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The problem asks to evaluate \( f'(1) \) and \( f'(6) \), which represent the slopes of the function \( f(x) \) at \( x = 1 \) and \( x = 6 \). This is essentially asking for the derivative at these points, which corresponds to the slope of the tangent line to the graph at each respective point.

### Step 1: Identify the points on the graph
1. At \( x = 1 \), it appears that \( f(1) \approx 2 \).
2. At \( x = 6 \), it seems that \( f(6) \approx 6 \).

### Step 2: Estimate the slopes
We can estimate the slope \( f'(x) \) using the rise-over-run formula:

\[
f'(x) = \frac{\Delta y}{\Delta x}
\]

#### For \( f'(1) \):
- The graph appears to be a straight line from \( x = 0 \) to \( x = 2 \), passing through points \( (0, 0) \) and \( (2, 4) \). 
- Slope calculation:
  \[
  f'(1) = \frac{4 - 0}{2 - 0} = 2
  \]

#### For \( f'(6) \):
- The graph between \( x = 4 \) and \( x = 8 \) appears to be another straight line. Using points \( (4, 4) \) and \( (8, 8) \):
  \[
  f'(6) = \frac{8 - 4}{8 - 4} = 1
  \]

### Final answers:
- \( f'(1) = 2 \)
- \( f'(6) = 1 \)

Let me know if you need more details, or if you have any other questions!

### Related questions:
1. How do you compute derivatives for non-linear graphs?
2. What is the significance of the derivative in terms of real-world applications?
3. Can the slope be negative? If so, how would that affect \( f'(x) \)?
4. What is the meaning of a zero derivative?
5. How can we approximate derivatives when the graph is not linear?

### Tip:
For linear portions of a graph, the derivative is constant. Always use the rise-over-run method when evaluating the slope for straight lines.

Use the graph below to evaluate f'(1) and f'(6).

Learn how to evaluate the derivatives f'(1) and f'(6) from a graph. This example explains the use of the slope formula to compute the derivative of the function at two specific points. Ideal for students in Grades 10-12 studying calculus and graph analysis.

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