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Manipulating Radical Functions

Manipulating Radical Functions 1.8 - 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)=x+2y=x+2\begin{gathered} f(x)=\sqrt{x+2} \quad \rightarrow \quad y=\sqrt{x+2} \end{gathered} Next step is to switch xx and yy in the function rule. y=x+2switchx=y+2 y=\sqrt{x+2} \quad \stackrel{\text{switch}}{\longrightarrow} \quad x=\sqrt{y+2} Now we need to solve for y.y. The resulting equation will be the inverse of the given function.
x=y+2x=\sqrt{y+2}
Solve for yy
x2=(y+2)2x^2=\left(\sqrt{y+2}\right)^2
x2=y+2x^2=y+2
x22=yx^2-2=y
y=x22y=x^2-2
Finally, to indicate that this is the inverse function of f(x),f(x), we will replace yy with f-1(x).f^{\text{-} 1}(x). y=x22f-1(x)=x22\begin{gathered} y=x^2-2 \quad \rightarrow \quad {\color{#0000FF}{f^{\text{-} 1}(x)}}=x^2-2 \end{gathered}