Answer:
#3 a. g(-1) = 2, g(0) = 3, and g(1) = 2
b. No restrictions for all real numbers

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;

By substitution, we get;







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>