Find the range. Write in interval notation. \( g(n)=\sqrt{2 n-5}+4 \) Show your work here Hint: To add infinity (o), type "infinity" \( \begin{array}{ll}\circ[28,7) & \bigcirc[4, \infty) \\ O[-4, \infty) & \bigcirc[4,7)\end{array} \)

To find the **range** of the function \( g(n) = \sqrt{2n - 5} + 4 \), let's follow these steps: ### 1. Determine the Domain Before finding the range, we need to identify the domain of the function, which is the set of all possible input values \( n \) for which the function is defined. - The expression inside the square root must be non-negative: \[ 2n - 5 \geq 0 \] - Solve for \( n \): \[ 2n \geq 5 \\ n \geq \frac{5}{2} \] **Domain:** \( n \geq \frac{5}{2} \) ### 2. Analyze the Function Now, let's analyze how \( g(n) \) behaves based on the domain. - **Minimum Value of \( \sqrt{2n - 5} \):** \[ \sqrt{2n - 5} \geq 0 \] The smallest value occurs when \( 2n - 5 = 0 \): \[ \sqrt{0} = 0 \] Therefore, the minimum value of \( g(n) \) is: \[ 0 + 4 = 4 \] - **Behavior as \( n \) Increases:** As \( n \) increases beyond \( \frac{5}{2} \), \( 2n - 5 \) becomes larger, making \( \sqrt{2n - 5} \) increase without bound. Consequently, \( g(n) \) also increases without bound. ### 3. Determine the Range Combining the above observations: - The smallest value of \( g(n) \) is **4**. - There is no upper limit to \( g(n) \); it can grow indefinitely. **Range in Interval Notation:** \([4, \text{infinity})\) ### Final Answer \[ \text{Range of } g(n): \quad [4, \text{infinity}) \]