Let's go through the image step-by-step to address the questions.

Information from the image:

We are given two functions:

• $f(x)$, represented in red.
• $g(x)$, represented in blue.

Let's address the sub-questions:

(a) What is $g(2)$?

From the graph, when $x = 2$, we can find the value of $g(x)$ (blue curve).

• From the blue graph, at $x = 2$, the value of $g(2)$ is approximately $1$.

Thus, $g(2) = 1$.

(b) Find all solutions to the equation $f(x) = -1$.

We need to find where the red graph intersects $y = -1$.

• From the graph, $f(x) = -1$ at $x = 4$ and $x = 5$.

Thus, the solutions are $x = 4$ and $x = 5$.

(c) Where does the line $x = 5$ intersect the graph of $f$?

The vertical line $x = 5$ intersects the red graph (function $f$) at the point where $f(x) = -1$.

Thus, the point of intersection is $(5, -1)$.

(d) Find the open intervals on which $g$ is increasing/decreasing.

• $g(x)$ (blue graph) is increasing on the interval $(0, 2)$.
• $g(x)$ is decreasing on the interval $(2, 6)$.

(e) Find the average rate of change of the function $g$ on the interval $[-4, 2]$.

The formula for the average rate of change of a function $g$ on an interval $[a, b]$ is:

$\text{Average rate of change} = \frac{g(b) - g(a)}{b - a}$

Here, $a = -4$ and $b = 2$. From the graph:

• $g(-4) = 0$ (the value of $g(x)$ at $x = -4$),
• $g(2) = 1$.

Thus, the average rate of change is:

$\frac{g(2) - g(-4)}{2 - (-4)} = \frac{1 - 0}{2 + 4} = \frac{1}{6}.$

So, the average rate of change is $\frac{1}{6}$.

Do you want any further details or have questions on any of the parts?

Here are 5 related questions to think about:

1. What is the domain and range of the function $g(x)$ based on the graph?
2. Find the x-intercepts of the graph of $f(x)$.
3. At which points do $f(x)$ and $g(x)$ intersect?
4. What are the maximum and minimum points of the function $f(x)$?
5. How can you determine if a function is concave up or concave down from a graph?

Tip:

To determine intervals of increase or decrease for any function, focus on the slope: increasing intervals occur where the function slopes upwards (positive slope), and decreasing intervals occur where it slopes downwards (negative slope).