Answer:
I think it's A
Step-by-step explanation:
Since the domain of f(x) is all real numbers except {-2,2}
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) [tex]\mathbf{D:\{x \in R|x \ne -2,2\}}[/tex] and [tex]\mathbf{R:\{(-\infty,1) \cup (2,\infty)\}}[/tex]
The functions are given as:
[tex]\mathbf{f(x) = \frac{4}{x^2 - 4}}[/tex]
[tex]\mathbf{g(x) = x + 2}[/tex]
(g o f)(x) is calculated as:
[tex]\mathbf{(g\ o\ f)(x) = g(f(x))}[/tex]
So, we have:
[tex]\mathbf{(g\ o\ f)(x) = \frac{4}{x^2 - 4} + 2}[/tex]
Take LCM
[tex]\mathbf{(g\ o\ f)(x) = \frac{4 + 2x^2 - 8}{x^2 - 4} }[/tex]
[tex]\mathbf{(g\ o\ f)(x) = \frac{2x^2 - 4}{x^2 - 4} }[/tex]
Represent the denominator as follows, to calculate the domain
[tex]\mathbf{x^2 - 4 \ne 0 }[/tex]
Add 4 to both sides
[tex]\mathbf{x^2 \ne 4 }[/tex]
Take square roots of bot sides
[tex]\mathbf{x \ne \±2 }[/tex]
Hence, the domain of the function is:
[tex]\mathbf{D:\{x \in R|x \ne -2,2\}}[/tex]
On the graph of [tex]\mathbf{(g\ o\ f)(x) = \frac{2x^2 - 4}{x^2 - 4} }[/tex] (see attachment), the function does not have a value from y = 1 to 2.
Hence, the range is:
[tex]\mathbf{R:\{(-\infty,1) \cup (2,\infty)\}}[/tex]
Read more about domain and range at:
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