\(\text{Compute the domain of the function $f(x)=\frac{1}{\lfloor x^2-7x+13\rfloor}.$}\)

Guest Jul 19, 2019

#1**+2 **

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.

Davis Jul 19, 2019

#2**+5 **

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 = x^{2} - 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 = x^{2} - 7x + 13

Subtract 1 from both sides of the equation.

0 = x^{2} - 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 \((-\infty,3)\cup(4,\infty)\)_

hectictar Jul 20, 2019