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

To find the values for \( f(0) \), we can simply substitute \( 0 \) into the function \( f(x) \): \[ f(0) = 0^2 - 5 = -5 \] Now, let’s consider the range of \( g(x) = x + 2 \). Since \( g(x) \) is a linear function with a slope of 1, the domain can be any real number, meaning the range of \( g \) is all real numbers as well. Due to this linearity, as \( x \) approaches both positive and negative infinity, \( g(x) \) will cover all values from negative infinity to positive infinity. Next, looking at the function \( f(x) = x^2 - 5 \), it is a quadratic function that opens upwards. The lowest point or vertex occurs at \( x = 0 \), and when evaluating \( f(0) \), we find this minimum value is \(-5\). Therefore, the range of \( f \) when the domain is all real numbers is \([-5, \infty)\). So here are your results: - \( f(0) = -5 \) - The range of \( g \) is all real numbers. - The range of \( f \) is \([-5, \infty)\).