If f(x) = `[{:( x^(2) + sin x , 0 le x lt 1) , (x + e^(-x) , x ge 1):}` then extend the definition of f(x) for `x in (-oo , 0)` such that f(x) becomes
(a) An even function (b) An odd function

Answer

To extend the definition of the function \( f(x) \) for \( x \in (-\infty, 0) \) such that \( f(x) \) becomes:

(a) An even function
(b) An odd function

Let's solve each case step by step.

### Part (a): Making \( f(x) \) an Even Function

1. **Definition of an Even Function**: A function \( f(x) \) is even if \( f(-x) = f(x) \) for all \( x \).

2. **Given Function**:
\[ f(x) = \begin{cases} x^2 + \sin x & \text{for } 0 \leq x < 1 \\ x + e^{-x} & \text{for } x \geq 1 \end{cases} \]

3. **Finding \( f(-x) \)**: We need to define \( f(-x) \) for \( x \in (0, \infty) \):
- For \( x \in (0, 1) \), \( -x \in (-1, 0) \). We can define \( f(-x) = x^2 + \sin(-x) = x^2 - \sin x \).
- For \( x \geq 1 \), \( -x \leq -1 \). We can define \( f(-x) = -x + e^{x} \).

4. **Defining \( f(x) \) for \( x < 0 \)**:
- For \( -x \in (-1, 0) \): \( f(x) = x^2 - \sin(-x) = x^2 + \sin x \).
- For \( -x \leq -1 \): \( f(x) = -x + e^{x} \).

5. **Final Definition for Even Function**:
\[ f(x) = \begin{cases} x^2 + \sin x & \text{for } x < 0 \\ x^2 + \sin x & \text{for } 0 \leq x < 1 \\ x + e^{-x} & \text{for } x \geq 1 \end{cases} \]

### Part (b): Making \( f(x) \) an Odd Function

1. **Definition of an Odd Function**: A function \( f(x) \) is odd if \( f(-x) = -f(x) \) for all \( x \).

2. **Finding \( f(-x) \)**:
- For \( x \in (0, 1) \), \( -x \in (-1, 0) \). We define \( f(-x) = -\left(x^2 + \sin x\right) = -x^2 - \sin x \).
- For \( x \geq 1 \), \( -x \leq -1 \). We define \( f(-x) = -\left(x + e^{-x}\right) = -x - e^{-x} \).

3. **Defining \( f(x) \) for \( x < 0 \)**:
- For \( -x \in (-1, 0) \): \( f(x) = -x^2 - \sin(-x) = -x^2 + \sin x \).
- For \( -x \leq -1 \): \( f(x) = -(-x) - e^{x} = x - e^{x} \).

4. **Final Definition for Odd Function**:
\[ f(x) = \begin{cases} -x^2 - \sin x & \text{for } x < 0 \\ x^2 + \sin x & \text{for } 0 \leq x < 1 \\ x + e^{-x} & \text{for } x \geq 1 \end{cases} \]

### Summary of Definitions
- For **Even Function**:
\[ f(x) = \begin{cases} x^2 + \sin x & \text{for } x < 0 \\ x^2 + \sin x & \text{for } 0 \leq x < 1 \\ x + e^{-x} & \text{for } x \geq 1 \end{cases} \]

- For **Odd Function**:
\[ f(x) = \begin{cases} -x^2 - \sin x & \text{for } x < 0 \\ x^2 + \sin x & \text{for } 0 \leq x < 1 \\ x + e^{-x} & \text{for } x \geq 1 \end{cases} \]