Question

Determine the domain and range of (g ○ f)(x) if f of x is equal to 4 over the quantity x squared minus 4 end quantity and g(x) = x + 2.
D: {x ∈ ℝ| x ≠ –2, 2} and R: (–∞, 1) ∪ (2, ∞)
D: {x ∈ ℝ| x ≠ –4, 0} and R: (–∞, –2) ∪ (2, ∞)
D: {x ∈ ℝ| x ≠ –4, –2, 0, 2} and R: (–∞, –1) ∪ (2, ∞)
D: {x ∈ ℝ} and R: (–∞, –2) ∪ (2, ∞)

Answer 1

Answer:

I think it's A

Step-by-step explanation:

Since the domain of f(x) is all real numbers except {-2,2}

Answer 2

The domain and range of a function are the possible x and y values of the function.

The domain and the range of the function is: (a) $\mathbf{D:\{x \in R|x \ne -2,2\}}$ and $\mathbf{R:\{(-\infty,1) \cup (2,\infty)\}}$

The functions are given as:

$\mathbf{f(x) = \frac{4}{x^2 - 4}}$

$\mathbf{g(x) = x + 2}$

(g o f)(x) is calculated as:

$\mathbf{(g\ o\ f)(x) = g(f(x))}$

So, we have:

$\mathbf{(g\ o\ f)(x) = \frac{4}{x^2 - 4} + 2}$

Take LCM

$\mathbf{(g\ o\ f)(x) = \frac{4 + 2x^2 - 8}{x^2 - 4} }$

$\mathbf{(g\ o\ f)(x) = \frac{2x^2 - 4}{x^2 - 4} }$

Represent the denominator as follows, to calculate the domain

$\mathbf{x^2 - 4 \ne 0 }$

Add 4 to both sides

$\mathbf{x^2 \ne 4 }$

Take square roots of bot sides

$\mathbf{x \ne \±2 }$

Hence, the domain of the function is:

$\mathbf{D:\{x \in R|x \ne -2,2\}}$

On the graph of $\mathbf{(g\ o\ f)(x) = \frac{2x^2 - 4}{x^2 - 4} }$ (see attachment), the function does not have a value from y = 1 to 2.

Hence, the range is:

$\mathbf{R:\{(-\infty,1) \cup (2,\infty)\}}$

Read more about domain and range at:

brainly.com/question/1632425