Answer
one-to-one function
inverse: $f^{-1}(x)=\dfrac{x^3+4}{6}$
Work Step by Step
Some of the ordered pairs of the given function, $
f(x)=\sqrt[3]{6x-4}
$, are $
\left\{(-2,-2.52)(-1,-2.15),(0,-1.59),(1,1.26),(2,2),...\right\}
$. Note that every $y$-coordinate from this function is unique. Hence, the given function is a one-to-one function.
To find the inverse, let $y=f(x)$. Then, interchange the $x$ and $y$ variables and solve for $y$. That is,
\begin{align*}\require{cancel}
y=&\sqrt[3]{6x-4}
\\\Rightarrow&
x=\sqrt[3]{6y-4}
&(\text{interchange $x$ and $y$})
\\&
(x)^3=\left(\sqrt[3]{6y-4}\right)^3
&(\text{solve for $y$})
\\\\&
x^3=6y-4
\\&
x^3+4=6y
\\\\&
\dfrac{x^3+4}{6}=\dfrac{\cancel6y}{\cancel6}
\\\\&
\dfrac{x^3+4}{6}=y
.\end{align*}
Hence, the inverse is $
f^{-1}(x)=\dfrac{x^3+4}{6}
$.
