Help with a calculus question?

Question

Given f(x)=thesquarerootof(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))

Answer 1

a)
f(x) = sqrt(x^2-9)
x^2-9 >= 0
x^2 >= 9
x >= 3 or x <= -3
when x = +/-3, f(x) = 0
when x -> infinity, f(x) -> +/-infinity
when x -> -infinity, f(x) ->+/-infinity

b)
f(g(x)) = sqrt((1/(x-4))^2-9)
(1/(x-4))^2-9 >=0
(1/(x-4))^2 >= 9
(x-4)^2 <= 1/9
x-4 <= 1/3 or x-4 >=-1/3
x <= 13/3 or x >= 11/3
but, x-4 <> 0 (for g(x))
x <> 4
so, 11/3 <= x <= 13/3 and x<>4
when x =11/3, f(x) = 0
when x -> 4, f(x) ->+/-infinity
when x = 13/3, f(x) = 0

c)
f(f(x)) = sqrt(sqrt(x^2-9)^2 -9)
f(f(x)) = sqrt(x^2-9-9)
f(f(x)) = sqrt(x^2-18)
x^2 -18>0
x^2 > 18
x > 3sqrt(2) or x < -3sqrt(2) ---(1)
but x^2 - 9 > 0 (for inner f(x))
x >= 3 or x <= -3 --- (2)
combine (1) and (2)
so, x >= 3sqrt(2) or x <= -3sqrt(2)
when x = +/-3sqrt(2), f(x) = 0
when x -> infinity, f(x) -> +/-infinity
when x -> -infinity, f(x) ->+/-infinity

Answer 2

Given f(x)=√(x^2-9) and g(x)=1/(x-4), find the domain and range of the following functions.
a. f(x)
x^2-9 ≥ 0
domain: x ≥ 3, x ≤ -3
range: f(x) ≥ 0

b. f(g(x))
x ≠ 4
1/(x-4)^2 - 9 ≥ 0
(x-4)^2 ≤ 1/9
4 -1/3 ≤ x ≤ 4+1/3
domain: 11/3 ≤ x < 4, 4 < x ≤ 13/3
range: f(x) ≥ 0

c. f(f(x))
x^2-9-9 ≥ 0
x^2 ≥ 18
domain:-√18 ≤ x ≤ √18
range: f(x) ≥ 0

Answer 3

Given f(x)=thesquarerootof(x^2-9) and g(x)=1/(x-4), find the domain and range of the following functions.
a. f(x) =Sqrt(x^2 -9) . As x^2 -9 should be positive or zero, so that f(x) is real, x should be greater or equal to 3 .
So: Domain =[3, infinte)
When x=3, f(x) =0 ; When x goes to infinity, f(x) goes to infinity.
So the range of f(x) =[0, infinity)

b. f(g(x)) = f[1/(x -4)] = Sqrt[ 1/(x-4)^2 -9)= Sqrt[1 - 9(x-4)^2] /(x-4)
etc. I am too tired.



c. f(f(x))