Qiven the fellowing functions: \[ g(x)=x+2 \quad f(x)=x^{2}-5 \] Find the fellewing 1. \( g(-2)=0 \) 2. \( g(0)= \) 3. \( f(-2)=1 \) 4. \( f(0)=-5 \) 5. The range of \( g \) when the domain is -1 , 6. The range of i when the domain is 1 .

Let's solve the problem step by step for each of the given functions \( g(x) \) and \( f(x) \). ### Given Functions: 1. \( g(x) = x + 2 \) 2. \( f(x) = x^2 - 5 \) ### 1. Find \( g(-2) \) To find \( g(-2) \), we substitute \(-2\) into the function \( g(x) \): \[ g(-2) = -2 + 2 \] Calculating this gives: \[ g(-2) = 0 \] ### 2. Find \( g(0) \) Next, we find \( g(0) \): \[ g(0) = 0 + 2 \] Calculating this gives: \[ g(0) = 2 \] ### 3. Find \( f(-2) \) Now, we substitute \(-2\) into the function \( f(x) \): \[ f(-2) = (-2)^2 - 5 \] Calculating this gives: \[ f(-2) = 4 - 5 = -1 \] ### 4. Find \( f(0) \) Next, we find \( f(0) \): \[ f(0) = 0^2 - 5 \] Calculating this gives: \[ f(0) = 0 - 5 = -5 \] ### 5. Find the range of \( g \) when the domain is \(-1\) Since the domain is a single value \(-1\), we find: \[ g(-1) = -1 + 2 \] Calculating this gives: \[ g(-1) = 1 \] Thus, the range of \( g \) when the domain is \(-1\) is \(\{1\}\). ### 6. Find the range of \( f \) when the domain is \(1\) Since the domain is a single value \(1\), we find: \[ f(1) = 1^2 - 5 \] Calculating this gives: \[ f(1) = 1 - 5 = -4 \] Thus, the range of \( f \) when the domain is \(1\) is \(\{-4\}\). ### Summary of Results: 1. \( g(-2) = 0 \) 2. \( g(0) = 2 \) 3. \( f(-2) = -1 \) 4. \( f(0) = -5 \) 5. The range of \( g \) when the domain is \(-1\) is \(\{1\}\). 6. The range of \( f \) when the domain is \(1\) is \(\{-4\}\).