The given problem asks you to sketch the graph of $f'(x)$, the derivative of $f(x)$, based on the graph of $f(x)$ provided.

Key points to observe:

1. Slope behavior: The derivative $f'(x)$ reflects the slope of $f(x)$. When $f(x)$ is increasing, $f'(x)$ is positive, and when $f(x)$ is decreasing, $f'(x)$ is negative.
2. Critical points: The points where the slope of $f(x)$ is zero (flat regions or peaks/valleys) will correspond to where $f'(x) = 0$.

Analyzing $f(x)$:

• From $x = -4$ to $x = -2$, $f(x)$ is increasing, so $f'(x) > 0$.
• At $x = -2$, there is a peak, so the slope is zero, meaning $f'(x) = 0$.
• From $x = -2$ to $x = 1$, $f(x)$ is decreasing, so $f'(x) < 0$.
• At $x = 1$, there is a valley, so the slope is zero, meaning $f'(x) = 0$.
• From $x = 1$ to $x = 3$, $f(x)$ is increasing again, so $f'(x) > 0$.
• From $x = 3$ to $x = 4$, $f(x)$ is flat, so $f'(x) = 0$.

Steps to sketch $f'(x)$:

1. Plot points where the slope of $f(x)$ is zero: $f'(x) = 0$ at $x = -2$, $x = 1$, and $x = 3$.
2. Identify regions of increasing and decreasing slopes:
   • $f'(x) > 0$ for $-4 < x < -2$ and $1 < x < 3$.
   • $f'(x) < 0$ for $-2 < x < 1$.
   • $f'(x) = 0$ at $x = 3$ because the slope is flat after that.

The resulting graph will have positive values for $f'(x)$ on intervals where $f(x)$ increases and negative values where $f(x)$ decreases, with zero values at the critical points.

Would you like a more detailed step-by-step graphing of $f'(x)$ or any specific clarifications?

Here are 5 related questions:

1. How does the graph of $f'(x)$ relate to the concavity of $f(x)$?
2. What happens to $f'(x)$ if $f(x)$ is a constant function?
3. How would the graph of $f''(x)$ behave based on the graph of $f(x)$?
4. Why is the derivative zero at local maximum and minimum points?
5. How can you interpret the slope of a graph in terms of real-world problems?

Tip: When analyzing the graph of $f(x)$, always look for points where the slope changes direction to identify critical points for $f'(x)$.