A function defined for all real numbers is defined for `x>-0` as follows `f(x)={x|x|, 0<=x<=1, 2x, x>=1}` How if f defined for `x<=0`. If (i) f is even ? (ii) f is odd ?

Answer

To solve the problem, we need to determine how the function \( f(x) \) behaves for \( x \leq 0 \) based on the given piecewise definition for \( x > 0 \). The function is defined as follows:

\[ f(x) = \begin{cases} x |x| & \text{for } 0 \leq x \leq 1 \\ 2x & \text{for } x \geq 1 \end{cases} \]

### Step 1: Define \( f(x) \) for \( x < 0 \)

To find \( f(x) \) for \( x < 0 \), we need to analyze the behavior of the function. We know that \( |x| = -x \) when \( x < 0 \). Therefore, we can extend the definition of \( f(x) \) for \( x < 0 \):

- For \( 0 \leq x \leq 1 \), \( f(x) = x |x| = x(-x) = -x^2 \).
- For \( x \geq 1 \), \( f(x) = 2x \) does not apply since we are considering \( x < 0 \).

Thus, we can define \( f(x) \) for \( x < 0 \) as follows:

\[ f(x) = -x^2 \quad \text{for } x < 0 \]

### Step 2: Check if \( f(x) \) is even

A function \( f(x) \) is even if \( f(-x) = f(x) \) for all \( x \).

- For \( x \geq 0 \):
- If \( 0 \leq x \leq 1 \), then \( f(x) = x |x| = -x^2 \).
- For \( x < 0 \), \( f(-x) = f(x) = -x^2 \).

- For \( x \geq 1 \):
- \( f(x) = 2x \).
- For \( x < 0 \), \( f(-x) = -(-x)^2 = -x^2 \) which does not equal \( 2x \).

Since \( f(-x) \neq f(x) \) for \( x \geq 1 \), the function is not even.

### Step 3: Check if \( f(x) \) is odd

A function \( f(x) \) is odd if \( f(-x) = -f(x) \) for all \( x \).

- For \( x \geq 0 \):
- If \( 0 \leq x \leq 1 \), then \( f(x) = -x^2 \) and \( f(-x) = -(-x)^2 = -x^2 \).
- Therefore, \( f(-x) = -f(x) \).

- For \( x \geq 1 \):
- \( f(x) = 2x \).
- For \( x < 0 \), \( f(-x) = -(-x)^2 = -x^2 \) which does not equal \( -2x \).

Since \( f(-x) \neq -f(x) \) for \( x \geq 1 \), the function is not odd.

### Conclusion

(i) \( f \) is not even.
(ii) \( f \) is not odd.