Answer:
#3 a. g(-1) = 2, g(0) = 3, and g(1) = 2
b. No restrictions for all real numbers
![\#4 \ a. \ h(-1) = \dfrac{1}{3}, \ h(0) =0, \ h(2) = \infty](https://tex.z-dn.net/?f=%5C%234%20%5C%20%20a.%20%5C%20h%28-1%29%20%3D%20%5Cdfrac%7B1%7D%7B3%7D%2C%20%5C%20%20h%280%29%20%3D0%2C%20%5C%20%20h%282%29%20%3D%20%20%5Cinfty)
b. Yes, <em>x ≠ 2</em>
Step-by-step explanation:
#3 The function is given as g(x) = -x² + 3
a. From the given function, by plugging in the value of 'x' in the bracket, we have;
g(-1) = -(-1)² + 3 = -1 + 3 = 2
g(-1) = 2
g(0) = -0² + 3 = 3
g(0) = 3
g(1) = -1² + 3 = -1 + 3 = 2
g(1) = 2
g(-1) = 2, g(0) = 3, and g(1) = 2
b. The given function g(x) = -x² + 3 for finding the value of <em>g</em> can take any value of <em>x</em> which is a real number
Therefore, therefore, there are no restrictions
#4 a. The given function is given as follows;
![h(x) = \dfrac{x}{x - 2}](https://tex.z-dn.net/?f=h%28x%29%20%3D%20%5Cdfrac%7Bx%7D%7Bx%20-%202%7D)
By substitution, we get;
![h(-1) = \dfrac{-1}{(-1) - 2} = \dfrac{-1}{-3} = \dfrac{1}{3}](https://tex.z-dn.net/?f=h%28-1%29%20%3D%20%5Cdfrac%7B-1%7D%7B%28-1%29%20-%202%7D%20%3D%20%5Cdfrac%7B-1%7D%7B-3%7D%20%3D%20%5Cdfrac%7B1%7D%7B3%7D)
![\therefore h(-1) = \dfrac{1}{3}](https://tex.z-dn.net/?f=%5Ctherefore%20h%28-1%29%20%3D%20%5Cdfrac%7B1%7D%7B3%7D)
![h(0) = \dfrac{0}{0 - 2} = \dfrac{0}{-2} =0](https://tex.z-dn.net/?f=h%280%29%20%3D%20%5Cdfrac%7B0%7D%7B0%20-%202%7D%20%3D%20%5Cdfrac%7B0%7D%7B-2%7D%20%3D0)
![\therefore h(0) =0](https://tex.z-dn.net/?f=%5Ctherefore%20h%280%29%20%3D0)
![h(2) = \dfrac{2}{2 - 2} = \dfrac{2}{0} = \infty](https://tex.z-dn.net/?f=h%282%29%20%3D%20%5Cdfrac%7B2%7D%7B2%20-%202%7D%20%3D%20%5Cdfrac%7B2%7D%7B0%7D%20%3D%20%5Cinfty)
![\therefore h(2) = \infty](https://tex.z-dn.net/?f=%5Ctherefore%20h%282%29%20%3D%20%20%5Cinfty)
![h(-1) = \dfrac{1}{3}, \ h(0) =0, \ h(2) = \infty](https://tex.z-dn.net/?f=h%28-1%29%20%3D%20%5Cdfrac%7B1%7D%7B3%7D%2C%20%5C%20%20h%280%29%20%3D0%2C%20%5C%20%20h%282%29%20%3D%20%20%5Cinfty)
b. From the values of the function, we have that h(x) is not defined at x = 2
Therefore, there is a restriction for <em>x</em> in the function, which is <em>x ≠ 2</em>