Answer
$ a.\quad f$ is one-to-one
$b.\quad f^{-1}(x)=\sqrt[3]{x-5}$
Work Step by Step
$ a.\quad$
The graph of $f$ is obtained by raising $y=x^{3}$ up by 5 units (similar to example 8, where it was raised 2 units).
Its graph passes the horizontal line test
(It is impossible to draw a horizontal line that intersects a function's graph more than once.)
It is one-to-one and has an inverse.
$ b.\quad$
To find a formula for the inverse,
1. Replace $f(x)$ with $y.$
$y=x^{3}+5$
2. Interchange $x$ and $y$. (This gives the inverse function.)
$x=y^{3}+5$
3. Solve for $y.$
... subtract $5$,
$ x-5=y^{3}\qquad$... take the cube root,
$\sqrt[3]{x-5}=y$
4. Replace $y$ with $f^{-1}(x)$ . (This is inverse function notation.)
$f^{-1}(x)=\sqrt[3]{x-5}$
