Find the range of f (x) = 1 + 2 cos(πx) + 1/2

We are given the function:

\[
f(x) = 1 + 2 \cos(\pi x) + \frac{1}{2}
\]

First, simplify the constant terms:

\[
f(x) = 1 + \frac{1}{2} + 2 \cos(\pi x) = \frac{3}{2} + 2 \cos(\pi x)
\]

Now, we analyze the term \( 2 \cos(\pi x) \). The cosine function \( \cos(\pi x) \) always has a range between \(-1\) and \(1\), i.e.,

\[
-1 \leq \cos(\pi x) \leq 1
\]

Multiply this inequality by 2:

\[
-2 \leq 2 \cos(\pi x) \leq 2
\]

Now, add \( \frac{3}{2} \) to the entire inequality:

\[
\frac{3}{2} - 2 \leq f(x) \leq \frac{3}{2} + 2
\]

Simplify both sides:

\[
-\frac{1}{2} \leq f(x) \leq \frac{7}{2}
\]

Thus, the range of \( f(x) \) is:

\[
\left[ -\frac{1}{2}, \frac{7}{2} \right]
\]

### Would you like further explanation or details on any of the steps?

### Related questions:
1. What is the period of \( \cos(\pi x) \)?
2. How do the constants affect the range of trigonometric functions?
3. What is the maximum and minimum value of \( f(x) \) for specific \( x \)?
4. How can we find the range of other transformed cosine functions?
5. Can the range change if we add more terms involving trigonometric functions?

**Tip:** When transforming trigonometric functions, always start by determining the range of the core function before applying any shifts or scaling.

Find the range of f(x) = 1 + 2 cos(πx) + 1/2

Explore how to determine the range of the function f(x) = 1 + 2 cos(πx) + 1/2. This trigonometry problem includes analyzing the range of the cosine function and applying transformations. Detailed step-by-step solution for students in Grades 10-12.

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