a. Answer The required intervals are [ − 2 , − 1 ] and [ 0 , 1 ] . Explanation Given: The polynomial function is g ( x ) = 3 x 4 + 4 x 3 − 3 . Formula/ concept used: The Intermediate Value Theorem states that f ( b ) Let “ a ” and “ b ”be real numbers such that a < b . If a polynomial function such that f ( a ) ≠ f ( b ) , then in interval [ a , b ] takes on every value between f ( a ) and f ( b ) . If f ( a ) and f ( b ) are of opposite sign then a zeros surely lie in interval [ a , b ] . Graph: The graph of function g ( x ) = 3 x 4 + 4 x 3 − 3 is shown in Figure-1 here. We have f ( a = − 2 ) = 13 and f ( b = − 1 ) = − 4 , thus f ( − 2 ) > 0 and f ( − 1 ) < 0 , hence, there exists a zero of g ( x ) = 3 x 4 + 4 x 3 − 3 in the intervalof unit length [ − 2 , − 1 ] .We see that f ( − 1.585 ) = 0 , i.e., x = − 1.585 is zero of polynomial function g ( x ) = 3 x 4 + 4 x 3 − 3 . We have f ( a = 0 ) = − 3 and f ( b = 1 ) = 4 , thus f ( 0 ) < 0 and f ( 1 ) > 0 , hence, there exists a zero of g ( x ) = 3 x 4 + 4 x 3 − 3 in the interval of unit length [ 0 , 1 ] . We see that f ( 0.779 ) = 0 , i.e., x = 0.779 is zero of polynomial function g ( x ) = 3 x 4 + 4 x 3 − 3 . Conclusion: There exists a zero of g ( x ) = 3 x 4 + 4 x 3 − 3 in each intervals [ − 2 , − 1 ] and [ 0 , 1 ] . b. To determine To find: The real zeros of polynomial function given in part (a) using graphing utility. Answer The zeros of function g ( x ) = 3 x 4 + 4 x 3 − 3 are x = − 1.585 and x = 0.779 . Explanation Given: The polynomial function and the graph of function g ( x ) = 3 x 4 + 4 x 3 − 3 in part (a). Concept used: The value of x where the graph of function intersect or touch the x -axis is the zero of function. Calculations: From graph Figure-1, in part (a) we see that graph of function f ( x ) intersect x -axis at points ( − 1.585 , 0 ) and ( 0.779 , 0 ) . Hence, x = − 1.585 and x = 0.779 are zeros of function g ( x ) = 3 x 4 + 4 x 3 − 3 . Conclusion: The points x = − 1.585 and x = 0.779 are zeros of function g ( x ) = 3 x 4 + 4 x 3 − 3 . c. To determine To verify: The answers of part (a) using table feature of graphing utility. Explanation Given: The zeros of function g ( x ) = 3 x 4 + 4 x 3 − 3 lie in intervals [ − 2 , − 1 ] and [ 0 , 1 ] . Verification: From the graph in Figure-1, we obtain a table of some solutions of the given polynomial function g ( x ) = 3 x 4 + 4 x 3 − 3 as: There is change of the function f ( x ) in intervals [ − 2 , − 1 ] and [ 0 , 1 ] each of length 1, these results are same as of part (a). Thus, the results of part (a) stands verified.

Precalculus with Limits: A Graphing Approach

7th Edition

ISBN: 9781305071711

Author: Ron Larson

Publisher: Brooks/Cole

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Question

Chapter 2.2, Problem 97E

To determine

To find:The interval of unit length in which zero of the given polynomial function lie definitely, using Intermediate Value Theorem and graphing utility.

Expert Solution

Answer to Problem 97E

The required intervals are [−2,−1] and [0,1] .

Explanation of Solution

Given:

The polynomial function is g(x)=3x4+4x3−3 .

Formula/ concept used:

The Intermediate Value Theorem states that f(b) Let “ a ” and “ b ”be real numbers such that a<b . If a polynomial function such that f(a)≠f(b) , then in interval [a,b] takes on every value between f(a) and f(b) . If f(a) and f(b) are of opposite sign then a zeros surely lie in interval [a,b] .

Graph:

The graph of function g(x)=3x4+4x3−3 is shown in Figure-1 here.

We have f(a=−2)=13 and f(b=−1)=−4 , thus f(−2)>0 and f(−1)<0 , hence, there exists a zero of g(x)=3x4+4x3−3 in the intervalof unit length [−2,−1] .We see that f(−1.585)=0 , i.e., x=−1.585 is zero of polynomial function g(x)=3x4+4x3−3 .

Precalculus with Limits: A Graphing Approach, Chapter 2.2, Problem 97E , additional homework tip 1

We have f(a=0)=−3 and f(b=1)=4 , thus f(0)<0 and f(1)>0 , hence, there exists a zero of g(x)=3x4+4x3−3 in the interval of unit length [0,1] . We see that f(0.779)=0 , i.e., x=0.779 is zero of polynomial function g(x)=3x4+4x3−3 .

Conclusion:

There exists a zero of g(x)=3x4+4x3−3 in each intervals [−2,−1] and [0,1] .

To determine

To find: The real zeros of polynomial function given in part (a) using graphing utility.

Expert Solution

Answer to Problem 97E

The zeros of function g(x)=3x4+4x3−3 are x=−1.585 and x=0.779 .

Explanation of Solution

Given:

The polynomial function and the graph of function g(x)=3x4+4x3−3 in part (a).

Concept used:

The value of x where the graph of function intersect or touch the x -axis is the zero of function.

Calculations:

From graph Figure-1, in part (a) we see that graph of function f(x) intersect x -axis at points (−1.585,0) and (0.779,0) .

Hence, x=−1.585 and x=0.779 are zeros of function g(x)=3x4+4x3−3 .

Conclusion:

The points x=−1.585 and x=0.779 are zeros of function g(x)=3x4+4x3−3 .

To determine

To verify: The answers of part (a) using table feature of graphing utility.

Expert Solution

Explanation of Solution

Given:

The zeros of function g(x)=3x4+4x3−3 lie in intervals [−2,−1] and [0,1] .

Verification:

From the graph in Figure-1, we obtain a table of some solutions of the given polynomial function g(x)=3x4+4x3−3 as:

Precalculus with Limits: A Graphing Approach, Chapter 2.2, Problem 97E , additional homework tip 2

There is change of the function f(x) in intervals [−2,−1] and [0,1] each of length 1, these results are same as of part (a).

Thus, the results of part (a) stands verified.

Chapter 2.2, Problem 96E

Chapter 2.2, Problem 98E

Chapter 2 Solutions

Precalculus with Limits: A Graphing Approach

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Prob. 68E Ch. 2.5 - Prob. 69E Ch. 2.5 - Prob. 70E Ch. 2.5 - Prob. 71E Ch. 2.5 - Prob. 72E Ch. 2.5 - Prob. 73E Ch. 2.5 - Prob. 74E Ch. 2.5 - Prob. 75E Ch. 2.5 - Prob. 76E Ch. 2.5 - Prob. 77E Ch. 2.5 - Prob. 78E Ch. 2.6 - Prob. 1E Ch. 2.6 - Prob. 2E Ch. 2.6 - Prob. 3E Ch. 2.6 - Prob. 4E Ch. 2.6 - Prob. 5E Ch. 2.6 - Prob. 6E Ch. 2.6 - Prob. 7E Ch. 2.6 - Prob. 8E Ch. 2.6 - Prob. 9E Ch. 2.6 - Prob. 10E Ch. 2.6 - Prob. 11E Ch. 2.6 - Prob. 12E Ch. 2.6 - Prob. 13E Ch. 2.6 - Prob. 14E Ch. 2.6 - Prob. 15E Ch. 2.6 - Prob. 16E Ch. 2.6 - Prob. 17E Ch. 2.6 - Prob. 18E Ch. 2.6 - Prob. 19E Ch. 2.6 - Prob. 20E Ch. 2.6 - Prob. 21E Ch. 2.6 - Prob. 22E Ch. 2.6 - Prob. 23E Ch. 2.6 - Prob. 24E Ch. 2.6 - Prob. 25E Ch. 2.6 - Prob. 26E Ch. 2.6 - Prob. 27E Ch. 2.6 - Prob. 28E Ch. 2.6 - Prob. 29E Ch. 2.6 - Prob. 30E Ch. 2.6 - Prob. 31E Ch. 2.6 - Prob. 32E Ch. 2.6 - Prob. 33E Ch. 2.6 - Prob. 34E Ch. 2.6 - Prob. 35E Ch. 2.6 - Prob. 36E Ch. 2.6 - Prob. 37E Ch. 2.6 - Prob. 38E Ch. 2.6 - Prob. 39E Ch. 2.6 - Prob. 40E Ch. 2.6 - Prob. 41E Ch. 2.6 - Prob. 42E Ch. 2.6 - Prob. 43E Ch. 2.6 - Prob. 44E Ch. 2.6 - Prob. 45E Ch. 2.6 - Prob. 46E Ch. 2.6 - Prob. 47E Ch. 2.6 - Prob. 48E Ch. 2.6 - Prob. 49E Ch. 2.6 - Prob. 50E Ch. 2.6 - Prob. 51E Ch. 2.6 - Prob. 52E Ch. 2.6 - Prob. 53E Ch. 2.6 - Prob. 54E Ch. 2.6 - Prob. 55E Ch. 2.6 - Prob. 56E Ch. 2.6 - Prob. 57E Ch. 2.6 - Prob. 58E Ch. 2.6 - Prob. 59E Ch. 2.6 - Prob. 60E Ch. 2.6 - Prob. 61E Ch. 2.6 - Prob. 62E Ch. 2.6 - Prob. 63E Ch. 2.6 - Prob. 64E Ch. 2.7 - Prob. 1E Ch. 2.7 - Prob. 2E Ch. 2.7 - Prob. 3E Ch. 2.7 - Prob. 4E Ch. 2.7 - Prob. 5E Ch. 2.7 - Prob. 6E Ch. 2.7 - Prob. 7E Ch. 2.7 - Prob. 8E Ch. 2.7 - Prob. 9E Ch. 2.7 - Prob. 10E Ch. 2.7 - Prob. 11E Ch. 2.7 - Prob. 12E Ch. 2.7 - Prob. 13E Ch. 2.7 - Prob. 14E Ch. 2.7 - Prob. 15E Ch. 2.7 - Prob. 16E Ch. 2.7 - Prob. 17E Ch. 2.7 - Prob. 18E Ch. 2.7 - Prob. 19E Ch. 2.7 - Prob. 20E Ch. 2.7 - Prob. 21E Ch. 2.7 - Prob. 22E Ch. 2.7 - Prob. 23E Ch. 2.7 - Prob. 24E Ch. 2.7 - Prob. 25E Ch. 2.7 - Prob. 26E Ch. 2.7 - Prob. 27E Ch. 2.7 - Prob. 28E Ch. 2.7 - Prob. 29E Ch. 2.7 - Prob. 30E Ch. 2.7 - Prob. 31E Ch. 2.7 - Prob. 32E Ch. 2.7 - Prob. 33E Ch. 2.7 - Prob. 34E Ch. 2.7 - Prob. 35E Ch. 2.7 - Prob. 36E Ch. 2.7 - Prob. 37E Ch. 2.7 - Prob. 38E Ch. 2.7 - Prob. 39E Ch. 2.7 - Prob. 40E Ch. 2.7 - Prob. 41E Ch. 2.7 - Prob. 42E Ch. 2.7 - Prob. 43E Ch. 2.7 - Prob. 44E Ch. 2.7 - Prob. 45E Ch. 2.7 - Prob. 46E Ch. 2.7 - Prob. 47E Ch. 2.7 - Prob. 48E Ch. 2.7 - Prob. 49E Ch. 2.7 - Prob. 50E Ch. 2.7 - Prob. 51E Ch. 2.7 - Prob. 52E Ch. 2.7 - Prob. 53E Ch. 2.7 - Prob. 54E Ch. 2.7 - Prob. 55E Ch. 2.7 - Prob. 56E Ch. 2.7 - Prob. 57E Ch. 2.7 - Prob. 58E Ch. 2.7 - Prob. 59E Ch. 2.7 - Prob. 60E Ch. 2.7 - Prob. 61E Ch. 2.7 - Prob. 62E Ch. 2.7 - Prob. 63E Ch. 2.7 - Prob. 64E Ch. 2.7 - Prob. 65E Ch. 2.7 - Prob. 66E Ch. 2.7 - Prob. 67E Ch. 2.7 - Prob. 68E Ch. 2.7 - Prob. 69E Ch. 2.7 - Prob. 70E Ch. 2.7 - Prob. 71E Ch. 2.7 - Prob. 72E Ch. 2.7 - Prob. 73E Ch. 2.7 - Prob. 74E Ch. 2.7 - Prob. 75E Ch. 2.7 - Prob. 76E Ch. 2.7 - Prob. 77E Ch. 2.7 - Prob. 78E Ch. 2.7 - Prob. 79E Ch. 2.7 - Prob. 80E Ch. 2.7 - Prob. 81E Ch. 2.7 - Prob. 82E Ch. 2.7 - Prob. 83E Ch. 2.7 - Prob. 84E Ch. 2.7 - Prob. 85E Ch. 2.7 - Prob. 86E Ch. 2.7 - Prob. 87E Ch. 2.7 - Prob. 88E Ch. 2.7 - Prob. 89E Ch. 2.7 - Prob. 90E Ch. 2.7 - Prob. 91E Ch. 2.7 - Prob. 92E Ch. 2.7 - Prob. 93E Ch. 2.7 - Prob. 94E Ch. 2.7 - Prob. 95E Ch. 2.7 - Prob. 96E Ch. 2.7 - Prob. 97E Ch. 2.7 - Prob. 98E Ch. 2.7 - Prob. 99E Ch. 2.7 - Prob. 100E Ch. 2.7 - Prob. 101E Ch. 2.7 - Prob. 102E Ch. 2.7 - Prob. 103E Ch. 2.7 - Prob. 104E Ch. 2.7 - Prob. 105E Ch. 2.7 - Prob. 106E Ch. 2.7 - Prob. 107E Ch. 2.8 - Prob. 1E Ch. 2.8 - Prob. 2E Ch. 2.8 - Prob. 3E Ch. 2.8 - Prob. 4E Ch. 2.8 - Prob. 5E Ch. 2.8 - Prob. 6E Ch. 2.8 - Prob. 7E Ch. 2.8 - Prob. 8E Ch. 2.8 - Prob. 9E Ch. 2.8 - Prob. 10E Ch. 2.8 - Prob. 11E Ch. 2.8 - Prob. 12E Ch. 2.8 - Prob. 13E Ch. 2.8 - Prob. 14E Ch. 2.8 - Prob. 15E Ch. 2.8 - Prob. 16E Ch. 2.8 - Prob. 17E Ch. 2.8 - Prob. 18E Ch. 2.8 - Prob. 19E Ch. 2.8 - Prob. 20E Ch. 2.8 - Prob. 21E Ch. 2.8 - Prob. 22E Ch. 2.8 - Prob. 23E Ch. 2.8 - Prob. 24E Ch. 2.8 - Prob. 25E Ch. 2.8 - Prob. 26E Ch. 2.8 - Prob. 27E Ch. 2.8 - Prob. 28E Ch. 2.8 - Prob. 29E Ch. 2.8 - Prob. 30E Ch. 2.8 - Prob. 31E Ch. 2.8 - Prob. 32E Ch. 2.8 - Prob. 33E Ch. 2.8 - Prob. 34E Ch. 2.8 - Prob. 35E Ch. 2.8 - Prob. 36E Ch. 2.8 - Prob. 37E Ch. 2.8 - Prob. 38E Ch. 2 - Prob. 1RE Ch. 2 - Prob. 2RE Ch. 2 - Prob. 3RE Ch. 2 - Prob. 4RE Ch. 2 - Prob. 5RE Ch. 2 - Prob. 6RE Ch. 2 - Prob. 7RE Ch. 2 - Prob. 8RE Ch. 2 - Prob. 9RE Ch. 2 - Prob. 10RE Ch. 2 - Prob. 11RE Ch. 2 - Prob. 12RE Ch. 2 - Prob. 13RE Ch. 2 - Prob. 14RE Ch. 2 - Prob. 15RE Ch. 2 - Prob. 16RE Ch. 2 - Prob. 17RE Ch. 2 - Prob. 18RE Ch. 2 - Prob. 19RE Ch. 2 - Prob. 20RE Ch. 2 - Prob. 21RE Ch. 2 - Prob. 22RE Ch. 2 - Prob. 23RE Ch. 2 - Prob. 24RE Ch. 2 - Prob. 25RE Ch. 2 - Prob. 26RE Ch. 2 - Prob. 27RE Ch. 2 - Prob. 28RE Ch. 2 - Prob. 29RE Ch. 2 - Prob. 30RE Ch. 2 - Prob. 31RE Ch. 2 - Prob. 32RE Ch. 2 - Prob. 33RE Ch. 2 - Prob. 34RE Ch. 2 - Prob. 35RE Ch. 2 - Prob. 36RE Ch. 2 - Prob. 37RE Ch. 2 - Prob. 38RE Ch. 2 - Prob. 39RE Ch. 2 - Prob. 40RE Ch. 2 - Prob. 41RE Ch. 2 - Prob. 42RE Ch. 2 - Prob. 43RE Ch. 2 - Prob. 44RE Ch. 2 - Prob. 45RE Ch. 2 - Prob. 46RE Ch. 2 - Prob. 47RE Ch. 2 - Prob. 48RE Ch. 2 - Prob. 49RE Ch. 2 - Prob. 50RE Ch. 2 - Prob. 51RE Ch. 2 - Prob. 52RE Ch. 2 - Prob. 53RE Ch. 2 - Prob. 54RE Ch. 2 - Prob. 55RE Ch. 2 - Prob. 56RE Ch. 2 - Prob. 57RE Ch. 2 - Prob. 58RE Ch. 2 - Prob. 59RE Ch. 2 - Prob. 60RE Ch. 2 - Prob. 61RE Ch. 2 - Prob. 62RE Ch. 2 - Prob. 63RE Ch. 2 - Prob. 64RE Ch. 2 - Prob. 65RE Ch. 2 - Prob. 66RE Ch. 2 - Prob. 67RE Ch. 2 - Prob. 68RE Ch. 2 - Prob. 69RE Ch. 2 - Prob. 70RE Ch. 2 - Prob. 71RE Ch. 2 - Prob. 72RE Ch. 2 - Prob. 73RE Ch. 2 - Prob. 74RE Ch. 2 - Prob. 75RE Ch. 2 - Prob. 76RE Ch. 2 - Prob. 77RE Ch. 2 - Prob. 78RE Ch. 2 - Prob. 79RE Ch. 2 - Prob. 80RE Ch. 2 - Prob. 81RE Ch. 2 - Prob. 82RE Ch. 2 - Prob. 83RE Ch. 2 - Prob. 84RE Ch. 2 - Prob. 85RE Ch. 2 - Prob. 86RE Ch. 2 - Prob. 87RE Ch. 2 - Prob. 88RE Ch. 2 - Prob. 89RE Ch. 2 - Prob. 90RE Ch. 2 - Prob. 91RE Ch. 2 - Prob. 92RE Ch. 2 - Prob. 93RE Ch. 2 - Prob. 94RE Ch. 2 - Prob. 95RE Ch. 2 - Prob. 96RE Ch. 2 - Prob. 97RE Ch. 2 - Prob. 98RE Ch. 2 - Prob. 99RE Ch. 2 - Prob. 100RE Ch. 2 - Prob. 101RE Ch. 2 - Prob. 102RE Ch. 2 - Prob. 103RE Ch. 2 - Prob. 104RE Ch. 2 - Prob. 105RE Ch. 2 - Prob. 106RE Ch. 2 - Prob. 107RE Ch. 2 - Prob. 108RE Ch. 2 - Prob. 109RE Ch. 2 - Prob. 110RE Ch. 2 - Prob. 111RE Ch. 2 - Prob. 112RE Ch. 2 - Prob. 113RE Ch. 2 - Prob. 114RE Ch. 2 - Prob. 115RE Ch. 2 - Prob. 116RE Ch. 2 - Prob. 117RE Ch. 2 - Prob. 118RE Ch. 2 - Prob. 119RE Ch. 2 - Prob. 120RE Ch. 2 - Prob. 121RE Ch. 2 - Prob. 122RE Ch. 2 - Prob. 123RE Ch. 2 - Prob. 124RE Ch. 2 - Prob. 125RE Ch. 2 - Prob. 126RE Ch. 2 - Prob. 127RE Ch. 2 - Prob. 128RE Ch. 2 - Prob. 129RE Ch. 2 - Prob. 130RE Ch. 2 - Prob. 131RE Ch. 2 - Prob. 132RE Ch. 2 - Prob. 133RE Ch. 2 - Prob. 134RE Ch. 2 - Prob. 135RE Ch. 2 - Prob. 136RE Ch. 2 - Prob. 137RE Ch. 2 - Prob. 138RE Ch. 2 - Prob. 139RE Ch. 2 - Prob. 140RE Ch. 2 - Prob. 141RE Ch. 2 - Prob. 142RE Ch. 2 - Prob. 143RE Ch. 2 - Prob. 144RE Ch. 2 - Prob. 145RE Ch. 2 - Prob. 146RE Ch. 2 - Prob. 147RE Ch. 2 - Prob. 148RE Ch. 2 - Prob. 149RE Ch. 2 - Prob. 150RE Ch. 2 - Prob. 151RE Ch. 2 - Prob. 152RE Ch. 2 - Prob. 153RE Ch. 2 - Prob. 154RE Ch. 2 - Prob. 155RE Ch. 2 - Prob. 156RE Ch. 2 - Prob. 157RE Ch. 2 - Prob. 158RE Ch. 2 - Prob. 159RE Ch. 2 - Prob. 160RE Ch. 2 - Prob. 161RE Ch. 2 - Prob. 162RE Ch. 2 - Prob. 163RE Ch. 2 - Prob. 164RE Ch. 2 - Prob. 165RE Ch. 2 - Prob. 166RE Ch. 2 - Prob. 167RE Ch. 2 - Prob. 168RE Ch. 2 - Prob. 1CT Ch. 2 - Prob. 2CT Ch. 2 - Prob. 3CT Ch. 2 - Prob. 4CT Ch. 2 - Prob. 5CT Ch. 2 - Prob. 6CT Ch. 2 - Prob. 7CT Ch. 2 - Prob. 8CT Ch. 2 - Prob. 9CT Ch. 2 - Prob. 10CT Ch. 2 - Prob. 11CT Ch. 2 - Prob. 12CT Ch. 2 - Prob. 13CT Ch. 2 - Prob. 14CT Ch. 2 - Prob. 15CT Ch. 2 - Prob. 16CT Ch. 2 - Prob. 17CT Ch. 2 - Prob. 18CT Ch. 2 - Prob. 19CT Ch. 2 - Prob. 20CT Ch. 2 - Prob. 21CT Ch. 2 - Prob. 22CT Ch. 2 - Prob. 23CT

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