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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Question
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Chapter 5.3, Problem 102E

a.

To determine

To find: The zero of f in the interval [0,6] .

a.

Expert Solution
Check Mark

Answer to Problem 102E

The zero of f in the interval [0,6] is at x=3.333¯ .

Explanation of Solution

Given information:

The following function:

  f(x)=3sin(0.6x2)

The following interval:

  [0,6]

Calculation:

The zero of f in the interval [0,6] is found by solving the equation as follows:

  3sin(0.6x2)=0             [Setting the function to 0]sin(0.6x2)=0               [Divide both sides by 3](0.6x2)=arcsin(0)       [Inverse trigonometric function]0.6x2=0                      [Value of arcsin(0)]0.6x=2                            [Adding 2 to both sides]x=20.6                             [Divide both sides by 0.6]x=3.333¯                          [Simplify]

b.

To determine

To graph: The given functions and describe the result.

b.

Expert Solution
Check Mark

Explanation of Solution

Given information:

The following functions:

  f(x)=3sin(0.6x2)g(x)=0.45x2+5.52x13.70

Near x=4 .

Graph:

Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press Y= button.

Step 3: Enter the expressions Y1=3sin(0.6x2) and Y2=0.45x2+5.52x13.70 which is required to graph.

Step 4: Press GRAPH button to graph the function.

The graph is obtained as:

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 5.3, Problem 102E

Interpretation:

As x approaches 4 , both the curves converge towards each other and then intersect at approximately x=5 .

c.

To determine

To find: The zero of g in the interval [0,6] and compare it with the result in part (a).

c.

Expert Solution
Check Mark

Answer to Problem 102E

The zeros of g is at x=3.455 and x=8.811

Explanation of Solution

Given information:

The following functions:

  g(x)=0.45x2+5.52x13.70

The following interval:

  [0,6]

Formula used:

The solutions of a quadratic equation in the form ax2+bx+c=0 is given by,

  x=b±b24ac2a

Calculation:

The zero of g in the interval [0,6] is found by solving the equation as follows:

  0.45x2+5.52x13.70=0                             [Setting the function to 0]x=5.52±(5.52)24(13.70)(0.45)2(0.45)      [Using quadratic formula]x=5.52±30.470424.660.9                        [Find square, multiply]x=5.52±5.81040.9                                       [Subtract] x=5.52±2.41050.9                                          [Find root]x=5.52+2.41050.9,5.522.41050.9                [Seperate the signs]x=3.455,8.811                                                [Simplify]

The zeros in the interval [0,6] is given by x=3.455

On comparing this with the result in part (a), which is x=3.333¯ , it ca be said that f(x) approaches 0 before g(x) .

Chapter 5 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

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Finding Local Maxima and Minima by Differentiation; Author: Professor Dave Explains;https://www.youtube.com/watch?v=pvLj1s7SOtk;License: Standard YouTube License, CC-BY