The task here is to find the derivative $f'(x_0)$ at the point $x_0$ using the graph provided. This derivative corresponds to the slope of the tangent line drawn at $x = x_0$ on the curve $y = f(x)$.

Here’s how to approach it:

1. Identify the Point: From the graph, $x_0$ is marked at $x = -3$.
2. Determine the Tangent’s Slope: The slope of the tangent line at this point is the derivative $f'(x_0)$.
   • The tangent line is shown in blue and appears to have a negative slope.
3. Calculate the Slope:
   • To determine the slope, pick two points on the tangent line. For example:
     • Point 1: Approximately at $(-5, -1)$
     • Point 2: Approximately at $(-3, -2)$
   • Using the slope formula $m = \frac{y_2 - y_1}{x_2 - x_1}$: $m = \frac{-2 - (-1)}{-3 - (-5)} = \frac{-1}{2} = -0.5$
   • Therefore, $f'(x_0) = -0.5$.

Answer:

The value of $f'(x_0)$ in decimal form is: $\text{Atsakymas: } -0.5$

Would you like further details on interpreting graphs for derivatives?

Further Related Questions

1. How do you find the slope of a line between two points?
2. What does the slope of a tangent line represent in calculus?
3. How can you approximate slopes if points aren't clearly marked?
4. How does a derivative relate to the function’s rate of change?
5. What are the steps to find the derivative at a specific point on any graph?

Tip

When interpreting tangent lines on graphs, always check for clear points to estimate the slope accurately.