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Chapter 2: Problem 14

Solve the inequality. Then graph the solution set. $$x^{2} \leq 16$$

Short Answer

Expert verified
The solution to the inequality \(x^{2} \leq 16\) is the interval \(-4 \leq x \leq 4\).

Step by step solution

01

Solve the quadratic equation

First, solve the equation \( x^2 = 16 \). This can be done by taking the square root of both sides of the equation, yielding \( x = 4 \) and \( x = -4 \). These are the critical points that will be used to partition the number line.
02

Partition the number line

Mark -4 and 4 on the number line. This partitions the number line into three intervals: \(-\infty, -4\), \(-4, 4\), and \(4, \infty\).
03

Test the intervals

Choose a test point from each interval, and substitute it into the original inequality \(x^{2} \leq 16\) to check which intervals make the inequality true. For interval \(-\infty, -4\), choose -5, which gives \((-5)^2 \leq 16\), which is false. For interval \(-4, 4\), choose 0, which gives \(0^2 \leq 16\), which is true. For interval \(4, \infty\), choose 5, which gives \(5^2 \leq 16\), which is false.
04

Write the interval notation for the solution set

The solution set is the interval \(-4 \leq x \leq 4\) in interval notation. This includes the endpoints -4 and 4, because the original inequality was less than or equal to.
05

Graph the solution set

Draw a number line, mark -4 and 4 as closed dots (since the inequality includes these values), and shade the region between them. This is the graph of the solution set.

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Most popular questions from this chapter

Find all the zeros of the function. When there is an extended list of possible rational zeros, use a graphing utility to graph the function in order to disregard any of the possible rational zeros that are obviously not zeros of the function. $$f(s)=2 s^{3}-5 s^{2}+12 s-5$$

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