Calculate domain of $f(x) = \sqrt{\frac{1}{x-4} + 1 }^2 -9$ I ran this through the MathWorld domain/range widget, and got $x \leq 3$ & x > 4. I understand that x=4 is undefined, therefore the domain of x includes x > 4, but I cannot for the life of me figure out how to determine the $x \leq 3$. When I set up the inequality $\dfrac{1}{x-4} + 1 \leq 0$, I get $x \geq 3$. Anyone know what I'm doing wrong?

Note that $$\frac{1}{x-4} + 1 \geq 0 \\ \Rightarrow \frac{1+(x-4)}{x-4} \geq 0 \\ \Rightarrow \frac{x-3}{x-4} \geq 0$$
Can you figure out the rest? (Sign chart)
