Let's solve the inequality [tex]\( 1 - x^2 \leq 0 \)[/tex].
### Step-by-Step Solution
1. Rewrite the Inequality:
First, rewrite the inequality for easier manipulation:
[tex]\[
1 - x^2 \leq 0
\][/tex]
This can be rearranged as:
[tex]\[
-x^2 \leq -1
\][/tex]
2. Reverse Inequality by Dividing by -1:
Dividing both sides by [tex]\(-1\)[/tex] will reverse the inequality:
[tex]\[
x^2 \geq 1
\][/tex]
3. Solve the Resulting Equation:
Now we need to find the values of [tex]\( x \)[/tex] such that:
[tex]\[
x^2 \geq 1
\][/tex]
To do this, recognize that [tex]\( x^2 = 1 \)[/tex] gives us the boundary points:
[tex]\[
x = \pm 1
\][/tex]
4. Consider the Intervals:
To find the solution set for [tex]\( x^2 \geq 1 \)[/tex], consider the intervals created by [tex]\( x = \pm 1 \)[/tex]:
- When [tex]\( x > 1 \)[/tex], [tex]\( x^2 > 1 \)[/tex]
- When [tex]\( x < -1 \)[/tex], [tex]\( x^2 > 1 \)[/tex]
- When [tex]\(-1 \leq x \leq 1\)[/tex], [tex]\( x^2 \leq 1\)[/tex]
Therefore, the intervals where [tex]\( x^2 \geq 1 \)[/tex] will be:
[tex]\[
x \in (-\infty, -1] \cup [1, \infty)
\][/tex]
### Summary
The solution set for the inequality [tex]\( 1 - x^2 \leq 0 \)[/tex] is:
[tex]\[
x \in (-\infty, -1] \cup [1, \infty)
\][/tex]
So, the solution set is:
[tex]\[
\boxed{\text{Union}(\text{Interval}(-\infty, -1], \text{Interval}[1, \infty))}
\][/tex]
