College Algebra

Write each of the following using interval notation:

1. All numbers between \(-1\) and \(3\text{,}\) including both endpoints. Since we want the numbers between \(-1\) and \(3\text{,}\) those are our enpoints. Since we are including both endpoints, they are going to have square brackets []. We always write the smaller endpoint first, so in interval notation, we would have \([-1,3]\text{.}\)

2. \(0\le x \lt 9\text{.}\) Since we want the numbers between 0 and 9, those are our enpoints. Since we are including 0, it's going to have a square bracket [. Since we are not incluidng 9, it is going to have a round parentheses ). We always write the smaller endpoint first, so in interval notation, we would have \([0,9)\text{.}\)

3. \(-10 \lt x \lt 10\text{.}\) Since we want the numbers between \(-10\) and \(10\text{,}\) those are our enpoints. Since we are not including either endpoint, they are going to have round parentheses ()). We always write the smaller endpoint first, so in interval notation, we would have \((-10,10)\text{.}\)

Write each of the following in interval notation:

1. All numbers less than 7. Here is what that would look like on a number line:

   

   Since the interval goes on forever in the negative direction, we can think of \(-\infty\) as the other endpoint. However, since \(-\infty\) isn't an actual number, it will always get a round parentheses (. Since the endpoint 7 is not included, it will also get a round parentheses ). So, in interval notation, our answer would be \((-\infty,7)\text{.}\)

2. \(x\le -3.7\text{,}\) here is what that would look like on a number line:

   

   Since the interval goes on forever in the negative direction, we can think of \(-\infty\) as the other endpoint. However, since \(-\infty\) isn't an actual number, it will always get a round parentheses (. Since the endpoint \(-3.7\) is included, it will get a square bracket ]. So, in interval notation, our answer would be \((-\infty, -3.7])\text{.}\)

3. All positive numbers means all numbers greater than 0, so here is what that would look like on a number line:

   

   Since the interval goes on forever in the positive direction, we can think of \(\infty\) as the other endpoint. However, since \(\infty\) isn't an actual number, it will always get a round parentheses ). Since the endpoint 0 is not included, it will also get a round parentheses (. So, in interval notation, our answer would be \((0,\infty)\text{.}\)

Write each of the following in interval notation:

1. \(x \lt -1\) or \(x \ge 4\text{:}\) First, we know that \(x \lt -1\) is written as \((-\infty, -1)\text{,}\) and that \(x \ge 4\) is written as \([4,\infty)\text{.}\) Since our answer needs to include both possibilities, we combine them with a union to get our final answer of \((-\infty,-1)\cup[4,\infty)\text{.}\)

2. \(x \neq 42\text{:}\) That means we want to include all the numbers less than 42 and all the numbers greater than 42, just not 42 itself. That's two intervals that we'll combine with a union. All the numbers less than 42 is written as \((-\infty, 42)\text{,}\) and all the numbers greater than 42 is written as \((42,\infty)\text{.}\) Therefore, our final answer is \((-\infty, 42)\cup(42,\infty)\text{.}\)

3. Everything except \(\pm 2\text{.}\) Since this one is everything except two numbers (\(-2\) and \(2\)), it is easiest to start by drawing a number line:

   

   From the picture, we see that there are three intervals we need to work with: everything less than \(-2\text{,}\) everything between \(-2\) and \(2\text{,}\) and everything greater than \(2\text{.}\) In interval notation, those would each be \((-\infty, -2)\text{,}\) \((-2,2)\text{,}\) and \((2,\infty)\text{,}\) respectively. Therefore, our final answer to this question is \((-\infty, -2)\cup(-2,2)\cup(2,\infty)\text{.}\)