[tex]\[
\begin{array}{l}
f(x)=\sqrt[3]{3 x} \\
g(x)=3 x+2
\end{array}
\][/tex]

Find [tex]\(\left(\frac{f}{g}\right)(x)\)[/tex]. Include any restrictions on the domain.

A. [tex]\(\left(\frac{f}{g}\right)(x)=\frac{\sqrt[3]{3 x}}{3 x+2}, x \neq -\frac{2}{3}\)[/tex]

B. [tex]\(\left(\frac{f}{g}\right)(x)=\frac{\sqrt[3]{3 x}}{3 x+2}, x \neq \frac{3}{2}\)[/tex]

C. [tex]\(\left(\frac{f}{g}\right)(x)=\frac{3 x+2}{\sqrt[3]{3 x}}, x \neq 0\)[/tex]

D. [tex]\(\left(\frac{f}{g}\right)(x)=\frac{3 x+2}{\sqrt[3]{3 x}}, x \geq 0\)[/tex]



Answer :

To solve the given problem, we need to find [tex]\(\left(\frac{f}{g}\right)(x)\)[/tex] for the given functions [tex]\( f(x) = \sqrt[3]{3x} \)[/tex] and [tex]\( g(x) = 3x + 2 \)[/tex]. Additionally, we need to determine any restrictions on the domain.

Let's go through the solution step-by-step:

### Step 1: Define the Functions

Given:
[tex]\[ f(x) = \sqrt[3]{3x} \][/tex]
[tex]\[ g(x) = 3x + 2 \][/tex]

### Step 2: Form the Function [tex]\(\left(\frac{f}{g}\right)(x)\)[/tex]

To find [tex]\(\left(\frac{f}{g}\right)(x)\)[/tex], we divide [tex]\( f(x) \)[/tex] by [tex]\( g(x) \)[/tex]:
[tex]\[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt[3]{3x}}{3x + 2} \][/tex]

### Step 3: Determine Restrictions on the Domain

For the fraction [tex]\(\frac{\sqrt[3]{3x}}{3x + 2}\)[/tex] to be defined, the denominator [tex]\( g(x) \)[/tex] must not be zero. Therefore, we set:
[tex]\[ g(x) \neq 0 \][/tex]
[tex]\[ 3x + 2 \neq 0 \][/tex]

Now, solve for [tex]\( x \)[/tex]:
[tex]\[ 3x + 2 = 0 \][/tex]
[tex]\[ 3x = -2 \][/tex]
[tex]\[ x = -\frac{2}{3} \][/tex]

So, [tex]\( x \)[/tex] must not be [tex]\(-\frac{2}{3}\)[/tex].

### Conclusion

Putting it all together:
[tex]\[ \left(\frac{f}{g}\right)(x) = \frac{\sqrt[3]{3x}}{3x + 2} \][/tex]
with the restriction [tex]\( x \neq -\frac{2}{3} \)[/tex].

Thus, the correct answer is:
[tex]\[ \boxed{\text{A. } \left(\frac{f}{g}\right)(x)=\frac{\sqrt[3]{3 x}}{3 x+2}, x \neq -\frac{2}{3}} \][/tex]