Precalculus with Limits
Precalculus with Limits
3rd Edition
ISBN: 9781133947202
Author: Ron Larson
Publisher: Cengage Learning
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Optimization
Optimization comes from the same root as "optimal". "Optimal" means the highest. When you do the optimization process, that is when you are "making it best" to maximize everything and to achieve optimal results, a set of parameters is the base for the se…
Integration
Integration means to sum the things. In mathematics, it is the branch of Calculus which is used to find the area under the curve. The operation subtraction is the inverse of addition, division is the inverse of multiplication. In the same way, integratio…
Application of Integration
In mathematics, the process of integration is used to compute complex area related problems. With the application of integration, solving area related problems, whether they are a curve, or a curve between lines, can be done easily.
Volume
In mathematics, we describe the term volume as a quantity that can express the total space that an object occupies at any point in time. Usually, volumes can only be calculated for 3-dimensional objects. By 3-dimensional or 3D objects, we mean objects th…
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Area refers to the amount of space a figure encloses and the number of square units that cover a shape. It is two-dimensional and is measured in square units.
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Question
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Chapter 2, Problem 63RE
To determine

To find: The solution set of the given inequality.

Expert Solution & Answer
Check Mark

Answer to Problem 63RE

Solution set:  (−23,14) .

  Precalculus with Limits, Chapter 2, Problem 63RE , additional homework tip 1

Explanation of Solution

Given information:

An inequality is given as - 12x2+5x<2 .

Concept used:

Key numbers of a polynomial are its zeros. Real zeros of a polynomial divides real line into intervals in which the polynomial does not change its sign. A test value is taken from each interval and corresponding value of inequality is calculated (whether positive or negative). The inequality maintains same sign for whole interval.

Calculation:

Given inequality is - 12x2+5x<2

  ⇒12x2+5x−2<0⇒12x2+8x−3x−2<0⇒4x(3x+2)−1(3x+2)<0⇒(3x+2)(4x−1)<0

Key numbers are x=−23,14 .

    Test intervalTest x− value Polynomial value

      12x2+5x−2<0

    		Conclusion
    (−∞,−23)x=−412(−4)2+5(−4)−2=+vePositive
    (−23,14)x=012(0)2+5(0)−2=−veNegative
    (14,∞)x=412(4)2+5(4)−2=+vePositive

From above table, it can be concluded that inequality is satisfied in the interval (−23,14) .

Hence, solution set for the given inequality will be (−23,14) .

  Precalculus with Limits, Chapter 2, Problem 63RE , additional homework tip 2

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Chapter 2 Solutions

Precalculus with Limits

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Prob. 52ECh. 2.6 - Prob. 53ECh. 2.6 - Prob. 54ECh. 2.6 - Prob. 55ECh. 2.6 - Prob. 56ECh. 2.6 - Prob. 57ECh. 2.6 - Prob. 58ECh. 2.6 - Prob. 59ECh. 2.6 - Prob. 60ECh. 2.6 - Prob. 61ECh. 2.6 - Prob. 62ECh. 2.6 - Prob. 63ECh. 2.6 - Prob. 64ECh. 2.6 - Prob. 65ECh. 2.6 - Prob. 66ECh. 2.6 - Prob. 67ECh. 2.6 - Prob. 68ECh. 2.6 - Prob. 69ECh. 2.6 - Prob. 70ECh. 2.6 - Prob. 71ECh. 2.6 - Prob. 72ECh. 2.6 - Prob. 73ECh. 2.6 - Prob. 74ECh. 2.6 - Prob. 75ECh. 2.6 - Prob. 76ECh. 2.6 - Prob. 77ECh. 2.6 - Prob. 78ECh. 2.6 - Prob. 79ECh. 2.6 - Prob. 80ECh. 2.6 - Prob. 81ECh. 2.6 - Prob. 82ECh. 2.7 - Prob. 1ECh. 2.7 - Prob. 2ECh. 2.7 - Prob. 3ECh. 2.7 - Prob. 4ECh. 2.7 - Prob. 5ECh. 2.7 - Prob. 6ECh. 2.7 - Prob. 7ECh. 2.7 - Prob. 8ECh. 2.7 - Prob. 9ECh. 2.7 - Prob. 10ECh. 2.7 - Prob. 11ECh. 2.7 - Prob. 12ECh. 2.7 - Prob. 13ECh. 2.7 - Prob. 14ECh. 2.7 - Prob. 15ECh. 2.7 - Prob. 16ECh. 2.7 - Prob. 17ECh. 2.7 - Prob. 18ECh. 2.7 - Prob. 19ECh. 2.7 - Prob. 20ECh. 2.7 - Prob. 21ECh. 2.7 - Prob. 22ECh. 2.7 - Prob. 23ECh. 2.7 - Prob. 24ECh. 2.7 - Prob. 25ECh. 2.7 - Prob. 26ECh. 2.7 - Prob. 27ECh. 2.7 - Prob. 28ECh. 2.7 - Prob. 29ECh. 2.7 - Prob. 30ECh. 2.7 - Prob. 31ECh. 2.7 - Prob. 32ECh. 2.7 - Prob. 33ECh. 2.7 - Prob. 34ECh. 2.7 - Prob. 35ECh. 2.7 - Prob. 36ECh. 2.7 - Prob. 37ECh. 2.7 - Prob. 38ECh. 2.7 - Prob. 39ECh. 2.7 - Prob. 40ECh. 2.7 - Prob. 41ECh. 2.7 - Prob. 42ECh. 2.7 - Prob. 43ECh. 2.7 - Prob. 44ECh. 2.7 - Prob. 45ECh. 2.7 - Prob. 46ECh. 2.7 - Prob. 47ECh. 2.7 - Prob. 48ECh. 2.7 - Prob. 49ECh. 2.7 - Prob. 50ECh. 2.7 - Prob. 51ECh. 2.7 - Prob. 52ECh. 2.7 - Prob. 53ECh. 2.7 - Prob. 54ECh. 2.7 - Prob. 55ECh. 2.7 - Prob. 56ECh. 2.7 - Prob. 57ECh. 2.7 - Prob. 58ECh. 2.7 - Prob. 59ECh. 2.7 - Prob. 60ECh. 2.7 - Prob. 61ECh. 2.7 - Prob. 62ECh. 2.7 - Prob. 63ECh. 2.7 - Prob. 64ECh. 2.7 - Prob. 65ECh. 2.7 - Prob. 66ECh. 2.7 - Prob. 67ECh. 2.7 - Prob. 68ECh. 2.7 - Prob. 69ECh. 2.7 - Prob. 70ECh. 2.7 - Prob. 71ECh. 2.7 - Prob. 72ECh. 2.7 - Prob. 73ECh. 2.7 - Prob. 74ECh. 2.7 - Prob. 75ECh. 2.7 - Prob. 76ECh. 2.7 - Prob. 77ECh. 2.7 - Prob. 78ECh. 2.7 - Prob. 79ECh. 2.7 - Prob. 80ECh. 2.7 - Prob. 81ECh. 2.7 - Prob. 82ECh. 2.7 - Prob. 83ECh. 2.7 - Prob. 84ECh. 2.7 - Prob. 85ECh. 2.7 - Prob. 86ECh. 2.7 - Prob. 87ECh. 2.7 - Prob. 88ECh. 2.7 - Prob. 89ECh. 2.7 - Prob. 90ECh. 2 - Prob. 1RECh. 2 - Prob. 2RECh. 2 - Prob. 3RECh. 2 - Prob. 4RECh. 2 - Prob. 5RECh. 2 - Prob. 6RECh. 2 - Prob. 7RECh. 2 - Prob. 8RECh. 2 - Prob. 9RECh. 2 - Prob. 10RECh. 2 - Prob. 11RECh. 2 - Prob. 12RECh. 2 - Prob. 13RECh. 2 - Prob. 14RECh. 2 - Prob. 15RECh. 2 - Prob. 16RECh. 2 - Prob. 17RECh. 2 - Prob. 18RECh. 2 - Prob. 19RECh. 2 - Prob. 20RECh. 2 - Prob. 21RECh. 2 - Prob. 22RECh. 2 - Prob. 23RECh. 2 - Prob. 24RECh. 2 - Prob. 25RECh. 2 - Prob. 26RECh. 2 - Prob. 27RECh. 2 - Prob. 28RECh. 2 - Prob. 29RECh. 2 - Prob. 30RECh. 2 - Prob. 31RECh. 2 - Prob. 32RECh. 2 - Prob. 33RECh. 2 - Prob. 34RECh. 2 - Prob. 35RECh. 2 - Prob. 36RECh. 2 - Prob. 37RECh. 2 - Prob. 38RECh. 2 - Prob. 39RECh. 2 - Prob. 40RECh. 2 - Prob. 41RECh. 2 - Prob. 42RECh. 2 - Prob. 43RECh. 2 - Prob. 44RECh. 2 - Prob. 45RECh. 2 - Prob. 46RECh. 2 - Prob. 47RECh. 2 - Prob. 48RECh. 2 - Prob. 49RECh. 2 - Prob. 50RECh. 2 - Prob. 51RECh. 2 - Prob. 52RECh. 2 - Prob. 53RECh. 2 - Prob. 54RECh. 2 - Prob. 55RECh. 2 - Prob. 56RECh. 2 - Prob. 57RECh. 2 - Prob. 58RECh. 2 - Prob. 59RECh. 2 - Prob. 60RECh. 2 - Prob. 61RECh. 2 - Prob. 62RECh. 2 - Prob. 63RECh. 2 - Prob. 64RECh. 2 - Prob. 65RECh. 2 - Prob. 66RECh. 2 - Prob. 1CTCh. 2 - Prob. 2CTCh. 2 - Prob. 3CTCh. 2 - Prob. 4CTCh. 2 - Prob. 5CTCh. 2 - Prob. 6CTCh. 2 - Prob. 7CTCh. 2 - Prob. 8CTCh. 2 - Prob. 9CTCh. 2 - Prob. 10CTCh. 2 - Prob. 11CTCh. 2 - Prob. 12CTCh. 2 - Prob. 13CTCh. 2 - Prob. 14CTCh. 2 - Prob. 15CTCh. 2 - Prob. 16CTCh. 2 - Prob. 17CTCh. 2 - Prob. 18CTCh. 2 - Prob. 1PSCh. 2 - Prob. 2PSCh. 2 - Prob. 3PSCh. 2 - Prob. 4PSCh. 2 - Prob. 5PSCh. 2 - Prob. 6PSCh. 2 - Prob. 7PSCh. 2 - Prob. 8PSCh. 2 - Prob. 9PSCh. 2 - Prob. 10PSCh. 2 - Prob. 11PSCh. 2 - Prob. 12PSCh. 2 - Prob. 13PS
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Area Between The Curve Problem No 1 - Applications Of Definite Integration - Diploma Maths II; Author: Ekeeda;https://www.youtube.com/watch?v=q3ZU0GnGaxA;License: Standard YouTube License, CC-BY