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## Manipulating Radical Functions 1.11 - Solution

Given that $f(x)=\sqrt[3]{2x}$ and $g(x)=\text{-}11\sqrt[3]{2x},$ we will define $(f+g)(x)$ and $(f-g)(x).$ Let's begin with the sum of the functions.

Note that $(f+g)(x)=f(x)+g(x).$ From here, we will substitute the given function rules and find the sum.
$f(x)+g(x)$
${\color{#0000FF}{\sqrt[3]{2x}}}+({\color{#009600}{\text{-}11\sqrt[3]{2x}}})$
$\sqrt[3]{2x}-11\sqrt[3]{2x}$
$(1-11)\sqrt[3]{2x}$
$\text{-}10\sqrt[3]{2x}$

### Subtracting Two Functions

Note that $(f-g)(x)=f(x)-g(x),$ so we can proceed in the same way as we add two functions.
$f(x)-g(x)$
${\color{#0000FF}{\sqrt[3]{2x}}}-({\color{#009600}{\text{-}11\sqrt[3]{2x}}})$
$\sqrt[3]{2x}+11\sqrt[3]{2x}$
$(1+11)\sqrt[3]{2x}$
$12\sqrt[3]{2x}$