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

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

(a) Identify:

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

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

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

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

(b) Sketch for One Period:

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

1. The general cosine function $\cos(x)$ 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 $f(x) = 3 \cos\left(\frac{x}{2}\right) - 1$:

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

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

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

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

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

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

Key Features:

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

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 $f(x) = 3 \sin\left(\frac{x}{2}\right) - 1$?
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.