To solve the inequality [tex]\( x + 1 \geq x^2 + x \)[/tex], follow these steps:
1. Subtract [tex]\( x + 1 \)[/tex] from both sides of the inequality:
[tex]\[
x + 1 - (x^2 + x) \geq 0
\][/tex]
2. Simplify the expression:
[tex]\[
1 - x^2 \geq 0
\][/tex]
3. Rewrite the inequality in a more standard form:
[tex]\[
-x^2 + 1 \geq 0
\][/tex]
This can be rewritten as:
[tex]\[
1 - x^2 \geq 0
\][/tex]
4. Factor the inequality:
Notice that [tex]\(1 - x^2\)[/tex] is a difference of squares, which can be factored as:
[tex]\[
(1 - x)(1 + x) \geq 0
\][/tex]
5. Determine the critical points:
Solve [tex]\( (1 - x)(1 + x) = 0 \)[/tex]:
[tex]\[
1 - x = 0 \quad \text{or} \quad 1 + x = 0
\][/tex]
This gives the critical points:
[tex]\[
x = 1 \quad \text{and} \quad x = -1
\][/tex]
6. Analyze the intervals defined by the critical points:
The critical points divide the number line into three intervals: [tex]\( (-\infty, -1) \)[/tex], [tex]\( (-1, 1) \)[/tex], and [tex]\( (1, \infty) \)[/tex].
Let's test a point from each interval to determine where the inequality holds:
- For [tex]\( x \in (-\infty, -1) \)[/tex], choose [tex]\( x = -2 \)[/tex]:
[tex]\[
(1 - (-2))(1 + (-2)) = (1 + 2)(1 - 2) = 3 \cdot (-1) = -3 \quad (\text{negative})
\][/tex]
- For [tex]\( x \in (-1, 1) \)[/tex], choose [tex]\( x = 0 \)[/tex]:
[tex]\[
(1 - 0)(1 + 0) = 1 \cdot 1 = 1 \quad (\text{positive})
\][/tex]
- For [tex]\( x \in (1, \infty) \)[/tex], choose [tex]\( x = 2 \)[/tex]:
[tex]\[
(1 - 2)(1 + 2) = (-1) \cdot 3 = -3 \quad (\text{negative})
\][/tex]
Also, check the critical points [tex]\(-1\)[/tex] and [tex]\(1\)[/tex]:
- At [tex]\( x = -1 \)[/tex]:
[tex]\[
(1 - (-1))(1 + (-1)) = (1 + 1)(1 - 1) = 2 \cdot 0 = 0 \quad (\text{zero, which is non-negative})
\][/tex]
- At [tex]\( x = 1 \)[/tex]:
[tex]\[
(1 - 1)(1 + 1) = 0 \cdot 2 = 0 \quad (\text{zero, which is non-negative})
\][/tex]
7. Conclusion:
Based on the test points, we find that the expression [tex]\((1 - x)(1 + x)\)[/tex] is non-negative on the interval [tex]\([-1, 1]\)[/tex].
Therefore, the solution set is:
[tex]\[
x \in [-1, 1]
\][/tex]
Thus, the solution to the inequality [tex]\( x + 1 \geq x^2 + x \)[/tex] is:
[tex]\[
[-1, 1]
\][/tex]
