For the function f : R → R defined by f (x) = 2 loge(2x − 1) + 4
(a) Find the domain of f (x).
(b) Describe the full set of transformations, which when applied to the graph of
y = loge(x) produces the graph of f .

Let's break this down step by step.

### (a) Finding the domain of \( f(x) = 2 \ln(2x - 1) + 4 \)

The logarithmic function \( \ln(x) \) (or \( \log_e(x) \)) is only defined when its argument is positive, i.e., \( x > 0 \).

For the function \( f(x) = 2 \ln(2x - 1) + 4 \), we need the expression inside the logarithm, \( 2x - 1 \), to be greater than 0:

\[
2x - 1 > 0
\]

Solving this inequality:

\[
2x > 1 \quad \Rightarrow \quad x > \frac{1}{2}
\]

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

\[
\boxed{x > \frac{1}{2}}
\]

### (b) Describing the transformations

The basic logarithmic function is \( y = \ln(x) \), and we need to describe the transformations applied to this function to obtain \( f(x) = 2 \ln(2x - 1) + 4 \).

Let’s identify each transformation step by step:

1. **Horizontal scaling**: 
   - The term \( \ln(2x - 1) \) involves \( 2x - 1 \). First, consider the factor of 2 in front of \( x \). This causes a horizontal compression by a factor of \( \frac{1}{2} \) because for \( \ln(kx) \), the graph is compressed horizontally by \( k \).
   
2. **Horizontal translation**: 
   - The \( -1 \) in \( 2x - 1 \) indicates a horizontal shift to the right by \( \frac{1}{2} \). This is because solving \( 2x - 1 = 0 \) gives \( x = \frac{1}{2} \), meaning the graph is shifted right by \( \frac{1}{2} \).

3. **Vertical scaling**: 
   - The factor of 2 in front of \( \ln(2x - 1) \) causes a vertical stretching of the graph by a factor of 2.

4. **Vertical translation**: 
   - The \( +4 \) at the end indicates a vertical shift upwards by 4 units.

### Final set of transformations:

- **Horizontal compression** by a factor of \( \frac{1}{2} \).
- **Horizontal shift** right by \( \frac{1}{2} \).
- **Vertical stretch** by a factor of 2.
- **Vertical shift** up by 4 units.

These transformations, when applied to the graph of \( y = \ln(x) \), result in the graph of \( f(x) = 2 \ln(2x - 1) + 4 \).

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Would you like further details or have any questions about these transformations? Here are 5 related questions you might find useful:

1. How would you find the range of \( f(x) \)?
2. What would be the effect on the domain if the constant inside the logarithm was different, say \( \ln(3x - 1) \)?
3. How would the transformations change if the function was \( 3 \ln(2x - 1) + 2 \)?
4. Can you graph \( f(x) = 2 \ln(2x - 1) + 4 \) and describe its key features (intercepts, asymptotes)?
5. How would the graph behave as \( x \) approaches infinity?

**Tip**: Always ensure the argument inside the logarithmic function is positive when finding domains involving logarithms.

For the function f : R → R defined by f (x) = 2 loge(2x − 1) + 4 (a) Find the domain of f (x). (b) Describe the full set of transformations, which when applied to the graph of y = loge(x) produces the graph of f.

Explore the domain and transformations of the logarithmic function f(x) = 2 ln(2x - 1) + 4. Learn how to find the domain and apply graph transformations, including horizontal compression, translation, vertical stretch, and shift. Suitable for advanced high school students in Grades 11-12.

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