Answer:
see explanation
Step-by-step explanation:
(a)
[tex]\frac{g}{h} [/tex] = [tex]\frac{g(x)}{h(x)} [/tex] = [tex]\frac{x+5}{(x-2)(x+2)} [/tex]
Substitute x = - 1 into the expression
[tex]\frac{g}{h} [/tex] (- 1 ) = [tex]\frac{-1+5}{(-1-2)(-1+2)} [/tex] = [tex]\frac{4}{(-3)(1)} [/tex] = [tex]\frac{4}{-3} [/tex] = - [tex]\frac{4}{3} [/tex]
(b)
The denominator of [tex]\frac{g}{h} [/tex] cannot be zero as this would make [tex]\frac{g}{h} [/tex] undefined
To find the values that x cannot be , equate the denominator to zero and solve for x
(x - 2)(x + 2) = 0
x - 2 = 0 ⇒ x = 2
x + 2 = 0 ⇒ x = - 2
values that are not in the domain of [tex]\frac{g}{h} [/tex] : x = - 2, x = 2
