Classify the following function `f(x)` defined in `RtoR` as injective, surjective, both or none.
`f(x)=x|x|`

Explanation

To classify the function \( f(x) = x|x| \) defined from \( \mathbb{R} \) to \( \mathbb{R} \) as injective, surjective, both, or neither, we can follow these steps:

### Step 1: Understand the function
The function \( f(x) = x|x| \) can be expressed in piecewise form:
- For \( x \geq 0 \): \( f(x) = x^2 \)
- For \( x < 0 \): \( f(x) = -x^2 \)

### Step 2: Analyze injectivity
A function is injective (one-to-one) if different inputs produce different outputs.

1. **For \( x \geq 0 \)**:
- The function \( f(x) = x^2 \) is a parabola opening upwards. It is increasing for \( x \geq 0 \), meaning it is injective in this interval.

2. **For \( x < 0 \)**:
- The function \( f(x) = -x^2 \) is a parabola opening downwards. It is decreasing for \( x < 0 \), meaning it is also injective in this interval.

3. **Check for overlap**:
- For \( x_1 \geq 0 \) and \( x_2 < 0 \), \( f(x_1) = x_1^2 \) (which is non-negative) and \( f(x_2) = -x_2^2 \) (which is non-positive). Therefore, there is no overlap in outputs for \( x_1 \) and \( x_2 \).

Since \( f(x) \) is injective in both intervals and does not overlap, we conclude that \( f(x) \) is injective.

### Step 3: Analyze surjectivity
A function is surjective (onto) if every possible output in the codomain \( \mathbb{R} \) is achieved by some input from the domain.

1. **Range for \( x \geq 0 \)**:
- The output of \( f(x) = x^2 \) for \( x \geq 0 \) is \( [0, \infty) \).

2. **Range for \( x < 0 \)**:
- The output of \( f(x) = -x^2 \) for \( x < 0 \) is \( (-\infty, 0) \).

3. **Combine the ranges**:
- The combined range of \( f(x) \) is \( (-\infty, 0] \).

Since the function does not cover all real numbers (specifically, it does not cover any positive values), \( f(x) \) is not surjective.

### Conclusion
The function \( f(x) = x|x| \) is injective but not surjective.

### Final Classification
- **Injective**: Yes
- **Surjective**: No
- **Both**: No
- **None**: No

Thus, the function is classified as **injective**.

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