Given the description of the graph of the function $f(x)$, let's determine the values of various quantities based on the function's behavior.

1. $f(0)$:

   • $f(0)$ is the value of the function when $x = 0$. According to the description, the function crosses the y-axis at approximately $y = 3.2$. Therefore, $f(0) \approx 3.2$.

2. $f(2)$:

   • At $x = 2$, the graph has a closed point at $y = 3$, meaning $f(2) = 3$.

3. $\lim_{x \to 2^-} f(x)$:

   • This represents the left-hand limit as $x$ approaches 2. The function approaches $y = 3$ from the left side and stops at the closed point $(2, 3)$, so $\lim_{x \to 2^-} f(x) = 3$.

4. $\lim_{x \to 2^+} f(x)$:

   • This represents the right-hand limit as $x$ approaches 2. The function restarts at the open point $(2, 1)$, meaning the function value jumps to 1. Therefore, $\lim_{x \to 2^+} f(x) = 1$.

5. $f(4)$:

   • At $x = 4$, there is an open point at $y = 4$, so the function does not have a value here. Thus, $f(4)$ does not exist (DNE).

Summary:

• $f(0) \approx 3.2$
• $f(2) = 3$
• $\lim_{x \to 2^-} f(x) = 3$
• $\lim_{x \to 2^+} f(x) = 1$
• $f(4)$ does not exist (DNE)

Would you like more details on any of these calculations or have any questions?

Here are some related questions:

1. What is the significance of open and closed points on a graph?
2. How do you determine the left-hand and right-hand limits of a function?
3. Can a function have different left-hand and right-hand limits at the same point?
4. What happens to a function's continuity at a point where it has an open or closed point?
5. How can you find the value of $f(x)$ if the function has a jump discontinuity?

Tip: When analyzing a graph, always pay attention to open and closed points—they indicate where the function's value exists or does not exist, which is crucial for evaluating limits and function values.