Help with composite function and absolute value please

How do I find f(g(x)) if f(x) = sq root all of x-5 .. and g(x) equals absolute value of all of 4-x^2. The answer is suppose to be sqrt x^2-9

g(x)=|4-x^2| and f(x)=sqrt(x-5)
f(g(x)= =f(|4-x^2|) = sqrt(|4-x^2|-5)
If x^2<=4,g(x)=4-x^2 and f(g(x)=sqrt[4-x^2-5]=sqrt(-x^2-1) which is not real
If x^2>4 , g(x)=x^2-4 and f(g(x)=sqrt(x^2-4-5)=sqrt(x^2-9) which requires x^2>=9 for real values

g(x) = | 4 - x^2 |
f(x) = sqrt(x - 5)

For f(g(x)), substitute g(x) wherever you see x in f(x)'s definition:
f(g(x)) = sqrt(g(x) - 5)

Now expand out what g(x) means:
f(g(x)) = sqrt( |4 - x^2| - 5)

now to do the absolute value, consider two cases
(a) 4 - x^2 < 0, which happens when x^2 > 2, i.e., x < -sqrt(2) or x > sqrt(2)
(b) 4 - x^2 >= 0, which happens when x^2 < 0, i.e., -sqrt(2) <= x <= sqrt(2)

For case (a), f(g(x)) = sqrt( -4 + x^2 - 5) = sqrt(x^2 - 9)
For case (b), f(g(x)) = sqrt( 4 - x^2 - 5) = sqrt(x^2 - 1)

So the answer is
f(g(x)) =
sqrt(x^2 - 9) when x < -sqrt(2) or x > sqrt(2), and
sqrt(x^2 - 1) when -sqrt(2) <= x <= sqrt(2)

You better double check my algebra...