Question

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

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

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

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

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

Explanation:

Factor the first function

Using the Polynomial Factoring and Removable Discontinuities knowledge points

$$ f(x) = \frac{x-1}{x^2-1} = \frac{x-1}{(x-1)(x+1)} = \frac{1}{x+1}, \quad x eq 1 $$

Since the factor \(x-1\) cancels, there is a removable discontinuity (hole) at \(x = 1\).

Factor the second function

Using the Polynomial Factoring and Removable Discontinuities knowledge points

$$ f(x) = \frac{x^2-9}{x^2+7x+12} = \frac{(x-3)(x+3)}{(x+3)(x+4)} = \frac{x-3}{x+4}, \quad x eq -3 $$

Since the factor \(x+3\) cancels, there is a removable discontinuity (hole) at \(x = -3\).

Factor the third function

Using the Polynomial Factoring and Removable Discontinuities knowledge points

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

Since no common factors cancel between the numerator and denominator, there are no removable discontinuities (holes).

Factor the fourth function

Using the Polynomial Factoring and Removable Discontinuities knowledge points

$$ f(x) = \frac{x+7}{x^2+5x-14} = \frac{x+7}{(x+7)(x-2)} = \frac{1}{x-2}, \quad x eq -7 $$

Since the factor \(x+7\) cancels, there is a removable discontinuity (hole) at \(x = -7\).

Answer:

• (A) \(f(x) = \frac{x-1}{x^2-1}\) (Correct answer)
• (B) \(f(x) = \frac{x^2-9}{x^2+7x+12}\) (Correct answer)
• (C) \(f(x) = \frac{x^2+4x+4}{x^2+2x-8}\)
• (D) \(f(x) = \frac{x+7}{x^2+5x-14}\) (Correct answer)