Answer:
Here are the missing options for this question:
f(x) = (x - 1) ÷ (x² - 1)
f(x) = (x²- 9) ÷ (x² + 7x + 12)
f(x) = (x² + 4x + 4) ÷ (x² + 2x - 8)
f(x) = (x + 7) ÷ (x² + 5x - 14)
Here are the steps to be followed to find the holes:
To identify a removable discontinuity:
Factor the numerator and the denominator.
Identify factors that occur in both the numerator and the denominator.
Set the common factors equal to zero.
Solve for x.
Write your answers in the form x =
Solving equation 1;
f(x) = (x-1) ÷ ((x+1)(x-1))
Set the common factors equal to zero
0 = x-1
x = 1
Solving equation 2;
f(x) = ((x + 3)(x - 3)) ÷ ((x + 3)(x + 4))
Set the common factors equal to zero
0 = x + 3
x = - 3
Solving equation 3;
f(x) = (x² + 4x + 4) ÷ (x² + 2x - 8)
f(x) = ((x + 2)( x + 2)) ÷ (x² + 2x - 8)
This does not have a common factor. Therefore, there are no holes found in this equation.
Solving equation 4; f(x) = (x + 7) ÷ ((x + 7)(x - 2)) Set the common factors equal to zero 0 = x + 7 x = -7 Hence, functions 1, 2, and 4 have removable discontinuities (holes)