To solve this problem, we need to evaluate the expression:

$\lim_{{h \to 0}} \frac{{f(3 + h) - f(3)}}{h}$

This expression represents the derivative of $f(x)$ at $x = 3$. From the graph, let’s analyze the values of $f(3)$ and $f(3 + h)$ as $h$ approaches 0 to determine this derivative.

Step-by-Step Solution

1. Determine $f(3)$: Looking at the graph, the value of $f(3)$ is at $y = -1$.

   $f(3) = -1$

2. Determine $f(3 + h)$ as $h \to 0$: We need to look at the behavior of $f(x)$ around $x = 3$. Observing the slope of the function near $x = 3$ on the graph, it appears that there is a constant slope as $x$ increases from $x = 2$ to $x = 4$.

   From the graph, $f(2) = -3$ and $f(4) = 0$. So, the slope between these points is:

   $\frac{{f(4) - f(2)}}{{4 - 2}} = \frac{0 - (-3)}{2} = \frac{3}{2} = 1.5$

3. Conclusion: The derivative $f'(3)$ is equal to the slope between these points, which is $\frac{3}{2}$ or $1.5$.

Thus,

$\lim_{{h \to 0}} \frac{{f(3 + h) - f(3)}}{h} = 1.5$

Answer:

$\lim_{{h \to 0}} \frac{{f(3 + h) - f(3)}}{h} = 1.5$

Would you like further explanation, or do you have any questions about this solution?

Here are five related questions to deepen your understanding:

1. How is the derivative defined in terms of limits?
2. How can we find the slope of a curve at a specific point using the graph?
3. What are the implications if the limit in this problem does not exist?
4. How does a graph help us interpret the derivative as a rate of change?
5. What if the function had a sharp corner at $x = 3$; how would this affect the derivative?

Tip: When evaluating derivatives graphically, always ensure that the function is continuous and differentiable at the point of interest. Discontinuities or sharp corners indicate that the derivative may not exist.