Given \( f(x)=\left\{\begin{array}{cc}2 x+8 ; & x \leq-2 \\ x^{2}-3 ; & -23\end{array}\right. \) find each required value. Type answer in the blank. If the answer is no solution/does not exit, type DNE. \( f(-2)= \) \( \square \) \( f(3)= \) \( \square(0)= \) \( \square \) \( f(-5)= \) A

Let's evaluate the given piecewise function \( f(x) \) at the specified points. The function is defined as: \[ f(x) = \begin{cases} 2x + 8 & \text{if } x \leq -2 \\ x^2 - 3 & \text{if } -2 < x < 3 \\ \sqrt{x + 3} & \text{if } x > 3 \end{cases} \] ### 1. \( f(-2) \) Since \( x = -2 \) falls into the first case (\( x \leq -2 \)): \[ f(-2) = 2(-2) + 8 = -4 + 8 = 4 \] **Answer:** \( f(-2) = 4 \) ### 2. \( f(3) \) At \( x = 3 \), we need to determine which case applies: - \( x = 3 \) is **not** less than 3 (\( -2 < x < 3 \)) and **not** greater than 3 (\( x > 3 \)). - Therefore, \( x = 3 \) does **not** fall into any of the defined cases. **Answer:** \( f(3) = \text{DNE} \) (Does Not Exist) ### 3. \( f(0) \) Since \( x = 0 \) falls into the second case (\( -2 < x < 3 \)): \[ f(0) = 0^2 - 3 = 0 - 3 = -3 \] **Answer:** \( f(0) = -3 \) ### 4. \( f(-5) \) Since \( x = -5 \) falls into the first case (\( x \leq -2 \)): \[ f(-5) = 2(-5) + 8 = -10 + 8 = -2 \] **Answer:** \( f(-5) = -2 \) --- ### Summary of Answers: - \( f(-2) = 4 \) - \( f(3) = \text{DNE} \) - \( f(0) = -3 \) - \( f(-5) = -2 \)