Question

Which functions have removable discontinuities (holes)? Check all of the boxes that apply.
f(x)=x-1/x^2-1
f(x)=x^2-9/x^2+7x+12
f(x)=x^2+4x+4/x^2+2x-8 ,
f(x)=x+7/x^2+5x-14

Answer 1

Answer:

a b d

Step-by-step explanation:

Answer 2

The functions that have removable discontinuities (holes) are option A), B) and D).

Solution :

A hole is created when the function has the same factor in both the numerator and denominator. This factor can be canceled out but needs to still be considered when evaluating the function.

Now,

Option A) -

$f(x)=\dfrac{x-1}{x^2-1}$

$f(x)= \dfrac{(x-1)}{(x-1)(x+1)}=\dfrac{1}{(x+1)}$

Therefore according to definition this option is correct.

Option B) -

$f(x)=\dfrac{x^2-9}{x^2+7x+12}$

$f(x)= \dfrac{(x-3)(x+3)}{x^2+4x+3x+12}=\dfrac{(x-3)(x+3)}{(x+3)(x+4)}$

$f(x)=\dfrac{(x-3)}{(x+4)}$

Therefore according to definition this option is also correct.

Option C) -

$f(x)=\dfrac{x^2+4x+4}{x^2+2x-8}$

$f(x)=\dfrac{(x+2)^2}{x^2+4x-2x-8}$

$f(x)=\dfrac{(x+2)^2}{(x+4)(x-2)}$

Therefore according to definition this option is incorrect.

Option D) -

$f(x)=\dfrac{(x+7)}{x^2+5x-14}$

$f(x)=\dfrac{(x+7)}{(x^2+7x-2x-14)}=\dfrac{(x+7)}{(x+7)(x-2)}$

$f(x)=\dfrac{1}{(x-2)}$

Therefore according to definition this option is also correct.

The functions that have removable discontinuities (holes) are option A), B) and D).

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