<s>[INST] Finding the domain of $f(x)=\sqrt\frac{x^2+6}{x-4}$ 
Given $f(x)=\sqrt{\dfrac{x^2+6}{x-4}}$, find the domain for the function $f(x)$.

My attempt failed:
So the first constraint is $\frac{x^2+6}{x-4}\geq0$ because of the radical.
Now, multiplying both sides by $x-4$ I get $x^2+6\geq0, x\neq4$.
And now $x^2\geq-6$, so $x\in\mathbb{R}-\{4\}$
But if I were to plug in say 3 for x I encounter a problem because in this case I get $\frac{x^2+6}{x-4}\lt0$
 [/INST] $x^{2}+6 >0$ for all $x$ so $\frac {x^{2}+6} {x-4}$ exists and is $ \geq 0$ iff $x>4$.
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