Let's take this step by step to solve the given problem.
First, recall the functions [tex]\( f(x) = x^3 + 2 \)[/tex] and [tex]\( g(x) = x - 9 \)[/tex].
We are asked to perform the operation [tex]\(\left(\frac{f}{g}\right)(x)\)[/tex], which means we need to find:
[tex]\[
\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x^3 + 2}{x - 9}
\][/tex]
The domain restriction for this function occurs where the denominator is zero because division by zero is undefined. Therefore, we need to find the value of [tex]\( x \)[/tex] that makes the denominator [tex]\( g(x) = 0 \)[/tex]:
[tex]\[
x - 9 = 0 \implies x = 9
\][/tex]
Thus, we have the domain restriction [tex]\( x \neq 9 \)[/tex].
Now we can analyze the given options:
1. [tex]\(\frac{x^3 + 2}{x - 9}, x \neq -9\)[/tex]
2. [tex]\(\frac{x - 9}{x^3 + 2}, x \neq 9\)[/tex]
3. [tex]\(\frac{x^3 + 2}{x - 9}, x \neq 9\)[/tex]
4. [tex]\(\frac{x - 9}{x^3 + 2}, x \neq -9\)[/tex]
We already found that the correct form for [tex]\( \left(\frac{f}{g}\right)(x) \)[/tex] is:
[tex]\[
\left(\frac{f}{g}\right)(x) = \frac{x^3 + 2}{x - 9}
\][/tex]
with the domain restriction [tex]\( x \neq 9 \)[/tex].
Now, let's evaluate the results of different options with specific input values for clarity:
Option 1: [tex]\(\left(\frac{f}{g}\right)(-9)\)[/tex]
[tex]\[
\left(\frac{-9^3 + 2}{-9 - 9}\right) = \frac{-729 + 2}{-18} = \frac{-727}{-18} = 40.388888888888886
\][/tex]
Option 2: [tex]\(\left(\frac{g}{f}\right)(9)\)[/tex]
[tex]\[
\left( \frac{9 - 9}{9^3 + 2}\right) = \frac{0}{729 + 2} = 0.0
\][/tex]
Option 3: [tex]\(\left(\frac{f}{g}\right)(9)\)[/tex]
[tex]\[
\frac{9^3 + 2}{9 - 9} = \frac{729 + 2}{0} \quad \text{undefined, as division by zero is not possible}
\][/tex]
Option 4: [tex]\(\left(\frac{g}{f}\right)(-9)\)[/tex]
[tex]\[
\frac{-9 - 9}{(-9)^3 + 2} = \frac{-18}{-727} = 0.024759284731774415
\][/tex]
Given the correct domain restriction and the correct computed forms, the accurate options should be [tex]\(\frac{x^3 + 2}{x - 9}, x \neq 9\)[/tex].
Therefore, the correct answer is:
[tex]\[
\boxed{\frac{x^3 + 2}{x - 9}, x \neq 9}
\][/tex]
