Question

Evaluate each function following the specification.

Answer 1

Answer:

#3 a. g(-1) = 2, g(0) = 3, and g(1) = 2

b. No restrictions for all real numbers

$\#4 \ a. \ h(-1) = \dfrac{1}{3}, \ h(0) =0, \ h(2) = \infty$

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;

$h(x) = \dfrac{x}{x - 2}$

By substitution, we get;

$h(-1) = \dfrac{-1}{(-1) - 2} = \dfrac{-1}{-3} = \dfrac{1}{3}$

$\therefore h(-1) = \dfrac{1}{3}$

$h(0) = \dfrac{0}{0 - 2} = \dfrac{0}{-2} =0$

$\therefore h(0) =0$

$h(2) = \dfrac{2}{2 - 2} = \dfrac{2}{0} = \infty$

$\therefore h(2) = \infty$

$h(-1) = \dfrac{1}{3}, \ h(0) =0, \ h(2) = \infty$

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