Answer
$\text{a})$ $f^{-1}(x) = \sqrt[3]{\frac{x+1}{8}}$.
$\text{b})$ The original function is not invertible. By restricting the domain, one can obtain partial inverses $f^{-1}(x) = \pm\sqrt{x}+1$.
$\text{c})$ $f^{-1}(x) = 3/x-1$.
$\text{d})$ $f^{-1}(x) = (x+2)/(x-1)$
Work Step by Step
$\text{a})$ Set up the equation $x = 8y^3 - 1$ and solve, yielding $y = \sqrt[3]{\frac{x+1}{8}}$.
$\text{b})$ This function is not invertible as its graph is a parabola which fails the horizontal line test. However, one may either restrict the domain of $f$ to $x \geq 1$ or $x \leq 1$ and thus the range of the inverse to $y \geq 1$ or $y \leq 1$. If done, we may set up the equation $x = y^2-2y+1$ and simplify by completing the square to $x = (y-1)^2$. Then taking a positive or negative square root to match the sign of $y-1$, one obtains $y = \pm\sqrt{x}+1$
$\text{c})$ Set up the equation $x = 3/(y+1)$ and solve, yielding $y = 3/x-1$.
$\text{d})$ Set up the equation $x = (y+2)/(y-1)$. To solve, we multiply both sides by $y-1$, yielding $xy-x = y+2$. Rearranging yields $xy-y = x+2$. Factoring yields $y(x-1) = x+2$, and dividing by $x-1$ gives $y = (x+2)/(x-1)$.
