Find the domain and range of the following real functions:(i) `f(x)=-|x|` (ii) `f(x)=sqrt(9-x^2)`

Explanation

To find the domain and range of the given functions, we will analyze each function step by step.

### Part (i): \( f(x) = -|x| \)

**Step 1: Determine the Domain**
- The absolute value function \( |x| \) is defined for all real numbers \( x \).
- Therefore, the domain of \( f(x) = -|x| \) is all real numbers.

**Domain:**
\[ \text{Domain} = \mathbb{R} \quad \text{(all real numbers)} \]

**Step 2: Determine the Range**
- The absolute value \( |x| \) is always non-negative, meaning \( |x| \geq 0 \).
- Thus, \( -|x| \) will always be non-positive, meaning \( -|x| \leq 0 \).
- The maximum value of \( -|x| \) occurs when \( x = 0 \), giving \( f(0) = 0 \).
- As \( |x| \) increases, \( -|x| \) decreases without bound.

**Range:**
\[ \text{Range} = (-\infty, 0] \quad \text{(all values less than or equal to 0)} \]

### Part (ii): \( f(x) = \sqrt{9 - x^2} \)

**Step 1: Determine the Domain**
- For the square root function to be defined, the expression inside the square root must be non-negative:
\[ 9 - x^2 \geq 0 \]
- Rearranging gives:
\[ x^2 \leq 9 \]
- Taking the square root of both sides gives:
\[ |x| \leq 3 \]
- This implies:
\[ -3 \leq x \leq 3 \]

**Domain:**
\[ \text{Domain} = [-3, 3] \]

**Step 2: Determine the Range**
- The minimum value of \( 9 - x^2 \) occurs at the endpoints \( x = -3 \) and \( x = 3 \):
\[ f(-3) = \sqrt{9 - (-3)^2} = \sqrt{0} = 0 \]
\[ f(3) = \sqrt{9 - 3^2} = \sqrt{0} = 0 \]
- The maximum value occurs at \( x = 0 \):
\[ f(0) = \sqrt{9 - 0^2} = \sqrt{9} = 3 \]
- Therefore, \( f(x) \) varies from 0 to 3.

**Range:**
\[ \text{Range} = [0, 3] \]

### Summary of Results
- For \( f(x) = -|x| \):
- Domain: \( \mathbb{R} \)
- Range: \( (-\infty, 0] \)

- For \( f(x) = \sqrt{9 - x^2} \):
- Domain: \( [-3, 3] \)
- Range: \( [0, 3] \)