Math Problem Statement

Graph the function f(x) = 3 cos(x/2) - 1. (a) Identify the amplitude, period, phase shift, and vertical shift of the function. (b) Sketch the graph for one period, labeling key features (intercepts, maxima, minima).

Solution

Let's analyze the problem and solve it step by step.


Problem: The function is f(x)=3cos(x2)1f(x) = 3 \cos\left(\frac{x}{2}\right) - 1.

(a) Identify:

  1. Amplitude: The amplitude is the coefficient in front of the cosine function, which is 33. Amplitude=3\text{Amplitude} = 3

  2. Period: The period is determined by the coefficient of xx inside the cosine argument. Recall: Period=2πFrequency,Frequency=12.\text{Period} = \frac{2\pi}{\text{Frequency}}, \quad \text{Frequency} = \frac{1}{2}. Period=2π1/2=4π\text{Period} = \frac{2\pi}{1/2} = 4\pi

  3. Phase Shift: Since there is no additional constant added or subtracted to x2\frac{x}{2} inside the cosine function, there is no phase shift. Phase Shift=0\text{Phase Shift} = 0

  4. Vertical Shift: The constant 1-1 outside the cosine function shifts the graph vertically downward by 11 unit. Vertical Shift=1\text{Vertical Shift} = -1


(b) Sketch for One Period:

To sketch f(x)f(x) for one period [0,4π][0, 4\pi], we identify the key points of the cosine function:

  1. The general cosine function cos(x)\cos(x) starts at its maximum value.
  2. It moves to zero, then to its minimum, back to zero, and ends at its maximum over one period.

For f(x)=3cos(x2)1f(x) = 3 \cos\left(\frac{x}{2}\right) - 1:

  1. Start at x=0x = 0: f(0)=3cos(0)1=3(1)1=2f(0) = 3 \cos(0) - 1 = 3(1) - 1 = 2

  2. Quarter-period x=πx = \pi: f(π)=3cos(π2)1=3(0)1=1f(\pi) = 3 \cos\left(\frac{\pi}{2}\right) - 1 = 3(0) - 1 = -1

  3. Half-period x=2πx = 2\pi: f(2π)=3cos(π)1=3(1)1=4f(2\pi) = 3 \cos\left(\pi\right) - 1 = 3(-1) - 1 = -4

  4. Three-quarters-period x=3πx = 3\pi: f(3π)=3cos(3π2)1=3(0)1=1f(3\pi) = 3 \cos\left(\frac{3\pi}{2}\right) - 1 = 3(0) - 1 = -1

  5. End of period x=4πx = 4\pi: f(4π)=3cos(2π)1=3(1)1=2f(4\pi) = 3 \cos(2\pi) - 1 = 3(1) - 1 = 2

These points allow us to graph the function over one period.


Key Features:

  • Maximum Value: 22 at x=0,4πx = 0, 4\pi.
  • Minimum Value: 4-4 at x=2πx = 2\pi.
  • X-intercepts: At x=π,3πx = \pi, 3\pi, where f(x)=1f(x) = -1.

Would you like me to sketch the graph for you? Let me know if you want further clarifications or details.


Additional Questions:

  1. How do phase shifts affect the position of cosine graphs?
  2. What changes in the graph if the amplitude is altered?
  3. How is the vertical shift interpreted on the graph?
  4. How would the graph change for f(x)=3sin(x2)1f(x) = 3 \sin\left(\frac{x}{2}\right) - 1?
  5. How can you identify the period directly from the function?

Tip:

The amplitude determines the "height" of the graph, while the vertical shift moves it up or down without changing the shape.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Graphing of Functions
Cosine Function
Amplitude
Period
Phase Shift
Vertical Shift

Formulas

Amplitude = Coefficient of cos(x)
Period = 2π / Frequency
Phase Shift = Horizontal Shift
Vertical Shift = Constant added/subtracted to the function

Theorems

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Suitable Grade Level

Grades 10-12