PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
8th Edition
ISBN: 9781337904285
Author: Larson
Publisher: CENGAGE L
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Chapter 2.5, Problem 69E

a.

To determine

To find: The real zeros of given function using graphing utility.

a.

Expert Solution
Check Mark

Answer to Problem 69E

The real zero of h(x)=8x3−14x2+18x−9 is x=34 .

Explanation of Solution

Given:

The given function is h(x)=8x3−14x2+18x−9 .

Concept used:

If the graph of polynomial function f(x) intersect x -axis at point (a,0) , then x=a is the real zero of f(x) , also (x−a) is its one facto of f(x) .

Graph:

The graph of function h(x)=8x3−14x2+18x−9 is shown in Figure-1 here.

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 2.5, Problem 69E

The graph of the function h(x) intersect x-axis at point (34,0) , therefore, x=34 is the real zero of h(x)=8x3−14x2+18x−9 .

Conclusion:

The real zero of h(x)=8x3−14x2+18x−9 is x=34 .

b.

To determine

To find:The exact values of the imaginary zeros, using real zero.

b.

Expert Solution
Check Mark

Answer to Problem 69E

The imaginary zeros of h(x)=8x3−14x2+18x−9 are x=12−i52 and x=12+i52 .

Explanation of Solution

Given:

The real zero of h(x)=8x3−14x2+18x−9 is x=34 calculated in part (a)

Concept used:

The number of zeros of a polynomial equals its degree.

Calculations:

The given polynomial function h(x)=8x3−14x2+18x−9 is of degree 3, therefore there are at most 3 zeros of h(x) . As calculated in part (a) one real zero is x=34 and other two zeros are imaginary. Also, (x−34) is a factor of h(x)=8x3−14x2+18x−9 .

If we divide 8x3−14x2+18x−9 by (x−34) then quotient is 8x2−8x+12 that implies

  h(x)=(x−34)(8x2−8x+12)    … (1)

Now, we factor 8x2−8x+12 by solving equation 8x2−8x+12=0 using formula for equation ax2+bx+c=0 that is x=−b±b2−4ac2a .

Thus, x=8±64−4×8×122×8=8±−32016

  ⇒x=8±8−516

  ⇒x=1±i52

  ∵−5=i5,i=−1

The factors of 8x2−8x+12=0 are (x−12−i52) and (x−12+i52) .

Thus we have h(x)=(x−34)(x−12−i52)(x−12+i52)

Thus, the imaginary zeros of h(x)=8x3−14x2+18x−9 are x=12−i52 and x=12+i52

Conclusion:

The imaginary zeros of h(x)=8x3−14x2+18x−9 are x=12−i52 and x=12+i52 .

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Chapter 2.5, Problem 68E
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Chapter 2.5, Problem 70E

Chapter 2 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

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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. 2.7 - Prob. 91ECh. 2.7 - Prob. 92ECh. 2.7 - Prob. 93ECh. 2.7 - Prob. 94ECh. 2.7 - Prob. 95ECh. 2.7 - Prob. 96ECh. 2.7 - Prob. 97ECh. 2.7 - Prob. 98ECh. 2.7 - Prob. 99ECh. 2.7 - Prob. 100ECh. 2.7 - Prob. 101ECh. 2.8 - Prob. 1ECh. 2.8 - Prob. 2ECh. 2.8 - Prob. 3ECh. 2.8 - Prob. 4ECh. 2.8 - Prob. 5ECh. 2.8 - Prob. 6ECh. 2.8 - Prob. 7ECh. 2.8 - Prob. 8ECh. 2.8 - Prob. 9ECh. 2.8 - Prob. 10ECh. 2.8 - Prob. 11ECh. 2.8 - Prob. 12ECh. 2.8 - Prob. 13ECh. 2.8 - Prob. 14ECh. 2.8 - Prob. 15ECh. 2.8 - Prob. 16ECh. 2.8 - Prob. 17ECh. 2.8 - Prob. 18ECh. 2.8 - Prob. 19ECh. 2.8 - Prob. 20ECh. 2.8 - Prob. 21ECh. 2.8 - Prob. 22ECh. 2.8 - Prob. 23ECh. 2.8 - Prob. 24ECh. 2.8 - Prob. 25ECh. 2.8 - Prob. 26ECh. 2.8 - Prob. 27ECh. 2.8 - Prob. 28ECh. 2.8 - Prob. 29ECh. 2.8 - Prob. 30ECh. 2.8 - Prob. 31ECh. 2.8 - Prob. 32ECh. 2.8 - Prob. 33ECh. 2.8 - Prob. 34ECh. 2.8 - Prob. 35ECh. 2.8 - Prob. 36ECh. 2.8 - Prob. 37ECh. 2.8 - Prob. 38ECh. 2.8 - Prob. 39ECh. 2.8 - Prob. 40ECh. 2 - Prob. 1CRCh. 2 - Prob. 2CRCh. 2 - Prob. 3CRCh. 2 - Prob. 4CRCh. 2 - Prob. 5CRCh. 2 - Prob. 6CRCh. 2 - Prob. 7CRCh. 2 - Prob. 8CRCh. 2 - Prob. 9CRCh. 2 - Prob. 10CRCh. 2 - Prob. 11CRCh. 2 - Prob. 12CRCh. 2 - Prob. 13CRCh. 2 - Prob. 14CRCh. 2 - Prob. 15CRCh. 2 - Prob. 16CRCh. 2 - Prob. 17CRCh. 2 - Prob. 18CRCh. 2 - Prob. 19CRCh. 2 - Prob. 20CRCh. 2 - Prob. 21CRCh. 2 - Prob. 22CRCh. 2 - Prob. 23CRCh. 2 - Prob. 24CRCh. 2 - Prob. 25CRCh. 2 - Prob. 26CRCh. 2 - Prob. 27CRCh. 2 - Prob. 28CRCh. 2 - Prob. 29CRCh. 2 - Prob. 30CRCh. 2 - Prob. 31CRCh. 2 - Prob. 32CRCh. 2 - Prob. 33CRCh. 2 - Prob. 34CRCh. 2 - Prob. 35CRCh. 2 - Prob. 36CRCh. 2 - Prob. 37CRCh. 2 - Prob. 38CRCh. 2 - Prob. 39CRCh. 2 - Prob. 40CRCh. 2 - Prob. 41CRCh. 2 - Prob. 42CRCh. 2 - Prob. 43CRCh. 2 - Prob. 44CRCh. 2 - Prob. 45CRCh. 2 - Prob. 46CRCh. 2 - Prob. 47CRCh. 2 - Prob. 48CRCh. 2 - Prob. 49CRCh. 2 - Prob. 50CRCh. 2 - Prob. 51CRCh. 2 - Prob. 52CRCh. 2 - Prob. 53CRCh. 2 - Prob. 54CRCh. 2 - Prob. 55CRCh. 2 - Prob. 56CRCh. 2 - Prob. 57CRCh. 2 - Prob. 58CRCh. 2 - Prob. 59CRCh. 2 - Prob. 60CRCh. 2 - Prob. 61CRCh. 2 - Prob. 62CRCh. 2 - Prob. 63CRCh. 2 - Prob. 64CRCh. 2 - Prob. 65CRCh. 2 - Prob. 66CRCh. 2 - Prob. 67CRCh. 2 - Prob. 68CRCh. 2 - Prob. 69CRCh. 2 - Prob. 70CRCh. 2 - Prob. 71CRCh. 2 - Prob. 72CRCh. 2 - Prob. 73CRCh. 2 - Prob. 74CRCh. 2 - Prob. 75CRCh. 2 - Prob. 76CRCh. 2 - Prob. 77CRCh. 2 - Prob. 78CRCh. 2 - Prob. 79CRCh. 2 - Prob. 80CRCh. 2 - Prob. 81CRCh. 2 - Prob. 82CRCh. 2 - Prob. 83CRCh. 2 - Prob. 84CRCh. 2 - Prob. 85CRCh. 2 - Prob. 86CRCh. 2 - Prob. 87CRCh. 2 - Prob. 88CRCh. 2 - Prob. 89CRCh. 2 - Prob. 90CRCh. 2 - Prob. 91CRCh. 2 - Prob. 92CRCh. 2 - Prob. 93CRCh. 2 - Prob. 94CRCh. 2 - Prob. 95CRCh. 2 - Prob. 96CRCh. 2 - Prob. 97CRCh. 2 - Prob. 98CRCh. 2 - Prob. 99CRCh. 2 - Prob. 100CRCh. 2 - Prob. 101CRCh. 2 - Prob. 102CRCh. 2 - Prob. 103CRCh. 2 - Prob. 104CRCh. 2 - Prob. 105CRCh. 2 - Prob. 106CRCh. 2 - Prob. 107CRCh. 2 - Prob. 108CRCh. 2 - Prob. 109CRCh. 2 - Prob. 110CRCh. 2 - Prob. 111CRCh. 2 - Prob. 112CRCh. 2 - Prob. 113CRCh. 2 - Prob. 114CRCh. 2 - Prob. 115CRCh. 2 - Prob. 116CRCh. 2 - Prob. 117CRCh. 2 - Prob. 118CRCh. 2 - Prob. 119CRCh. 2 - Prob. 120CRCh. 2 - Prob. 121CRCh. 2 - Prob. 122CRCh. 2 - Prob. 123CRCh. 2 - Prob. 124CRCh. 2 - Prob. 125CRCh. 2 - Prob. 126CRCh. 2 - Prob. 127CRCh. 2 - Prob. 128CRCh. 2 - Prob. 129CRCh. 2 - Prob. 130CRCh. 2 - Prob. 131CRCh. 2 - Prob. 132CRCh. 2 - Prob. 133CRCh. 2 - Prob. 134CRCh. 2 - Prob. 135CRCh. 2 - Prob. 136CRCh. 2 - Prob. 137CRCh. 2 - Prob. 138CRCh. 2 - Prob. 139CRCh. 2 - Prob. 140CRCh. 2 - Prob. 141CRCh. 2 - Prob. 142CRCh. 2 - Prob. 143CRCh. 2 - Prob. 144CRCh. 2 - Prob. 145CRCh. 2 - Prob. 146CRCh. 2 - Prob. 147CRCh. 2 - Prob. 148CRCh. 2 - Prob. 149CRCh. 2 - Prob. 150CRCh. 2 - Prob. 151CRCh. 2 - Prob. 152CRCh. 2 - Prob. 153CRCh. 2 - Prob. 154CRCh. 2 - Prob. 155CRCh. 2 - Prob. 156CRCh. 2 - Prob. 157CRCh. 2 - Prob. 158CRCh. 2 - Prob. 159CRCh. 2 - Prob. 160CRCh. 2 - Prob. 161CRCh. 2 - Prob. 162CRCh. 2 - Prob. 163CRCh. 2 - Prob. 164CRCh. 2 - Prob. 165CRCh. 2 - Prob. 166CRCh. 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. 19CTCh. 2 - Prob. 20CTCh. 2 - Prob. 21CT
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Derivatives of Trigonometric Functions - Product Rule Quotient & Chain Rule - Calculus Tutorial; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=_niP0JaOgHY;License: Standard YouTube License, CC-BY