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## Manipulating Radical Functions 1.4 - Solution

Given that $f(x)=2x^3$ and $g(x)=\sqrt[3]{x},$ we will define $(fg)(x)$ and $\left(\frac{f}{g}\right)(x).$ Let's begin with the product of the functions.

### Multiplying Two Functions

Let's recall the definition we will use. $\begin{gathered} (fg)(x)=f(x)g(x) \end{gathered}$ From here, we will substitute the given function rules and simplify.
$(fg)(x)=f(x)g(x)$
$(fg)(x)={\color{#0000FF}{2x^3}}{\color{#009600}{\sqrt[3]{x}}}$
Simplify right-hand side
$(fg)(x)=2x^3x^{1/3}$
$(fg)(x)=2x^{3+1/3}$
$(fg)(x)=2x^{9/3+1/3}$
$(fg)(x)=2x^{10/3}$

### Dividing Two Functions

Let's recall the rule for dividing functions. $\begin{gathered} \left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)} \end{gathered}$ We can proceed in the same way as we did for multiplying the functions.
$\left(\dfrac{f}{g}\right)(x)=\dfrac{f(x)}{g(x)}$
$\left(\dfrac{f}{g}\right)(x)=\dfrac{{\color{#0000FF}{2x^3}}}{{\color{#009600}{\sqrt[3]{x}}}}$
Simplify right-hand side
$\left(\dfrac{f}{g}\right)(x)=2\dfrac{x^3}{\sqrt[3]{x}}$
$\left(\dfrac{f}{g}\right)(x)=2\dfrac{x^3}{x^{1/3}}$
$\left(\dfrac{f}{g}\right)(x)=2x^{3-1/3}$
$\left(\dfrac{f}{g}\right)(x)=2x^{9/3-1/3}$
$\left(\dfrac{f}{g}\right)(x)=2x^{8/3}$