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 will analyze the function step by step.

### Step 1: Rewrite the function
The function \( f(x) = x|x| \) can be expressed differently based on the value of \( x \):
- If \( x \geq 0 \), then \( |x| = x \), so \( f(x) = x \cdot x = x^2 \).
- If \( x < 0 \), then \( |x| = -x \), so \( f(x) = x \cdot (-x) = -x^2 \).

Thus, we can define the function piecewise:
\[ f(x) = \begin{cases} x^2 & \text{if } x \geq 0 \\ -x^2 & \text{if } x < 0 \end{cases} \]

### Step 2: Analyze injectivity (one-to-one)
To determine if \( f(x) \) is injective, we need to check if different inputs produce different outputs.

- For \( x \geq 0 \):
- The function \( f(x) = x^2 \) is a parabola opening upwards. It is not injective because \( f(a) = f(b) \) for \( a = 1 \) and \( b = -1 \) (both yield \( f(1) = 1 \) and \( f(-1) = -1 \)).

- For \( x < 0 \):
- The function \( f(x) = -x^2 \) is a parabola opening downwards. It is also not injective because \( f(a) = f(b) \) for \( a = -1 \) and \( b = -2 \) (both yield \( f(-1) = -1 \) and \( f(-2) = -4 \)).

Since the function is not one-to-one in both cases, we conclude that \( f(x) \) is **not injective**.

### Step 3: Analyze surjectivity (onto)
To determine if \( f(x) \) is surjective, we need to check if every element in the codomain \( \mathbb{R} \) is covered by the function.

- The range of \( f(x) \) when \( x \geq 0 \) is \( [0, \infty) \) (since \( f(x) = x^2 \)).
- The range of \( f(x) \) when \( x < 0 \) is \( (-\infty, 0] \) (since \( f(x) = -x^2 \)).

Combining both ranges, we see that the overall range of \( f(x) \) is \( (-\infty, 0] \cup [0, \infty) = \mathbb{R} \).

Since the range of \( f(x) \) covers all real numbers, we conclude that \( f(x) \) is **surjective**.

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