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Manipulating Radical Functions

Manipulating Radical Functions 1.4 - Solution

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Given that f(x)=2x3f(x)=2x^3 and g(x)=x3,g(x)=\sqrt[3]{x}, we will define (fg)(x)(fg)(x) and (fg)(x).\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. (fg)(x)=f(x)g(x)\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)=f(x)g(x)
(fg)(x)=2x3x3(fg)(x)={\color{#0000FF}{2x^3}}{\color{#009600}{\sqrt[3]{x}}}
Simplify right-hand side
(fg)(x)=2x3x1/3(fg)(x)=2x^3x^{1/3}
(fg)(x)=2x3+1/3(fg)(x)=2x^{3+1/3}
(fg)(x)=2x9/3+1/3(fg)(x)=2x^{9/3+1/3}
(fg)(x)=2x10/3(fg)(x)=2x^{10/3}

Dividing Two Functions

Let's recall the rule for dividing functions. (fg)(x)=f(x)g(x)\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.
(fg)(x)=f(x)g(x)\left(\dfrac{f}{g}\right)(x)=\dfrac{f(x)}{g(x)}
(fg)(x)=2x3x3\left(\dfrac{f}{g}\right)(x)=\dfrac{{\color{#0000FF}{2x^3}}}{{\color{#009600}{\sqrt[3]{x}}}}
Simplify right-hand side
(fg)(x)=2x3x3\left(\dfrac{f}{g}\right)(x)=2\dfrac{x^3}{\sqrt[3]{x}}
(fg)(x)=2x3x1/3\left(\dfrac{f}{g}\right)(x)=2\dfrac{x^3}{x^{1/3}}
(fg)(x)=2x31/3\left(\dfrac{f}{g}\right)(x)=2x^{3-1/3}
(fg)(x)=2x9/31/3\left(\dfrac{f}{g}\right)(x)=2x^{9/3-1/3}
(fg)(x)=2x8/3\left(\dfrac{f}{g}\right)(x)=2x^{8/3}