Answer
(a) $(-\infty,\infty)$, $(-\infty,\infty)$.
(b) see explanations.
(c) $f^{-1}(x)=\frac{x^3+7}{2}$
(d) $(-\infty,\infty)$, $(-\infty,\infty)$.
(e) See graph and explanations.
Work Step by Step
Given $f(x)=\sqrt[3] {2x-7}$, we have:
(a) the domain of $f(x)$ can be found as $(-\infty,\infty)$ and the range as $(-\infty,\infty)$.
(b) For each $x$ value, there is a unique $y$ value, which means that the function is one-to-one, thus the inverse $f^{-1}(x)$ exist for this function.
(c) (i) Let $y=\sqrt[3] {2x-7}$; (ii) exchange $x,y$, we have $x=\sqrt[3] {2y-7}$; (iii) solve for $y$ to get $y=\frac{x^3+7}{2}$; (iv) replace $y$ with $f^{-1}(x)$ to get $f^{-1}(x)=\frac{x^3+7}{2}$
(d) We can identify the domain of $f^{-1}(x)$ as $(-\infty,\infty)$, and range as $(-\infty,\infty)$.
(e) See graph for $f(x)$ (red curve) and $f^{-1}(x)$ (blue curve). We can see that they are symmetric with respect to the line $y=x$ (green).
