Great question! Let's walk through that step in more detail.

We start with the inequality:

$x^2 \geq 1$

Step 1: Understanding $x^2 \geq 1$

This inequality says that the square of $x$ must be greater than or equal to 1. To understand when this is true, let's break it down:

• Square Root Analysis: $x^2 = 1$ has two solutions: $x = 1$ and $x = -1$.
• Greater Than Analysis: For $x^2$ to be greater than 1, $x$ must either be larger than 1 or smaller than -1.

Step 2: Solve for $x$

To solve the inequality $x^2 \geq 1$, let's consider both possible cases:

1. When $x \geq 0$: $x^2 \geq 1 \implies x \geq 1$ Here, $x$ must be greater than or equal to 1.

2. When $x < 0$: $x^2 \geq 1 \implies x \leq -1$ Here, $x$ must be less than or equal to -1.

Step 3: Combine the Results

The solution to the inequality $x^2 \geq 1$ is the union of these two intervals:

• $x \geq 1$ or $x \leq -1$

In interval notation, this is expressed as:

$(-\infty, -1] \cup [1, \infty)$

Summary

The reason the inequality $x^2 \geq 1$ leads to $x \leq -1$ or $x \geq 1$ is because for the square of $x$ to be greater than or equal to 1, $x$ must either be sufficiently negative (less than or equal to -1) or sufficiently positive (greater than or equal to 1).

If you have any further questions or need clarification, feel free to ask! Here are some related questions you might explore:

1. What are the solutions to the inequality $x^2 < 4$?
2. How does the inequality $x^2 \leq 9$ differ from $x^2 \geq 9$ in terms of solutions?
3. Can you find the domain of the function $f(x) = \sqrt{4 - x^2}$?
4. What would be the domain of $g(x) = \sqrt{x^4 - 16}$?
5. How do absolute value inequalities relate to these kinds of problems?

Tip: Remember that when solving inequalities involving squares, the sign of $x$ is crucial for determining the correct solution intervals.