Let `f(x)=min{x,x^2}`, for every `x in R`. Then

Options

A. `f(x)={{:(x"," x ge1),(x^(2)","0 lex lt1),(x" ," x lt 0):}`
B. `f(x)={{:(x^(2)","x ge1),(x"," x lt1):}`
C. `f(x)={{:(x","x ge1),(x^(2)"," x lt1):}`
D. `f(x)={{:(x^(2)","x ge1),(x"," 0 lex lt1),(x^(2)"," x lt0):}`

Answer

To solve the problem, we need to analyze the function \( f(x) = \min\{x, x^2\} \) for all \( x \in \mathbb{R} \). We will determine the intervals where each expression is the minimum.

### Step-by-Step Solution:

1. **Identify the Functions**:
We have two functions to compare: \( y = x \) and \( y = x^2 \).

2. **Find Intersection Points**:
To find where these two functions intersect, we set them equal to each other:
\[ x = x^2 \]
Rearranging gives:
\[ x^2 - x = 0 \implies x(x - 1) = 0 \]
Thus, the solutions are:
\[ x = 0 \quad \text{and} \quad x = 1 \]

3. **Analyze Intervals**:
We need to analyze the behavior of \( f(x) \) in the intervals defined by the intersection points \( (-\infty, 0) \), \( [0, 1] \), and \( (1, \infty) \).

- **Interval \( (-\infty, 0) \)**:
For \( x < 0 \):
- \( x^2 > x \) (since \( x^2 \) is positive and \( x \) is negative).
- Therefore, \( f(x) = \min\{x, x^2\} = x \).

- **Interval \( [0, 1] \)**:
For \( 0 \leq x < 1 \):
- \( x^2 < x \) (since \( x^2 \) is less than \( x \) in this range).
- Therefore, \( f(x) = \min\{x, x^2\} = x^2 \).

- **Interval \( (1, \infty) \)**:
For \( x \geq 1 \):
- \( x^2 > x \) (since \( x^2 \) grows faster than \( x \)).
- Therefore, \( f(x) = \min\{x, x^2\} = x \).

4. **Combine Results**:
We can summarize the function \( f(x) \) as follows:
\[ f(x) = \begin{cases} x & \text{if } x < 0 \\ x^2 & \text{if } 0 \leq x < 1 \\ x & \text{if } x \geq 1 \end{cases} \]

### Conclusion:
The piecewise function can be expressed as:
\[ f(x) = \begin{cases} x & \text{if } x < 0 \\ x^2 & \text{if } 0 \leq x < 1 \\ x & \text{if } x \geq 1 \end{cases} \]