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Chapter 6: Problem 47

Graph each function over a two-period interval. $$y=-1-2 \cos 5 x$$

Short Answer

Expert verified
Graph the function over the interval \[0, \frac{4\pi}{5}\] by plotting the key points and repeating the pattern.

Step by step solution

01

- Identify the basic function

The given function is based on the cosine function. The basic form of the cosine function is given by \( y = \text{cos}(x) \).
02

- Identify the modifications to the basic function

Notice the given function is modified to \( y = -1 - 2 \text{cos}(5x) \). Here we have multiple modifications:- Amplitude is 2 (the coefficient of the cosine function is 2).- The cosine value is multiplied by -2, indicating a vertical flip and stretch.- The value is shifted vertically by -1.
03

- Determine the period

The cosine function with an angular frequency of 5, written as \( \text{cos}(5x) \), affects the period of the function. The period of the cosine function is given by \( \frac{2\pi}{B} \), where B is the coefficient of x. For \( B = 5 \), the period is \( \frac{2\pi}{5} \).
04

- Determine the two-period interval

One period of \( y = -1 - 2 \text{cos}(5x) \) is \( \frac{2\pi}{5} \). For a two-period interval, calculate it as \( 2 \times \frac{2\pi}{5} = \frac{4\pi}{5} \). Therefore, the function should be graphed over the interval \[0, \frac{4\pi}{5}\].
05

- Plot key points

Calculate and plot key points within one period based on the modifications:- Maximum point where \( \text{cos}(5x) = 1 \): \((0, -3)\)- Zero point where \( \text{cos}(5x) = 0 \): \( (\frac{\pi}{10}, -1)\)- Minimum point where \( \text{cos}(5x) = -1 \): \( (\frac{\pi}{5}, 1) \)
06

- Sketch the graph

Graph the points found in Step 5 and smoothly connect them to form one period. Repeat the pattern to complete the graph over the interval \[0, \frac{4\pi}{5}\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Amplitude
Understanding the amplitude of a trigonometric function is vital for graphing. The amplitude represents the height of the wave from its central axis. In the function given, \( y = -1 - 2 \cos{5x} \), the amplitude is the coefficient of the cosine term, which is 2. This indicates that the maximum value from the central axis to the peak of the wave is 2 units.Amplitude also affects the range:
  • For the cosine function \( y = \cos{(x)} \), the amplitude is 1, and its range is \([-1, 1]\).
  • When modified to \( -2 \cos{(5x)} \), the amplitude changes to 2, thus stretching the function vertically.
With an added vertical shift of -1, the range becomes \([-3, 1]\), accounting for both the amplitude and the shift.
Period of Trigonometric Functions
The period of a trigonometric function defines how long it takes for the function to repeat its pattern. For a cosine function \( y = \cos{(Bx)} \), the period is calculated as \( \frac{2\pi}{B} \), where B is the coefficient of x. In our example, B is 5, so the period is: \ \[ \frac{2\pi}{5} \] This shorter period means that the function completes a cycle more quickly compared to the standard cosine function. To graph over a two-period interval for our function, you'd plot from 0 to: \ \[ 2 \times \frac{2\pi}{5} = \frac{4\pi}{5} \]Understanding the period helps in spacing out key points uniformly along the x-axis while plotting the graph.
Vertical Shifts
Vertical shifts move the graph up or down along the y-axis. For the function \( y = -1 - 2 \cos{(5x)} \), the term '-1' represents a vertical shift downward by 1 unit.Here's how it affects the graph:
  • Without any shifts, the function \( -2 \cos{(5x)} \) ranges from -2 to 2.
  • Shifting the function down by 1 unit changes the range to [-3, 1].
When combined with amplitude changes, vertical shifts modify the location of peaks and troughs of the function, which is crucial for accurate graphing.
Cosine Function
The basic cosine function, \( y = \cos{(x)} \), forms a wave-like pattern with:
  • Amplitude 1
  • Period \( 2\pi \)
  • Peaks at \( y = 1 \) and troughs at \( y = -1 \)
For the given function, \( y = -1 - 2 \cos{(5x)} \), the general characteristics of the cosine wave still apply. However, due to modifications:
  • The amplitude changes to 2, making the waves taller.
  • The period shrinks to \( \frac{2\pi}{5} \), making the waves repeat more often.
  • The entire wave shifts down 1 unit.
These transformations form the basis for plotting the modified cosine function accurately.
Graphing Transformations
Graphing transformations include changes in amplitude, period, vertical shifts, and flips. Our function $$y = -1 - 2 \cos{(5x)}$$ undergoes multiple transformations.
  • Amplitude: The function's amplitude is modified to 2, resulting in taller waves.
  • Period: The period is shortened due to the coefficient of 5, giving a new period of \( \frac{2\pi}{5} \).
  • Vertical Flip: The negative sign before 2 indicates a vertical flip, turning peaks into troughs and vice versa.
  • Vertical Shift: The '-1' moves the function down by 1 unit.
When graphing, start by plotting the midline, add key points considering amplitude and period, and adjust for the vertical shift. Connect the points smoothly for one period and repeat for the desired interval to complete the graph.

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Most popular questions from this chapter

The position of a weight attached to a spring is \(s(t)=-5 \cos 4 \pi t\) inches after \(t\) seconds. (a) What is the maximum height that the weight rises above the equilibrium position? (b) What are the frequency and period? (c) When does the weight first reach its maximum height? (d) Calculate and interpret \(s(1.3)\)

Graph each function over a two-period interval. $$y=3 \sin \left(x-\frac{3 \pi}{2}\right)$$

Graph each function over a one-period interval. $$y=\frac{1}{2}+\sin \left[2\left(x+\frac{\pi}{4}\right)\right]$$

Find the value of \(s\) in the interval \(\left[0, \frac{\pi}{2}\right]\) that makes each statement true. $$\tan s=0.2126$$

Solve each problem. The voltage \(E\) in an electrical circuit is modeled by $$ E=5 \cos 120 \pi t $$ where \(t\) is time measured in seconds. (a) Find the amplitude and the period. (b) How many cycles are completed in 1 sec? (The number of cycles, or periods, completed in 1 sec is the frequency of the function.) (c) Find \(E\) when \(t=0,0.03,0.06,0.09,0.12\). (d) Graph \(E\) for \(0 \leq t \leq \frac{1}{30}\).
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