The problem is asking to find the sum of the abscissas (x-coordinates) of the extremum points of $f(x)$. The graph provided is of $f'(x)$, the derivative of $f(x)$.

Key Concept:

• Critical points (potential extrema) occur when $f'(x) = 0$ or $f'(x)$ changes sign.
• From the graph of $f'(x)$, we identify where the curve intersects the x-axis, because at these points $f'(x) = 0$. These are the candidates for extremum points of $f(x)$.

Step-by-Step Solution:

1. Observe the graph of $f'(x)$ and find where it intersects the x-axis:

   • The graph intersects the x-axis at three points:
     • $x = -2$
     • $x = 0$
     • $x = 2$

2. Sum the x-coordinates of these points: $-2 + 0 + 2 = 0$

Final Answer:

The sum of the abscissas of the extremum points is 0.

Let me know if you'd like further clarification or explanation! Here are some related questions to expand your understanding:

1. What does the sign change of $f'(x)$ tell us about the nature of extremum points?
2. How can you determine whether a critical point is a minimum or maximum?
3. What happens if $f'(x)$ does not cross the x-axis?
4. How would the second derivative $f''(x)$ help in classifying extrema?
5. What role does concavity play in identifying extrema?

Tip: Always check where $f'(x)$ changes sign to confirm if a critical point is a local maximum or minimum!