Answers: 1
is determined to be ![{f^{-1}(x)=\sqrt[3]{\frac{1}{4}x-8}](/tpl/images/0927/9305/f94b2.png)
.
.![\displaystyle{x=4y^3+8}\\\\4y^3=x-8\\\\y^3=\frac{x}{4}-8\\\\\sqrt[3]{y^3}=\sqrt[3]{\frac{x}{4}-8} \\\\y = \sqrt[3]{\frac{1}{4}x-8}\\\\f^{-1}(x)= \sqrt[3]{\frac{1}{4}x-8}](/tpl/images/0927/9305/a8512.png)
![\displaystyle{f^{-1}(x)=\sqrt[3]{\frac{1}{4}x-8}](/tpl/images/0927/9305/f35fa.png)
.![\huge\boxed{f^{-1}(x) = \sqrt[3]{\frac{x-8}{4}}}](/tpl/images/0927/9305/c14d2.png)
form
as 
.
.
Divide both sides by 4:
Take the cube root of each side: ![y = \sqrt[3]{\frac{x-8}{4}}](/tpl/images/0927/9305/8e9d7.png)
Convert into ![f^{-1}(x) = \sqrt[3]{\frac{x-8}{4}}](/tpl/images/0927/9305/725c1.png)
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7. Find the inverse of f(x) = 4x^3 + 8....
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