The problem provides a graph of a function $f$. We need to: 1. Determine the domain $D_f$ of the function $f$.

1. Problem Description

The problem provides a graph of a function $f$. We need to:

1. Determine the domain $D_f$ of the function $f$.

2. Determine the images of the following numbers: -5, -4, -3, -2, 0,

4. This means finding the values $f(-5)$, $f(-4)$, $f(-3)$, $f(-2)$, $f(0)$, $f(4)$.

3. Determine the antecedents (preimages) of 5 and

3. This means finding the values of $x$ such that $f(x) = 5$ and $f(x) = 3$.

2. Solution Steps

1. Determining the domain $D_f$ of the function $f$:

From the graph, we see that the function is defined for $x$ values starting from -5 (inclusive) and extends to approximately $x=4$ (inclusive). Thus, the domain is the closed interval $[-5, 4]$.

2. Determining the images of the given numbers:

- $f(-5) = 0$

- $f(-4) \approx 2$

- $f(-3) \approx 3$

- $f(-2) \approx 4$

- $f(0) \approx 3$

- $f(4) = 0$

3. Determining the antecedents of 5 and 3:

- For $f(x) = 5$, we look for points on the graph where $y = 5$. There are no such points. Thus, 5 has no antecedents.

- For $f(x) = 3$, we look for points on the graph where $y = 3$. From the graph, we see that $f(x) = 3$ for approximately $x = -3$ and $x=0$. Therefore, the antecedents of 3 are -3 and

0.

3. Final Answer

1. $D_f = [-5, 4]$

2. $f(-5) = 0$, $f(-4) \approx 2$, $f(-3) \approx 3$, $f(-2) \approx 4$, $f(0) \approx 3$, $f(4) = 0$

3. 5 has no antecedents. The antecedents of 3 are -3 and

0.