A function `f : R rarr R` defined by `f(x) = x^(2)`. Determine
`(i)` range of `f`
`(ii). {x: f(x) = 4}`
`(iii). {y: f(y) = –1}`

Answer

To solve the given problem step by step, we will address each part of the question regarding the function \( f(x) = x^2 \).

### (i) Determine the range of \( f \)

1. **Understanding the Function**: The function \( f(x) = x^2 \) takes any real number \( x \) and squares it.
2. **Identifying Output Values**: The square of any real number is always non-negative. This means \( f(x) \geq 0 \) for all \( x \in \mathbb{R} \).
3. **Finding Minimum Value**: The minimum value of \( f(x) \) occurs at \( x = 0 \), where \( f(0) = 0^2 = 0 \).
4. **Behavior as \( x \) Increases**: As \( x \) moves away from 0 in either direction (positive or negative), \( f(x) \) increases without bound. For example, \( f(1) = 1^2 = 1 \), \( f(2) = 2^2 = 4 \), and so on.
5. **Conclusion**: Therefore, the range of \( f \) is all non-negative real numbers, which can be expressed as:
\[ \text{Range of } f = [0, \infty) \]

### (ii) Determine the set \( \{ x : f(x) = 4 \} \)

1. **Setting Up the Equation**: We need to solve the equation \( f(x) = 4 \), which translates to:
\[ x^2 = 4 \]
2. **Solving for \( x \)**: Taking the square root of both sides gives:
\[ x = \pm 2 \]
3. **Conclusion**: Therefore, the set of \( x \) such that \( f(x) = 4 \) is:
\[ \{ x : f(x) = 4 \} = \{-2, 2\} \]

### (iii) Determine the set \( \{ y : f(y) = -1 \} \)

1. **Setting Up the Equation**: We need to solve the equation \( f(y) = -1 \), which translates to:
\[ y^2 = -1 \]
2. **Analyzing the Equation**: The equation \( y^2 = -1 \) has no real solutions because the square of a real number cannot be negative.
3. **Conclusion**: Therefore, the set of \( y \) such that \( f(y) = -1 \) is:
\[ \{ y : f(y) = -1 \} = \emptyset \]

### Summary of Solutions

1. **Range of \( f \)**: \([0, \infty)\)
2. **Set \( \{ x : f(x) = 4 \} \)**: \(\{-2, 2\}\)
3. **Set \( \{ y : f(y) = -1 \} \)**: \(\emptyset\)