Answer:
#3 a. g(-1) = 2, g(0) = 3, and g(1) = 2
b. No restrictions for all real numbers
[tex]\#4 \ a. \ h(-1) = \dfrac{1}{3}, \ h(0) =0, \ h(2) = \infty[/tex]
b. Yes, x ≠ 2
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 g can take any value of x which is a real number
Therefore, therefore, there are no restrictions
#4 a. The given function is given as follows;
[tex]h(x) = \dfrac{x}{x - 2}[/tex]
By substitution, we get;
[tex]h(-1) = \dfrac{-1}{(-1) - 2} = \dfrac{-1}{-3} = \dfrac{1}{3}[/tex]
[tex]\therefore h(-1) = \dfrac{1}{3}[/tex]
[tex]h(0) = \dfrac{0}{0 - 2} = \dfrac{0}{-2} =0[/tex]
[tex]\therefore h(0) =0[/tex]
[tex]h(2) = \dfrac{2}{2 - 2} = \dfrac{2}{0} = \infty[/tex]
[tex]\therefore h(2) = \infty[/tex]
[tex]h(-1) = \dfrac{1}{3}, \ h(0) =0, \ h(2) = \infty[/tex]
b. From the values of the function, we have that h(x) is not defined at x = 2
Therefore, there is a restriction for x in the function, which is x ≠ 2
