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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Limits
When it comes to calculus, limits are considered to be a very important topic of discussion. The main importance of limit lies in expressing the value of a function as it gets closer to a certain value. Also, limits can help you to understand other topic…
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Chapter 5.5, Problem 119E

(a)

To determine

To find:The x− intercepts of the graph of f algebraically.

(a)

Expert Solution
Check Mark

Answer to Problem 119E

The x− intercepts of the graph of f algebraically are {0,π3,π,5π3,2π} .

Explanation of Solution

Given information:

The given trigonometric equation is 2cos2x−cosx=0 .

The given function is f(x)=sin2x−sinx .

The given graph is shown in figure (1).

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 5.5, Problem 119E , additional homework tip 1

Figure (1)

Calculation:

The x− intercepts of the curve are those points where the curve crosses the x− axis. Calculate the x− intercepts of the graph algebraically.

  f(x)=0sin2x−sinx=02cos(2x+x2)sin(2x−x2)=02cos(3x2)sin(x2)=0

Making the factor sin(x2)=0 .

  sin(x2)=0x=0,2π

Making the factor cos(3x2)=0 .

  cos(3x2)=0x=π3,π,5π3

The set of the solution is {0,2π}∪{π3,π,5π3}={0,π3,π,5π3,2π} .

From the above graph, the curve crosses the x− axis at the points x={0,π3,π,5π3,2π} .

Therefore, the x− intercepts of the graph of f algebraically are {0,π3,π,5π3,2π} .

(b)

To determine

To find:The extremes of the function f .

(b)

Expert Solution
Check Mark

Answer to Problem 119E

The extremes of the function f is x=1.009, 1.8046 .

Explanation of Solution

Given information:

The given trigonometric equation is 2cos2x−cosx=0 .

The given function is f(x)=sin2x−sinx .

The given graph is shown in figure (1).

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 5.5, Problem 119E , additional homework tip 2

Figure (1)

Calculation:

Calculate the critical points.

  f'=dfdx=ddx(sin2x−sinx)=ddxsin2x−ddxsinx=2cos2x−cosx

Make the derivative function equal to zero.

  f'=02cos2x−cosx=02(2cos2x−1)−cosx=04cos2x−cosx−2=0

Calculate the roots of the quadratic equation ax2+bx+c=0 is x=−b±b2−4ac2a .

  cosx=−(−1)±(−1)2−4⋅4⋅(−2)2⋅4=1±1+328=1±338x=1.009, 1.8046

Therefore, the extremes of the function f is x=1.009, 1.8046 .

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Chapter 5.5, Problem 118E
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Chapter 5.5, Problem 120E

Chapter 5 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

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Limits and Continuity; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=9brk313DjV8;License: Standard YouTube License, CC-BY