2. [KU \( \quad \) /6] For each of the following functions below i) determine the domain and range. \( \begin{array}{ll}\text { a) } g(x)=\frac{-1}{3} \sqrt{-3 x+9}+2 \\ \text { ii) evaluate. } \\ \text { Range: } \\ \text { Domain: } \\ \text { Evaluate. Show all steps for full marks: } \\ \text { ii) } 0(0)\end{array} \)

To solve the problem, we will analyze the function \( g(x) = \frac{-1}{3} \sqrt{-3x + 9} + 2 \) step by step. ### Step 1: Determine the Domain The domain of a function is the set of all possible input values (x-values) for which the function is defined. 1. **Identify restrictions**: The function contains a square root, which means the expression inside the square root must be non-negative: \[ -3x + 9 \geq 0 \] 2. **Solve the inequality**: \[ -3x + 9 \geq 0 \implies -3x \geq -9 \implies x \leq 3 \] Thus, the domain of \( g(x) \) is: \[ \text{Domain: } (-\infty, 3] \] ### Step 2: Determine the Range Next, we will find the range of the function. The range is the set of all possible output values (y-values) of the function. 1. **Analyze the function**: The square root function \( \sqrt{-3x + 9} \) will yield non-negative values. The maximum value occurs when \( x = 3 \): \[ g(3) = \frac{-1}{3} \sqrt{-3(3) + 9} + 2 = \frac{-1}{3} \sqrt{0} + 2 = 2 \] 2. **Minimum value**: As \( x \) decreases (moving towards \(-\infty\)), \( -3x + 9 \) increases, and thus \( \sqrt{-3x + 9} \) increases. Therefore, \( g(x) \) will decrease without bound: \[ \text{As } x \to -\infty, g(x) \to -\infty \] Thus, the range of \( g(x) \) is: \[ \text{Range: } (-\infty, 2] \] ### Step 3: Evaluate \( g(0) \) Now, we will evaluate \( g(0) \): \[ g(0) = \frac{-1}{3} \sqrt{-3(0) + 9} + 2 = \frac{-1}{3} \sqrt{9} + 2 = \frac{-1}{3} \cdot 3 + 2 = -1 + 2 = 1 \] ### Summary of Results - **Domain**: \( (-\infty, 3] \) - **Range**: \( (-\infty, 2] \) - **Evaluation**: \( g(0) = 1 \) This completes the analysis of the function \( g(x) \).