To solve this problem, we need to analyze the graph of the function $g$ and determine the order of its derivatives $g'(-2)$, $g'(0)$, $g'(2)$, and $g'(4)$.

Steps:

1. Understand $g'(x)$:
   The derivative $g'(x)$ represents the slope of the tangent line to the graph of $g(x)$ at point $x$.

2. Identify key points on the graph:

   • $x = -2$: The slope is very negative because the graph is steeply decreasing. $g'(-2)$ is a large negative value.
   • $x = 0$: This is a local minimum. At local extrema (minimum or maximum), the slope $g'(x)$ is zero. Hence, $g'(0) = 0$.
   • $x = 2$: The graph is increasing steeply, so the slope $g'(2)$ is positive.
   • $x = 4$: The graph is flatter and decreasing, so $g'(4)$ is negative, but not as steep as at $x = -2$.

Compare the slopes:

• $g'(-2)$: Large negative value (very steep downwards).
• $g'(4)$: Negative value, but smaller in magnitude than $g'(-2)$.
• $g'(0)$: Zero (flat tangent at the local minimum).
• $g'(2)$: Positive slope (increasing steeply).

Order the values in increasing order:

From most negative to most positive: $g'(-2) < g'(4) < g'(0) < g'(2)$ Thus, the order is A, C, D, B.

Final Answer:

$\textbf{A, C, D, B}$

Let me know if you need further clarification or detailed explanations! 😊

Related Questions:

1. What does the derivative tell us about the shape of a function's graph?
2. How do we identify critical points from a graph?
3. Why is the derivative zero at local minima or maxima?
4. What is the difference between increasing and decreasing intervals on a graph?
5. How can we estimate the slope of a tangent line visually?

Tip: At critical points (local minima or maxima), the slope of the tangent line is zero, meaning $g'(x) = 0$.