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Manipulating Radical Functions

Manipulating Radical Functions 1.10 - 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)=42xy=42x\begin{gathered} f(x)=4-2\sqrt{x} \quad \rightarrow \quad y=4-2\sqrt{x} \end{gathered} Next step is to switch xx and yy in the function rule. y=42xswitchx=42y y=4-2\sqrt{x} \quad \stackrel{\text{switch}}{\longrightarrow} \quad x=4-2\sqrt{y} Now we need to solve for y.y. The resulting equation will be the inverse of the given function.
x=42yx=4-2\sqrt{y}
Solve for yy
x4=-2yx-4=\text{-} 2\sqrt{y}
x4-2=y\dfrac{x-4}{\text{-} 2}=\sqrt{y}
(x4-2)2=(y)2\left(\dfrac{x-4}{\text{-} 2}\right)^2=\left(\sqrt{y}\right)^2
(x4)2(-2)2=(y)2\dfrac{(x-4)^2}{(\text{-} 2)^2}=\left(\sqrt{y}\right)^2
(x4)24=(y)2\dfrac{(x-4)^2}{4}=\left(\sqrt{y}\right)^2
14(x4)2=(y)2\dfrac{1}{4}(x-4)^2=\left(\sqrt{y}\right)^2
14(x4)2=y\dfrac{1}{4}(x-4)^2=y
y=14(x4)2y=\dfrac{1}{4}(x-4)^2
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=14(x4)2f-1(x)=14(x4)2\begin{gathered} y=\dfrac{1}{4}(x-4)^2 \quad \rightarrow \quad {\color{#0000FF}{f^{\text{-} 1}(x)}}=\dfrac{1}{4}(x-4)^2 \end{gathered}