Question

which functions have removable discontinuities (holes)? choose three correct answers.
$f(x) = \frac{x + 7}{x^2 + 5x - 14}$
$f(x) = \frac{x^2 - 9}{x^2 + 7x + 12}$
$f(x) = \frac{x^2 + 4x + 4}{x^2 + 2x - 8}$
$f(x) = \frac{x - 1}{x^2 - 1}$

Explanation:

Step1: Recall Removable Discontinuity

A function has a removable discontinuity (hole) if the numerator and denominator have a common factor (other than a constant), which can be canceled out. So we factor numerator and denominator for each function.

Step2: Analyze \( f(x)=\frac{x + 7}{x^{2}+5x - 14} \)

Factor denominator: \( x^{2}+5x - 14=(x + 7)(x - 2) \). So \( f(x)=\frac{x + 7}{(x + 7)(x - 2)} \), \( x
eq - 7,2 \). Cancel \( (x + 7) \), so removable discontinuity at \( x=-7 \).

Step3: Analyze \( f(x)=\frac{x^{2}-9}{x^{2}+7x + 12} \)

Factor numerator: \( x^{2}-9=(x + 3)(x - 3) \). Factor denominator: \( x^{2}+7x + 12=(x + 3)(x + 4) \). So \( f(x)=\frac{(x + 3)(x - 3)}{(x + 3)(x + 4)} \), \( x
eq - 3,-4 \). Cancel \( (x + 3) \), removable discontinuity at \( x=-3 \).

Step4: Analyze \( f(x)=\frac{x^{2}+4x + 4}{x^{2}+2x - 8} \)

Factor numerator: \( x^{2}+4x + 4=(x + 2)^{2} \). Factor denominator: \( x^{2}+2x - 8=(x + 4)(x - 2) \). No common factors, so no removable discontinuity.

Step5: Analyze \( f(x)=\frac{x - 1}{x^{2}-1} \)

Factor denominator: \( x^{2}-1=(x - 1)(x + 1) \). So \( f(x)=\frac{x - 1}{(x - 1)(x + 1)} \), \( x
eq 1,-1 \). Cancel \( (x - 1) \), removable discontinuity at \( x = 1 \).

Answer:

• \( f(x)=\frac{x + 7}{x^{2}+5x - 14} \)
• \( f(x)=\frac{x^{2}-9}{x^{2}+7x + 12} \)
• \( f(x)=\frac{x - 1}{x^{2}-1} \)