To evaluate $f'(1)$ and $f'(6)$ using the graph, we need to determine the slope of the function $f(x)$ at these points because $f'(x)$ represents the derivative, which corresponds to the slope of the tangent line at a given $x$-value.

Looking at the graph:

1. At $x = 1$:

   • From the graph, between $x = 0$ and $x = 2$, the function appears to be a straight line.
   • The slope of the line can be calculated by the rise over run formula: $\text{slope} = \frac{\Delta y}{\Delta x}$.
   • Between $x = 0$ and $x = 2$, the graph moves from $y = 0$ to $y = 4$, giving us: $f'(1) = \frac{4 - 0}{2 - 0} = \frac{4}{2} = 2.$ So, $f'(1) = 2$.

2. At $x = 6$:

   • For $x \geq 2$, the function remains a straight line. Let's check the slope between $x = 2$ and $x = 8$.
   • The graph moves from $y = 4$ at $x = 2$ to $y = 8$ at $x = 8$, so: $f'(6) = \frac{8 - 4}{8 - 2} = \frac{4}{6} = \frac{2}{3}.$ So, $f'(6) = \frac{2}{3}$.

Summary:

• $f'(1) = 2$
• $f'(6) = \frac{2}{3}$

Would you like further details or have any questions?

Here are 5 related questions to deepen your understanding:

1. How do we find the derivative for a non-linear part of a function graphically?
2. What does it mean if the derivative is zero at a certain point on the graph?
3. How do piecewise functions affect the behavior of their derivatives?
4. How can we estimate derivatives if the graph is not perfectly straight?
5. What is the physical interpretation of the derivative in real-world contexts?

Tip: The slope of a curve at any point gives the instantaneous rate of change, which is what the derivative represents!