Let's break down each part of this problem based on the graph of $g(x)$:

(a) State the value of $g(2)$:

From the graph, observe the value of the function when $x = 2$. It appears that the graph intersects $y = 2$ at $x = 2$, so: $g(2) \approx 2$

(b) Why is $g$ one-to-one?

A function is one-to-one if it passes the horizontal line test, meaning no horizontal line intersects the graph more than once. From the given graph, every horizontal line intersects the graph at most once. Therefore, $g$ is one-to-one because it passes the horizontal line test.

(c) Estimate the value of $g^{-1}(2)$:

The inverse function $g^{-1}(x)$ gives the value of $x$ for a given $y$. Here, we are looking for $g^{-1}(2)$, which means we need to find $x$ such that $g(x) = 2$. From part (a), we know that: $g(2) = 2 \quad \text{so} \quad g^{-1}(2) = 2$

(d) Estimate the domain of $g^{-1}$:

The domain of the inverse function is the range of the original function. From the graph, the function $g(x)$ starts at $y = 0$ and increases without bound. Therefore, the range of $g(x)$ is $[0, \infty)$, meaning the domain of $g^{-1}$ is: $[0, \infty)$

(e) Sketch the graph of $g^{-1}$:

To sketch the graph of $g^{-1}$, reflect the graph of $g(x)$ across the line $y = x$. The new graph will pass through points like $(0, 0)$, $(1, 1)$, and $(2, 2)$, and it will maintain the general shape of the original graph but swapped across the line $y = x$.

Would you like further details on any of these steps or a sketch of the graph for $g^{-1}$?

Here are 5 related questions to expand your understanding:

1. How would the graph of $g^{-1}$ change if the range of $g(x)$ was restricted?
2. How can you algebraically determine if a function is one-to-one without graphing?
3. What would be the effect on $g^{-1}(x)$ if $g(x)$ had a horizontal asymptote?
4. How does the domain of a function affect its inverse?
5. Can a function have an inverse if it is not one-to-one?

Tip: Remember, a function's graph must pass both the vertical and horizontal line tests for it to have an inverse that is also a function.