Answer:
D
Step-by-step explanation:
Given
f(x) = [tex]\frac{(x+2)(x-1)}{(x-3)(x+2)}[/tex] ← cancel (x + 2) on the numerator/ denominator
This creates a discontinuity ( a hole ) at x = - 2
f(x) = [tex]\frac{x-1}{x-3}[/tex]
The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be.
x - 3 = 0 ⇒ x = 3 ← excluded value
Thus domain is { x | x ≠ - 2, 3 } → D
