Question

Which functions have removable discontinuities (holes)? check all of the boxes that apply?

Answer 1

Unfortunately, this question is incomplete. A point on a graph that is undefined is what is known as a removable discontinuity. To identify a removable discontinuity:

1.    Factor the numerator and the denominator.

2.    Identify factors that occur in both the numerator and the denominator.

3.    Set the common factors equal to zero.

4.    Solve for x.

Write your answers in the form x =

Answer 2

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)