Question

Determine the domain and range of (g ○ f)(x) if f of x is equal to 9 over the quantity x squared minus 9 end quantity and g(x) = x + 3.

A. D: {x ∈ ℝ} and R: (–∞, –3) ∪ (3, ∞)
B. D: {x ∈ ℝ| x ≠ –6, 0} and R: (–∞, –3) ∪ (3, ∞)
C. D: {x ∈ ℝ| x ≠ –3, 3} and R: (–∞, 2) ∪ (3, ∞)
D. D: {x ∈ ℝ| x ≠ –6, –3, 0, 3} and R: (–∞, –2) ∪ (2, ∞)

Answer 1

The domain is all real numbers except -6 and 0. Hence, domain D: {x ∈ ℝ| x ≠ –6, 0}

Hence the range is all real numbers except -3 and 3. Hence, range R: (–∞, –3) ∪ (3, ∞)

Given the following functions:

$f(x)=\frac{9}{x^2-9}\\g(x)=x+3$

First we need to get the composite function f(g(x))

$f(g(x)) = f(x+3)\\f(x+3)=\frac{9}{(x+3)^2-9} \\f(x+3)=\frac{9}{x^2+6x+9-9} \\f(x+3)=\frac{9}{x^2+6x}$

Get the domain

The domain the values of x for which the function exists. The function cannot exists at when x = -6 and x = 0

• Hence the domain is all real numbers except -6 and 0. Hence, domain D: {x ∈ ℝ| x ≠ –6, 0}

The range is the value of y for which the function exists. The function cannot exists at when x = -6 and x = 0

• Hence the range is all real numbers except -3 and 3. Hence, range R: (–∞, –3) ∪ (3, ∞)

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