To solve these inequalities by analyzing the graph of $y = f(x)$, let's go through each part by observing where the function $f(x)$ is positive, negative, or zero.

Part 1: $f(x) < 0$

This inequality asks for the intervals where $f(x)$ is negative. Looking at the graph, $f(x)$ is below the x-axis in specific intervals.

From the graph:

• $f(x) < 0$ on the interval $(-\infty, a) \cup (b, c)$ where $a$, $b$, and $c$ are points where $f(x)$ crosses the x-axis. Let’s assume:
  • $a \approx -2$,
  • $b \approx 0.5$,
  • $c \approx 1.5$.

So, the interval for $f(x) < 0$ is approximately: $(-\infty, -2) \cup (0.5, 1.5)$

Part 2: $f(x) \leq 0$

This inequality includes the points where $f(x) = 0$ as well as where $f(x) < 0$. From the graph, $f(x) = 0$ at $x \approx -2$, $x \approx 0.5$, and $x \approx 1.5$.

Thus, the interval for $f(x) \leq 0$ is: $(-\infty, -2] \cup [0.5, 1.5]$

Part 3: $f(x) > 0$

This inequality asks for the intervals where $f(x)$ is positive, meaning where the graph is above the x-axis.

From the graph:

• $f(x) > 0$ on the intervals $(a, b) \cup (c, \infty)$.

Using our approximate values for these crossing points: $(-2, 0.5) \cup (1.5, \infty)$

Part 4: $f(x) \geq 0$

This inequality includes the points where $f(x) = 0$ as well as where $f(x) > 0$.

Thus, the interval for $f(x) \geq 0$ is: $[-2, 0.5] \cup [1.5, \infty)$

Let me know if you would like further clarification on any of these solutions or have questions.

Related Questions:

1. How can we determine intervals of positivity and negativity by looking at a function's graph?
2. What does it mean when a function’s value is zero at certain points?
3. How does interval notation represent solutions to inequalities?
4. Why are certain intervals open or closed when solving inequalities?
5. How can you use critical points to solve inequalities on a graph?

Tip:

When analyzing a function's graph for inequalities, pay close attention to where the graph intersects the x-axis, as these points often define the boundaries of intervals where the function changes sign.