This question involves determining the derivative of the function $f(x)$ at the point $x_0$, where the derivative represents the slope of the tangent line to the curve $y = f(x)$ at $x_0$.

Solution:

1. Slope of the Tangent: From the graph, the tangent at $x_0$ is drawn, and the slope can be determined using two points on this tangent line. These points are:

   • $(x_0, f(x_0))$ (the red dot), which appears to be $(0, -4)$,
   • Another point on the tangent, $(1, -3)$.

2. Compute the Slope: The slope of a line is given by: $m = \frac{y_2 - y_1}{x_2 - x_1},$ where $(x_1, y_1) = (0, -4)$ and $(x_2, y_2) = (1, -3)$.

   Substituting the coordinates: $m = \frac{-3 - (-4)}{1 - 0} = \frac{-3 + 4}{1} = 1.$

3. Interpretation: The slope of the tangent at $x_0 = 0$ is 1. Therefore: $f'(x_0) = f'(0) = 1.$

Final Answer:

$f'(0) = 1$

Do you need further clarification or a deeper explanation? 😊

Here are 5 questions to deepen understanding:

1. What does the derivative of a function represent geometrically?
2. How can you identify points on a tangent line if not given explicitly?
3. Can the slope of a tangent line be negative? When would this occur?
4. How does the derivative relate to the rate of change of a function?
5. How would you compute the derivative numerically if the graph was not provided?

Tip: Always double-check the coordinates of points used to calculate the slope from the graph to ensure accuracy!