We are given the two functions:

$f(x) = \frac{3}{2}x \quad \text{and} \quad g(x) = \frac{1}{8}x^2$

The problem mentions a graph where:

• $f(x)$ is a line (a linear function) passing through the points $(0,0)$ and $(2,3)$.
• $g(x)$ is a parabola, meeting $f(x)$ first at $(0,0)$, then at $(12,18)$.
• For $x$ between 0 and 12, $f(x) > g(x)$, and after $x = 12$, $g(x) > f(x)$.

Now, let's break down the average rate of change for both functions.

Average Rate of Change for $f(x)$:

For a linear function like $f(x) = \frac{3}{2}x$, the average rate of change is constant. It's simply the slope of the line. Since the slope of $f(x)$ is $\frac{3}{2}$, the average rate of change of $f(x)$ is:

$\text{Average rate of change of } f(x) = \frac{3}{2}$

Average Rate of Change for $g(x)$:

For a quadratic function like $g(x) = \frac{1}{8}x^2$, the average rate of change depends on the interval considered. We can calculate the average rate of change of $g(x)$ over an interval $[a, b]$ as:

$\frac{g(b) - g(a)}{b - a}$

Let's first consider the interval $[0, 12]$, as given in the problem.

$g(0) = 0 \quad \text{and} \quad g(12) = \frac{1}{8} \times 12^2 = \frac{1}{8} \times 144 = 18$

The average rate of change of $g(x)$ from $x = 0$ to $x = 12$ is:

$\frac{g(12) - g(0)}{12 - 0} = \frac{18 - 0}{12} = \frac{18}{12} = 1.5$

For $x$ between 0 and 12, the average rate of change of $g(x)$ is $1.5$.

As $x$ increases beyond 12:

For $x > 12$, the average rate of change of $g(x)$ increases since the function is quadratic, and the slope of a parabola increases as $x$ grows.

Comparison:

• For $x$ between 0 and 12, the average rate of change of $f(x)$ is $\frac{3}{2} = 1.5$, which is exactly equal to the average rate of change of $g(x)$ in this interval.
• For $x > 12$, the average rate of change of $g(x)$ increases (since $g(x)$ is quadratic), while the average rate of change of $f(x)$ remains constant at $\frac{3}{2}$.

Thus, as $x$ increases beyond 12, the average rate of change of $g(x)$ will eventually exceed the constant rate of change of $f(x)$.

Conclusion:

The correct statement is:

B. As x increases, the average rate of change of g(x) eventually exceeds the average rate of change of f(x).

Additional Questions:

1. How does the average rate of change of $g(x)$ compare to that of $f(x)$ over the interval $[12, 20]$?
2. What happens to the graph of $g(x)$ as $x$ becomes very large (i.e., $x \to \infty$)?
3. How can the average rate of change of a function be interpreted geometrically on a graph?
4. How does the slope of the tangent line relate to the average rate of change for non-linear functions?
5. What happens if we change the coefficient in $g(x)$ to a larger value, say $g(x) = \frac{1}{4}x^2$?

Tip:

To visualize how the average rate of change changes for different functions, plot both $f(x)$ and $g(x)$ on a graph and compare the slopes of secant lines (for average rate of change) for various intervals.