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 120E

(a)

To determine

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

(a)

Expert Solution
Check Mark

Answer to Problem 120E

The x− intercepts of the graph of f algebraically are {0,7π6,π2,3π2} .

Explanation of Solution

Given information:

The given trigonometric equation is −2sin2x+cosx=0 .

The given function is f(x)=cos2x−sinx .

The given graph is shown in figure (1).

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 5.5, Problem 120E , 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)=0cos2x+sinx=01−2sin2x+sinx=02sin2x−sinx−1=0

  2sin2x−2sinx+sinx−1=02sinx(sinx−1)+(sinx−1)=0(sinx−1)+(2sinx+1)=0

Making the factor sinx−1=0 .

  sinx−1=0sinx=1x=0,π2,3π2

Making the factor (2sinx+1)=0 .

  2sinx+1=0sinx=−12x=7π6

The set of the solution is {0,π2,3π2}∪{7π6}={0,7π6,π2,3π2} .

From the above graph, the curve crosses the x− axis at the points x={0,7π6,π2,3π2} .

Therefore, the x− intercepts of the graph of f algebraically are {0,7π6,π2,3π2} .

(b)

To determine

To find:The extremes of the function f .

(b)

Expert Solution
Check Mark

Answer to Problem 120E

The extremes of the function f is x=π2,3π2,π6,5π6 .

Explanation of Solution

Given information:

The given trigonometric equation is −2sin2x+cosx=0 .

The given function is f(x)=cos2x−sinx .

The given graph is shown in figure (1).

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 5.5, Problem 120E , additional homework tip 2

Figure (1)

Calculation:

Calculate the critical points.

  f'=dfdx=ddx(cos2x+sinx)=ddxcos2x+ddxsinx=−2sin2x+cosx

Make the derivative function equal to zero.

  f'=0−2sin2x+cosx=0−4sinxcosx+cosx=0cosx(−4sinx+1)=0

Making the factor cosx=0 .

  cosx=0x=π2,3π2

Making the factor (−4sinx+1)=0 .

  −4sinx+1=0sinx=12x=2πN+π6 or 2πN+5π6

Therefore, the extremes of the function f is x=π2,3π2,π6,5π6 .

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Chapter 5.5, Problem 119E
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Chapter 5.5, Problem 121E

Chapter 5 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

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