If `f(x)=[4x]-2x` with domain ` 0 le x le 2`, then f(x) can also be written as

Options

A. 2x
B. `-x`
C. `-2x`
D. none of the above

Explanation

To solve the problem, we need to analyze the function \( f(x) = [4x] - 2x \) over the domain \( 0 \leq x \leq 2 \). Here, \( [4x] \) denotes the greatest integer less than or equal to \( 4x \).

### Step-by-step Solution:

1. **Identify the intervals**: Since \( 4x \) varies from \( 0 \) to \( 8 \) as \( x \) goes from \( 0 \) to \( 2 \), we need to determine the intervals where \( [4x] \) remains constant. The integer values of \( 4x \) will change at \( x = \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \) and \( 1, \) and so on, up to \( x = 2 \).

2. **Calculate \( f(x) \) for different intervals**:
- For \( 0 \leq x < \frac{1}{4} \):
- \( 4x < 1 \) so \( [4x] = 0 \)
- \( f(x) = 0 - 2x = -2x \)
- For \( \frac{1}{4} \leq x < \frac{1}{2} \):
- \( 1 \leq 4x < 2 \) so \( [4x] = 1 \)
- \( f(x) = 1 - 2x \)
- For \( \frac{1}{2} \leq x < \frac{3}{4} \):
- \( 2 \leq 4x < 3 \) so \( [4x] = 2 \)
- \( f(x) = 2 - 2x \)
- For \( \frac{3}{4} \leq x < 1 \):
- \( 3 \leq 4x < 4 \) so \( [4x] = 3 \)
- \( f(x) = 3 - 2x \)
- For \( 1 \leq x < \frac{5}{4} \):
- \( 4 \leq 4x < 5 \) so \( [4x] = 4 \)
- \( f(x) = 4 - 2x \)
- For \( \frac{5}{4} \leq x < \frac{3}{2} \):
- \( 5 \leq 4x < 6 \) so \( [4x] = 5 \)
- \( f(x) = 5 - 2x \)
- For \( \frac{3}{2} \leq x < \frac{7}{4} \):
- \( 6 \leq 4x < 7 \) so \( [4x] = 6 \)
- \( f(x) = 6 - 2x \)
- For \( \frac{7}{4} \leq x \leq 2 \):
- \( 7 \leq 4x < 8 \) so \( [4x] = 7 \)
- \( f(x) = 7 - 2x \)

3. **Combine the results**: We can summarize the piecewise function as follows:
\[ f(x) = \begin{cases} -2x & \text{for } 0 \leq x < \frac{1}{4} \\ 1 - 2x & \text{for } \frac{1}{4} \leq x < \frac{1}{2} \\ 2 - 2x & \text{for } \frac{1}{2} \leq x < \frac{3}{4} \\ 3 - 2x & \text{for } \frac{3}{4} \leq x < 1 \\ 4 - 2x & \text{for } 1 \leq x < \frac{5}{4} \\ 5 - 2x & \text{for } \frac{5}{4} \leq x < \frac{3}{2} \\ 6 - 2x & \text{for } \frac{3}{2} \leq x < \frac{7}{4} \\ 7 - 2x & \text{for } \frac{7}{4} \leq x \leq 2 \end{cases} \]

4. **Conclusion**: Since the function \( f(x) \) changes at multiple intervals and does not simplify to a single expression over the entire domain, the correct answer is that \( f(x) \) cannot be expressed as \( 2x - x - 2x \) or any other single function. Thus, the answer is "none of the above."