To determine $f'(-4)$, we need to find the slope of the graph at $x = -4$, as $f'(x)$ represents the derivative of $f(x)$, which is the slope of the tangent line to the curve at a given point.

From the graph, between $x = -5$ and $x = -3$, the function appears to be a straight line. Specifically:

• At $x = -5$, $f(x) = 3$
• At $x = -3$, $f(x) = 3$

This means the line segment is horizontal between these points, so the slope of the line between $x = -5$ and $x = -3$ is 0. Therefore, the derivative at $x = -4$, which lies on this horizontal segment, is:

$f'(-4) = 0$

Would you like more details or have any further questions?

Here are some follow-up questions related to this graph:

1. What is $f'(1)$?
2. What is the value of $f(-2)$?
3. Where is the function $f(x)$ increasing?
4. Identify any discontinuities in the graph of $f(x)$.
5. What is the maximum value of $f(x)$?

Tip: The slope of a line segment is calculated by dividing the difference in $y$-values by the difference in $x$-values for two points on the line.