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.