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$What is the inverse of the function g(x)=(x^(3))/(8)+16 ? {:[g^(-1)(x)=(root(3)(x-16))/(2)],[g^(-1)(x)=2root(3)(x)-16],[g^(-1)(x)=2root(3)(x+16)],[g^(-1)(x)=2root(3)(x-16)]:} Progress$

What is the inverse of the function $g(x)=\frac{x^{3}}{8}+16$ ? $\newline$ $\begin{array}{l} g^{-1}(x)=\frac{\sqrt[3]{x-16}}{2} \\ g^{-1}(x)=2 \sqrt[3]{x}-16 \\ g^{-1}(x)=2 \sqrt[3]{x+16} \\ g^{-1}(x)=2 \sqrt[3]{x-16} \end{array}$ $\newline$ Progress

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Find the vertex of the transformed function

Full solution

Q. What is the inverse of the function

g(x)=\frac{x^{3}}{8}+16

?

\newline

\begin{array}{l} g^{-1}(x)=\frac{\sqrt[3]{x-16}}{2} \\ g^{-1}(x)=2 \sqrt[3]{x}-16 \\ g^{-1}(x)=2 \sqrt[3]{x+16} \\ g^{-1}(x)=2 \sqrt[3]{x-16} \end{array}

\newline

Progress

Replace with $y$ : To find the inverse, first replace $g(x)$ with $y$ : $y = \frac{x^3}{8} + 16$
Swap x and y: Now, swap x and y to find the inverse: $x = \frac{y^3}{8} + 16$
Isolate cubic term: Subtract $16$ from both sides to isolate the cubic term: $\newline$ $x - 16 = \frac{y^3}{8}$
Multiply by $8$ : Multiply both sides by $8$ to get rid of the denominator: $8(x - 16) = y^3$
Take cube root: Now take the cube root of both sides to solve for $y$ : $y = \sqrt[3]{8(x - 16)}$
Simplify expression: Simplify the cube root expression: $y = 2 \times \sqrt[3]{x - 16}$
Replace with inverse function: Replace $y$ with $g^{-1}(x)$ to denote the inverse function: $\newline$ $g^{-1}(x) = 2 \cdot \sqrt[3]{x - 16}$

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