Tricky Domain and Range Question

Hello. I have encountered a question that begins simply enough and then ends up confusing me.
Here it is:

Given f(x) = (sqrt)(x^2 - 9) and g(x) = 1/(x-4), find the domain and range of the following functions:
a) f(x)
b) f(g(x))
c. f(f(x))

I solved "a" with a domain of (- infinity, to -3) (3, infinity) and range of all real #'s.

Part "b" I get stuck though. I know f(g(x)) = (sqrt)((1/(x-4))^2 - 9) but then I do not know where to go. Help Please!

Also, I do not get part "c" at all. f(f((x)) = (sqrt)((sqrt)(x^2 -9))^2 -9) ??

For b) you have $\displaystyle f(g(x))=\sqrt{\left(\frac{1}{x-4}\right)^2 - 9}$

The square root function only is defined when the argument is positive, so all you have to do is find x which makes

$\displaystyle \left(\frac{1}{x-4}\right)^2 - 9>0$ true.

Expand the denominator, multiply both sides by it then you'll end with a polynomial - find its roots and the interval in which it is positive. This is the domain of f(g(x)).

For c) you use the same approach.

OK so I did that and got:

(1/(x-2))^2 - 9 > 0
1 / (x^2 -8x +16) - 9 > 0
-9x^2 + 72x -144 > x^2 - 8x + 16
10x^2 - 80x + 160
10(x-4)^2
Isn't it always positive?

note that $\displaystyle y = \sqrt{x^2 - 9}$ has domain $\displaystyle x \le -3$ and $\displaystyle x \ge 3$

for $\displaystyle y = \sqrt{[g(x)]^2 - 9}$ ... domain is $\displaystyle g(x) \le -3$ and $\displaystyle g(x) \ge 3$

substitute $\displaystyle \frac{1}{x-4}$ in for $\displaystyle g(x)$ and finish solving the inequalities.