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To solve this problem, we need to determine the values of the derivative \( f'(a) \) at specific points \( a = 1, 2, 4, 7 \). The derivative at a point is the slope of the tangent line to the graph of \( f(x) \) at that point.

From the graph you provided, we can make the following observations:

1. **\( f'(1) \)**: 
   - At \( x = 1 \), the graph has a peak, meaning the slope of the tangent line is zero.
   - Thus, \( f'(1) = 0 \).

2. **\( f'(2) \)**:
   - At \( x = 2 \), the graph appears to have a negative slope.
   - So, \( f'(2) \) will be a negative value. Since the slope doesn't appear to be too steep, we'll approximate \( f'(2) \approx -1 \).

3. **\( f'(4) \)**:
   - At \( x = 4 \), the graph has a minimum (valley), meaning the slope of the tangent line is zero.
   - Thus, \( f'(4) = 0 \).

4. **\( f'(7) \)**:
   - At \( x = 7 \), the graph has a positive slope as it is increasing again.
   - We'll approximate \( f'(7) \approx 1 \) as the slope looks to be around that value.

### Summary of answers:
- \( f'(1) = 0 \)
- \( f'(2) \approx -1 \)
- \( f'(4) = 0 \)
- \( f'(7) \approx 1 \)

Would you like further details on any of the approximations, or do you have any questions?

### Related Questions:
1. What is the significance of the derivative being zero at a point?
2. How can the second derivative be used to analyze the concavity at points like \( x = 1 \) or \( x = 4 \)?
3. What happens to the derivative at inflection points?
4. How do you find the derivative graphically if the curve is more complex?
5. Can a point have a derivative if it has a sharp corner or cusp?

**Tip:** The slope of a function's tangent line at a point gives the best linear approximation to the function near that point.

Let f(x) be the function whose graph is shown. Determine f'(a) for a = 1, 2, 4, 7.

Learn how to find the derivative f'(a) at points 1, 2, 4, and 7 by analyzing the graph of the function f(x). This problem involves calculating the slope of the tangent line and interpreting critical points.

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