Question

(35 POINTS!!) Given f of x is equal to 1 over the quantity x minus 3 and g of x is equal to the square root of the quantity x plus 3, what is the domain of f (g(x))?

[–3, ∞)
[–3, 6) ∪ (6, ∞)
(–∞, 3) ∪ (3, ∞)
ℝ

Answer 1

the second option

[–3, 6) ∪ (6, ∞)

Explanation:

f(x) = 1/(x - 3)

the domain is every value of x that does not turn the denominator to 0. so, x <> 3.

g(x) = sqrt(x + 3)

the domain is every value of x that does not turn (x+3) into a negative value. so, x >= -3.

f(g(x)) = 1/(sqrt(x + 3) - 3)

the domain is all values of x that are valid for g(x) AND that do not generate range values of g(x) that violate the domain of f(x).

so, it is (x >= -3) AND (sqrt(x + 3) <> 3).

and sqrt(x + 3) = 3, if (x + 3) = 9.

that means x = 6.

so, 6 is forbidden, every other value of x is allowed here.

therefore, the domain of f(g(x)) is

(x >= -3) AND (x <> 6)

and that is equivalent to

[-3, 6) u (6, infinity)

everything larger or equal to -3 and different from 6.