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Manipulating Radical Functions 1.9 - Solution

Given that $f(x)=\text{-}5\sqrt[4]{x}$ and $g(x)=19\sqrt[4]{x},$ we will find $(f+g)(x)$ and $(f-g)(x).$ Let's begin with the sum of the functions.

Let's recall the definition we will use. $\begin{gathered} (f+g)(x)=f(x)+g(x) \end{gathered}$ From here, we will substitute the given function rules and simplify.
$(f+g)(x)=f(x)+g(x)$
$(f+g)(x)={\color{#0000FF}{\text{-}5\sqrt[4]{x}}}+{\color{#009600}{19\sqrt[4]{x}}}$
$(f+g)(x)=(\text{-}5+19)\sqrt[4]{x}$
$(f+g)(x)=14\sqrt[4]{x}$

Subtracting Two Functions

Let's recall the rule for subtracting functions. $\begin{gathered} (f-g)(x)=f(x)-g(x) \end{gathered}$ We can proceed in the same way as we did for adding the functions.
$(f-g)(x)=f(x)-g(x)$
$(f-g)(x)={\color{#0000FF}{\text{-}5\sqrt[4]{x}}}-{\color{#009600}{19\sqrt[4]{x}}}$
$(f-g)(x)=(\text{-}5-19)\sqrt[4]{x}$
$(f-g)(x)=\text{-}24\sqrt[4]{x}$