For the function f(x)=8root(3)(x)+4, find f^(-1)(x). f^(-1)(x)=((x)/(8))^(3)-4 f^(-1)(x)=((x-4)/(8))^(3) f^(-1)(x)=(x^(3))/(8)-4 f^(-1)(x)=((x)/(8)-4)^(3)

Subtract and isolate y: Subtract 4 from both sides to isolate the term with y: x - 4 = 8√3(y)Divide to isolate y: Divide both sides by 8 to further isolate the term with y: (x - 4) / 8 = √3(y)Eliminate cube root: Now, we need to get rid of the cube root. To do this, we raise both sides to the power of 3: ((x - 4) / 8)^3 = (√3(y))^3Final inverse function: Since (√3(y))^3 is just y, we have: y = ((x - 4) / 8)^3 This is the inverse function, f^(-1)(x).

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For the function
f(x)=8root(3)(x)+4, find
f^(-1)(x).

f^(-1)(x)=((x)/(8))^(3)-4

f^(-1)(x)=((x-4)/(8))^(3)

f^(-1)(x)=(x^(3))/(8)-4

f^(-1)(x)=((x)/(8)-4)^(3)

For the function $f(x)=8 \sqrt[3]{x}+4$ , find $f^{-1}(x)$ . $\newline$ $f^{-1}(x)=\left(\frac{x}{8}\right)^{3}-4$ $\newline$ $f^{-1}(x)=\left(\frac{x-4}{8}\right)^{3}$ $\newline$ $f^{-1}(x)=\frac{x^{3}}{8}-4$ $\newline$ $f^{-1}(x)=\left(\frac{x}{8}-4\right)^{3}$

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Q. For the function

f(x)=8 \sqrt[3]{x}+4

, find

f^{-1}(x)

\newline

f^{-1}(x)=\left(\frac{x}{8}\right)^{3}-4

\newline

f^{-1}(x)=\left(\frac{x-4}{8}\right)^{3}

\newline

f^{-1}(x)=\frac{x^{3}}{8}-4

\newline

f^{-1}(x)=\left(\frac{x}{8}-4\right)^{3}

Subtract and isolate $y$ : Subtract $4$ from both sides to isolate the term with $y$ : $\newline$ $x - 4 = 8\sqrt{3}(y)$
Divide to isolate $y$ : Divide both sides by $8$ to further isolate the term with $y$ : $\newline$ $(x - 4) / 8 = \sqrt{3}(y)$
Eliminate cube root: Now, we need to get rid of the cube root. To do this, we raise both sides to the power of $3$ : $\left(\frac{x - 4}{8}\right)^3 = (\sqrt[3]{y})^3$
Final inverse function: Since $(\sqrt{3}(y))^3$ is just $y$ , we have: $\newline$ $y = \left(\frac{x - 4}{8}\right)^3$ $\newline$ This is the inverse function, $f^{-1}(x)$ .