that there are no  x  values which make  y  equal  0 , just like Davis found.

But we can also see that there are some  x  values which make  y  fall between  0  and  1

Let's find what  x  values make  y  equal to  1

1  =  x² - 7x + 13

                                               Subtract  1  from both sides of the equation.

0  =  x² - 7x + 12

                                               Factor the right side of the equation.

0   =  (x - 3)(x - 4)

                                               Set each factor equal to zero and solve for  x

x - 3  =  0     or     x - 4  =  0

  x  =  3        or       x  =  4

Now we can see that if  x  is between  3  and  4,  y  will be between  0  and  1

So the domain is all real numbers except those exclusively between  3  and  4

In interval notation, the domain is  $(-\infty,3)\cup(4,\infty)$_