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.