Answer
$ a.\quad g$ is one-to-one
$b.\quad g^{-1}(x)=\sqrt[3]{x}-7$
Work Step by Step
$ a.\quad$
The graph of $g$ is obtained by shifting $y=x^{3}$ by 2 units to the right.
(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 $g(x)$ with $y.$
$y=(x+7)^{3}$
2. Interchange $x$ and $y$. (This gives the inverse function.)
$x=(y+7)^{3}$
3. Solve for $y.$
... take the cube root,
$\sqrt[3]{x}=y+7\qquad$... subtract $7$,
$\sqrt[3]{x}-7=y$
4. Replace $y$ with $g^{-1}(x)$ . (This is inverse function notation.)
$g^{-1}(x)=\sqrt[3]{x}-7$
