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Q54.

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Precalculus Mathematics for Calculus

Found in: Page 11

Book edition 7th Edition

Author(s) James Stewart, Lothar Redlin, Saleem Watson

Pages 948 pages

ISBN 9781337067508

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Short Answer

Express the inequality in interval notation, and then graph the corresponding interval.
$1 \leq x \leq 2$

The inequality form of the given interval $[1, 2]$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Apply the concept of interval and inequality.

If $a < b$ , then the open interval from a to b consists of all numbers between a and b and is denoted by $(a, b)$ . The closed interval from a to b includes the endpoints and is denoted by $[a, b]$ .

Using the set-builder form, we can write the intervals as inequality:

$(a, b) = \{x | a < x < b\} [a, b] = \{x | a \leq x \leq b\}$

Step 2. Express in terms of inequality.

Given inequality is $1 \leq x \leq 2$

This includes all the numbers between 1 and 2, including both.

Using set-builder form to express the inequality:

${x | 1 \leq x \leq 2}$

So, the interval is given by $[1, 2]$

Step 3. Graph the interval.

From step 2, we have $[1, 2]$

This represents all the numbers between 1 and 2, including both.

Since 1 and 2 are included in the interval, they will be represented by closed circles.

So, the graph will be:

Most popular questions for Math Textbooks

The rational expression $\frac{x^{2} - 9}{x - 3}$ is not defined for $x = 3$ . Complete the tables, and determine what value the expression approaches as x gets closer and closer to 3. Why is this reasonable? Factor the numerator of the expression and simplify to see why

Simplify the fractional expression.

$\frac{\frac{1}{{(x + h)}^{2}} - \frac{1}{x^{2}}}{h}$

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$(- 3, 0)$

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In Exercises 19-22, rewrite the expression using the given property of real numbers.

Commutative property of Addition $x + 3$

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