\(\text{Compute the domain of the function $f(x)=\frac{1}{\lfloor x^2-7x+13\rfloor}.$}\)
+501 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.
+9481 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 \((-\infty,3)\cup(4,\infty)\)_
