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Let's break down and solve both parts (a) and (b) of this problem step by step.

### (a) Find the range of \( f(x) = 1 + 2|\cos(\pi x) + \frac{1}{2}| \), algebraically.

#### Step 1: Start with the range of \( \cos x \)

We know that the cosine function, \( \cos x \), has a range of \( [-1, 1] \).

#### Step 2: Analyze \( \cos(\pi x) \)

Since \( \cos(\pi x) \) is simply a horizontal stretch of \( \cos x \), its range will also be \( [-1, 1] \). This is because multiplying the input of the cosine function by \( \pi \) does not affect its range.

#### Step 3: Apply the transformation \( \cos(\pi x) + \frac{1}{2} \)

The next step is to add \( \frac{1}{2} \) to \( \cos(\pi x) \). When you add \( \frac{1}{2} \) to any value in \( [-1, 1] \), the new range becomes:
\[
\left[ -1 + \frac{1}{2}, 1 + \frac{1}{2} \right] = \left[ -\frac{1}{2}, \frac{3}{2} \right]
\]

#### Step 4: Take the absolute value \( |\cos(\pi x) + \frac{1}{2}| \)

The absolute value operation affects negative values by making them positive. Thus, the range of \( |\cos(\pi x) + \frac{1}{2}| \) becomes:
\[
\left[ 0, \frac{3}{2} \right]
\]
because the smallest value \( -\frac{1}{2} \) is turned into 0.

#### Step 5: Multiply by 2

Now, multiplying by 2 stretches the range. So, we multiply each value in the range \( [0, \frac{3}{2}] \) by 2:
\[
\left[ 0, 3 \right]
\]

#### Step 6: Add 1

Finally, adding 1 shifts the entire range upward by 1, so the final range of \( f(x) \) becomes:
\[
\left[ 1, 4 \right]
\]

Thus, the range of \( f(x) = 1 + 2|\cos(\pi x) + \frac{1}{2}| \) is \( [1, 4] \).

### (b) Graph the function \( f(x) = 1 + 2|\cos(\pi x) + \frac{1}{2}| \)

You are asked to graph this function by starting with the graph of \( y = \cos x \) and applying appropriate transformations step by step. Let's outline the transformations:

1. **Start with the base graph \( y = \cos x \):**
   - The cosine graph oscillates between \( -1 \) and \( 1 \).

2. **Apply horizontal scaling to get \( \cos(\pi x) \):**
   - This stretches the cosine function horizontally by a factor of \( \pi \).

3. **Shift vertically by \( +\frac{1}{2} \) to get \( \cos(\pi x) + \frac{1}{2} \):**
   - This shifts the graph upward by 0.5 units, so it now oscillates between \( -\frac{1}{2} \) and \( \frac{3}{2} \).

4. **Apply absolute value to get \( |\cos(\pi x) + \frac{1}{2}| \):**
   - Any negative parts of the graph (those that dipped below 0) are reflected above the x-axis.

5. **Multiply by 2 to get \( 2|\cos(\pi x) + \frac{1}{2}| \):**
   - This stretches the graph vertically, so it now oscillates between 0 and 3.

6. **Add 1 to get \( 1 + 2|\cos(\pi x) + \frac{1}{2}| \):**
   - Finally, shift the entire graph upward by 1 unit, resulting in the final range from 1 to 4.

These transformations would produce the graph step-by-step. You are asked to draw and annotate each transformation, resulting in six graphs, including the original \( y = \cos x \).

---

Do you want a detailed drawing of each transformation? 

Here are five related questions for you:
1. How does the period of \( \cos(\pi x) \) compare to \( \cos(x) \)?
2. What effect does taking the absolute value of a function have on its graph?
3. Why does multiplying by 2 stretch the graph vertically?
4. How do you determine the amplitude of a cosine function after transformations?
5. What happens to the graph of \( \cos(x) \) when you shift it vertically by a constant?

**Tip:** When graphing functions, always apply transformations step by step in the correct order to avoid mistakes!

Find the range of f(x) = 1 + 2|cos(πx) + 1/2| algebraically. Graph the function f(x) = 1 + 2|cos(πx) + 1/2|.

This problem focuses on finding the range and graphing the function f(x) = 1 + 2|cos(πx) + 1/2|. It covers transformations applied to trigonometric functions such as cosine, absolute values, and vertical shifts. Suitable for students in Grades 10-12.

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