The function `f(x)="max" {(1-x), (1+x), 2}` is equivalent to

Options

A. `f(x)={(1-x",",x le -1),(2",", -1 lt x lt 1),(1+x"," ,x ge 1):}`
B. `f(x)={(1+x",",x le -1),(2",", -1 lt x lt 1),(1-x"," ,x ge 1):}`
C. `f(x)={(1-x",",x le -1),(1",", -1 lt x lt 1),(1+x"," ,x ge 1):}`
D. None of these

Answer

To solve the problem, we need to analyze the function \( f(x) = \max \{ 1 - x, 1 + x, 2 \} \) and determine its equivalent expression.

### Step-by-Step Solution:

1. **Identify the Functions**: We have three functions to consider:
- \( f_1(x) = 1 - x \)
- \( f_2(x) = 1 + x \)
- \( f_3(x) = 2 \)

2. **Find the Intersection Points**:
- To find where \( f_1(x) \) and \( f_2(x) \) intersect, set \( 1 - x = 1 + x \):
\[ 1 - x = 1 + x \implies -x - x = 0 \implies 2x = 0 \implies x = 0 \]
- To find where \( f_1(x) \) and \( f_3(x) \) intersect, set \( 1 - x = 2 \):
\[ 1 - x = 2 \implies -x = 1 \implies x = -1 \]
- To find where \( f_2(x) \) and \( f_3(x) \) intersect, set \( 1 + x = 2 \):
\[ 1 + x = 2 \implies x = 1 \]

3. **Evaluate the Functions at Key Points**:
- For \( x < -1 \):
- \( f_1(x) = 1 - x \) is greater than \( f_2(x) = 1 + x \) and \( f_3(x) = 2 \).
- For \( -1 \leq x < 0 \):
- At \( x = -1 \): \( f_1(-1) = 2 \), \( f_2(-1) = 0 \), \( f_3(-1) = 2 \) → \( f(x) = 2 \).
- At \( x = 0 \): \( f_1(0) = 1 \), \( f_2(0) = 1 \), \( f_3(0) = 2 \) → \( f(x) = 2 \).
- For \( 0 < x \leq 1 \):
- At \( x = 0.5 \): \( f_1(0.5) = 0.5 \), \( f_2(0.5) = 1.5 \), \( f_3(0.5) = 2 \) → \( f(x) = 2 \).
- For \( x > 1 \):
- \( f_1(x) \) will be less than \( f_2(x) \) and \( f_3(x) \).

4. **Conclusion**:
- The function \( f(x) \) is equal to \( 2 \) for \( -1 \leq x \leq 1 \).
- For \( x < -1 \), \( f(x) = 1 - x \).
- For \( x > 1 \), \( f(x) = 1 + x \).

Thus, the equivalent function can be expressed piecewise as:
\[ f(x) = \begin{cases} 1 - x & \text{if } x < -1 \\ 2 & \text{if } -1 \leq x \leq 1 \\ 1 + x & \text{if } x > 1 \end{cases} \]