To solve the inequality \( n^2 \leq 4 \), we start by rewriting it in a more manageable form:
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Rearranging the inequality: \[ n^2 - 4 \leq 0 \]
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Factoring: This can be factored as: \[ (n - 2)(n + 2) \leq 0 \]
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Finding the critical points: The critical points occur when the expression is equal to zero: \[ n - 2 = 0 \quad \Rightarrow \quad n = 2 \] \[ n + 2 = 0 \quad \Rightarrow \quad n = -2 \] So the critical points are \( n = -2 \) and \( n = 2 \).
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Testing intervals: We will test intervals around the critical points to determine where the product \((n - 2)(n + 2)\) is less than or equal to zero:
- Interval 1: \( n < -2 \) (e.g., \( n = -3 \)): \[ (-3 - 2)(-3 + 2) = (-5)(-1) = 5 \quad (\text{not } \leq 0) \]
- Interval 2: \( -2 < n < 2 \) (e.g., \( n = 0 \)): \[ (0 - 2)(0 + 2) = (-2)(2) = -4 \quad (\text{ } \leq 0) \]
- Interval 3: \( n > 2 \) (e.g., \( n = 3 \)): \[ (3 - 2)(3 + 2) = (1)(5) = 5 \quad (\text{not } \leq 0) \]
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Endpoints: The endpoints \( n = -2 \) and \( n = 2 \) are included in the inequality (because it is \( \leq 0 \)), so they are closed intervals.
Thus, the solution to \( n^2 \leq 4 \) is: \[ [-2, 2] \]
Graph of the Solution:
- The graph will show a closed interval between \(-2\) and \(2\), which includes both endpoints.
- Plot the points \(-2\) and \(2\) and shade the region in between.
- Since we are specifically told to "select an endpoint to change it from closed to open," let's choose the endpoint \(2\) to make it open:
The final solution thus changes to: \[ [-2, 2) \]
Representation on a Number Line:
- Mark \(-2\) with a closed dot (as it is included).
- Mark \(2\) with an open dot (as it is not included).
- Shade the region between \(-2\) and \(2\).
Here’s a textual representation:
<---|-----|-----|-----|-----|-----|-----|--->
-3 -2 -1 0 1 2 3
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This visual representation shows that the solution set includes all numbers from \(-2\) to \(2\), excluding \(2\).