If `f(x)=x and g(x)=|x|`, then define the following functions:
`(i) f+g " "(ii) f-g`
`(iii) f*g" "(iv) (f)/(g)`

Answer

To solve the problem, we need to define the functions \( f + g \), \( f - g \), \( f \cdot g \), and \( \frac{f}{g} \) given that \( f(x) = x \) and \( g(x) = |x| \).

### Step-by-Step Solution:

1. **Define \( f + g \)**:
\[ f + g = f(x) + g(x) = x + |x| \]
- For \( x \geq 0 \):
\[ |x| = x \implies f + g = x + x = 2x \]
- For \( x < 0 \):
\[ |x| = -x \implies f + g = x - x = 0 \]
Thus,
\[ f + g = \begin{cases} 2x & \text{if } x \geq 0 \\ 0 & \text{if } x < 0 \end{cases} \]

2. **Define \( f - g \)**:
\[ f - g = f(x) - g(x) = x - |x| \]
- For \( x \geq 0 \):
\[ |x| = x \implies f - g = x - x = 0 \]
- For \( x < 0 \):
\[ |x| = -x \implies f - g = x + x = 2x \]
Thus,
\[ f - g = \begin{cases} 0 & \text{if } x \geq 0 \\ 2x & \text{if } x < 0 \end{cases} \]

3. **Define \( f \cdot g \)**:
\[ f \cdot g = f(x) \cdot g(x) = x \cdot |x| \]
- For \( x \geq 0 \):
\[ |x| = x \implies f \cdot g = x \cdot x = x^2 \]
- For \( x < 0 \):
\[ |x| = -x \implies f \cdot g = x \cdot (-x) = -x^2 \]
Thus,
\[ f \cdot g = \begin{cases} x^2 & \text{if } x \geq 0 \\ -x^2 & \text{if } x < 0 \end{cases} \]

4. **Define \( \frac{f}{g} \)**:
\[ \frac{f}{g} = \frac{f(x)}{g(x)} = \frac{x}{|x|} \]
- For \( x > 0 \):
\[ |x| = x \implies \frac{f}{g} = \frac{x}{x} = 1 \]
- For \( x < 0 \):
\[ |x| = -x \implies \frac{f}{g} = \frac{x}{-x} = -1 \]
- Note: \( \frac{f}{g} \) is undefined at \( x = 0 \).
Thus,
\[ \frac{f}{g} = \begin{cases} 1 & \text{if } x > 0 \\ -1 & \text{if } x < 0 \end{cases} \]

### Final Results:
- \( f + g = \begin{cases} 2x & \text{if } x \geq 0 \\ 0 & \text{if } x < 0 \end{cases} \)
- \( f - g = \begin{cases} 0 & \text{if } x \geq 0 \\ 2x & \text{if } x < 0 \end{cases} \)
- \( f \cdot g = \begin{cases} x^2 & \text{if } x \geq 0 \\ -x^2 & \text{if } x < 0 \end{cases} \)
- \( \frac{f}{g} = \begin{cases} 1 & \text{if } x > 0 \\ -1 & \text{if } x < 0 \end{cases} \) (undefined at \( x = 0 \))