Questions: f(x)=sqrt[3]2 x g(x)=3 x+1 Find (f/g)(x). Include any restrictions on the domain. A. (f/g)(x)=(3 x+1)/sqrt[3]2 x, x ≠ 0 B. (f/g)(x)=(3 x+1)/sqrt[3]2 x, x ≥ 0 C. (f/g)(x)=sqrt[3]2 x/(3 x+1), x ≠ -3 D. (f/g)(x)=sqrt[3]2 x/(3 x+1), x ≠ -1/3

Solution Steps

To find \(\left(\frac{f}{g}\right)(x)\), we need to divide the function \(f(x)\) by \(g(x)\). This means we will compute \(\frac{\sqrt[3]{2x}}{3x+1}\). We also need to determine the domain restrictions. The denominator \(3x + 1\) cannot be zero, so we solve \(3x + 1 = 0\) to find the restriction on \(x\).

We have the functions defined as follows: \[ f(x) = \sqrt[3]{2x} \] \[ g(x) = 3x + 1 \]

To find \(\left(\frac{f}{g}\right)(x)\), we compute: \[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt[3]{2x}}{3x + 1} \]

The function \(\left(\frac{f}{g}\right)(x)\) is undefined when the denominator is zero. We solve for \(x\) in the equation: \[ 3x + 1 = 0 \implies x = -\frac{1}{3} \] Thus, the restriction on the domain is \(x \neq -\frac{1}{3}\).

Final Answer