If `f: R to R, g , R to R` be two funcitons, and `h(x) = 2 "min" {f(x) - g(x),0}` then `h(x)=`

Options

A. `f(x) + g(x) -|g(x)-f(x)|`
B. `f(x) + g(x) +|g(x) - f(x)|`
C. `f(x)-g(x)+|g(x) +|g(x) - f(x)|`
D. `f(x) -g(x) -|g(x) - f(x)|`

Explanation

To solve the problem, we need to analyze the function \( h(x) = 2 \min \{ f(x) - g(x), 0 \} \). We will consider two cases based on the relationship between \( f(x) \) and \( g(x) \).

### Step-by-Step Solution:

1. **Understanding the Function**:
We have \( h(x) = 2 \min \{ f(x) - g(x), 0 \} \). This means that \( h(x) \) takes the minimum value between \( f(x) - g(x) \) and \( 0 \), and then multiplies that minimum value by \( 2 \).

2. **Case 1: When \( f(x) - g(x) \geq 0 \)**:
- If \( f(x) - g(x) \) is greater than or equal to \( 0 \), then:
\[ \min \{ f(x) - g(x), 0 \} = 0 \]
- Therefore, substituting this into \( h(x) \):
\[ h(x) = 2 \cdot 0 = 0 \]

3. **Case 2: When \( f(x) - g(x) < 0 \)**:
- If \( f(x) - g(x) \) is less than \( 0 \), then:
\[ \min \{ f(x) - g(x), 0 \} = f(x) - g(x) \]
- Thus, substituting this into \( h(x) \):
\[ h(x) = 2(f(x) - g(x)) \]

4. **Final Expression for \( h(x) \)**:
- We can summarize the results from both cases:
\[ h(x) = \begin{cases} 0 & \text{if } f(x) - g(x) \geq 0 \\ 2(f(x) - g(x)) & \text{if } f(x) - g(x) < 0 \end{cases} \]

5. **Alternative Representation**:
- We can express \( h(x) \) in a single formula:
\[ h(x) = 2 \min \{ f(x) - g(x), 0 \} = 2(f(x) - g(x)) \cdot \mathbf{1}_{\{f(x) - g(x) < 0\}} \]
- Where \( \mathbf{1}_{\{f(x) - g(x) < 0\}} \) is an indicator function that is \( 1 \) when \( f(x) - g(x) < 0 \) and \( 0 \) otherwise.