Question

The function \( f(x) = \max\left\{(1 - x), (1 + x), 2\right\}, \, x \in (-\infty, \infty) \) is equivalent to

• A. \( f(x) = \left\{ \begin{array}{ll} 1 + x, & x \leq -1 2, & -1 1 - x, & x \geq 1 \end{array} \right. \)
• B. \( f(x) = \left\{ \begin{array}{ll} 2, & -1 1 + x, & x \geq 1 1 - x, & x \leq -1 \end{array} \right. \)
• C. \( f(x) = \left\{ \begin{array}{ll} 1, & -1 1 + x, & x \geq 1 1 - x, & x \leq -1 \end{array} \right. \)
• D. None of the above

Hint:

When working with maximum functions, analyze the behavior of each component function in different intervals to identify the maximum value for each range.

Answer 1

The Correct Option is B

Step 1: Analyze the given function.
We are given the function \( f(x) = \max\left\{(1 - x), (1 + x), 2\right\} \). The function returns the maximum value among the three components \( 1 - x \), \( 1 + x \), and 2 for any value of \( x \).

Step 2: Examine the behavior of each part of the function.
- \( 1 - x \) is a decreasing linear function.
- \( 1 + x \) is an increasing linear function.
- The constant 2 is obviously always greater than or equal to the two linear functions for certain ranges of \( x \).

Step 3: Break the function into intervals.
We now find which part of the function is greater than the others in different ranges of \( x \).
- When \( x \leq -1 \), \( 1 - x \) dominates.
- When \( -1<x<1 \), the maximum is 2.
- When \( x \geq 1 \), \( 1 - x \) is less than \( 1 + x \), so \( 1 + x \) dominates.

Step 4: Write the final equivalent function.
Thus, the function \( f(x) \) can be written as: \[ f(x) = \begin{cases} 1 + x, & \text{if } x \leq -1 2, & \text{if } -1 1 - x, & \text{if } x \geq 1 \end{cases} \]

Step 5: Conclusion.
The correct equivalent function corresponds to option (B).