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Manipulating Radical Functions

Manipulating Radical Functions 1.15 - Solution

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To find the inverse of f(x),f(x), we first need to replace f(x)f(x) with y.y. f(x)=2x53y=2x53\begin{gathered} f(x)=2\sqrt[3]{x-5} \quad \rightarrow \quad {\color{#0000FF}{y}}=2\sqrt[3]{x-5} \end{gathered} Next step is to switch xx and yy in the function rule. y=2x53switchx=2y53 {\color{#0000FF}{y}}=2\sqrt[3]{{\color{#009600}{x}}-5} \quad \stackrel{\text{switch}}{\longrightarrow} \quad {\color{#009600}{x}}=2\sqrt[3]{{\color{#0000FF}{y}}-5} Now we need to solve for y.y. The resulting equation will be the inverse of the given function.
x=2y53x=2\sqrt[3]{y-5}
Solve for yy
x2=y53\dfrac{x}{2}=\sqrt[3]{y-5}
x38=(y53)3\dfrac{x^3}{8}=\left(\sqrt[3]{y-5}\right)^3
(an)n=a\left ( \sqrt[n]{a}\right )^n = a
x38=y5\dfrac{x^3}{8}=y-5
x38+5=y\dfrac{x^3}{8}+5=y
y=x38+5y=\dfrac{x^3}{8}+5
Finally, to indicate that this is the inverse of f(x),f(x), we will replace yy with f-1(x).f^{\text{-} 1}(x). y=x38+5f-1(x)=x38+5\begin{gathered} y=\dfrac{x^3}{8}+5 \quad \rightarrow \quad {\color{#0000FF}{f^{\text{-} 1}(x)}}=\dfrac{x^3}{8}+5 \end{gathered}