EBK PRECALCULUS W/LIMITS
EBK PRECALCULUS W/LIMITS
4th Edition
ISBN: 9781337516853
Author: Larson
Publisher: CENGAGE CO
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Concept explainers

Minimization
In mathematics, traditional optimization problems are typically expressed in terms of minimization. When we talk about minimizing or maximizing a function, we refer to the maximum and minimum possible values of that function. This can be expressed in ter…
Maxima and Minima
The extreme points of a function are the maximum and the minimum points of the function. A maximum is attained when the function takes the maximum value and a minimum is attained when the function takes the minimum value.
Derivatives
A derivative means a change. Geometrically it can be represented as a line with some steepness. Imagine climbing a mountain which is very steep and 500 meters high. Is it easier to climb? Definitely not! Suppose walking on the road for 500 meters. Which …
Concavity
In calculus, concavity is a descriptor of mathematics that tells about the shape of the graph. It is the parameter that helps to estimate the maximum and minimum value of any of the functions and the concave nature using the graphical method. We use the …
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Question
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Chapter 2.2, Problem 34E

(a)

To determine

To find: all the real zeros of the given function.

(a)

Expert Solution
Check Mark

Answer to Problem 34E

The real zeros are x=−9,9

Explanation of Solution

Given information:

Consider f(x)=81−x2

Calculation:

Find the zero’s of the given function

The function

  f(x)=81−x2=0(x−9)(x+9)=0x=−9,9

Hence, the real zeros are x=−9,9

(b)

To determine

To find: whether the multiplicity of the each zero of the given function is even or odd .

(b)

Expert Solution
Check Mark

Answer to Problem 34E

Multiplicity of each zero is odd.

Explanation of Solution

Given information:

Consider f(x)=81−x2

Calculation:

From part a,

The real zeros are x=−9,9

Multiplicity of −9 is =1

Multiplicity of 9 is =1

Hence, Multiplicity of each zero is odd.

(c)

To determine

To find: the maximum possible number of turning points of the given function.

(c)

Expert Solution
Check Mark

Answer to Problem 34E

The maximum number of turning points =1

Explanation of Solution

Given information:

Consider f(x)=81−x2

Calculation:

The maximum number of turning points is one less than the degree of the function

Degree of the function is =2

Hence, the maximum number of turning points =1

(d)

To determine

To plot: the given function and to verify the results of previous parts.

(d)

Expert Solution
Check Mark

Explanation of Solution

Given information:

Consider f(x)=81−x2

Graph:

The graph of the f(x)=81−x2 is shown below

  EBK PRECALCULUS W/LIMITS, Chapter 2.2, Problem 34E

Hence, the results of previous parts are correct from the graph.

Previous
Chapter 2.2, Problem 33E
Next
Chapter 2.2, Problem 35E

Chapter 2 Solutions

EBK PRECALCULUS W/LIMITS

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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. 67RECh. 2 - Prob. 68RECh. 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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