QUESTION IMAGE
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}\\)
🆕 New Concept Discovered: Removable Discontinuities (Holes)
A hole occurs when a factor cancels out from both the numerator and denominator.
Step 1: Analyze the first function
Factor the numerator and denominator of \( f(x) = \frac{x - 1}{x^2 - 1} \):
The factor \( (x - 1) \) is common to both the numerator and denominator. Canceling this factor leaves a hole at \( x = 1 \).
This function has a removable discontinuity.
Step 2: Analyze the second function
Factor the numerator and denominator of \( f(x) = \frac{x^2 - 9}{x^2 + 7x + 12} \):
The factor \( (x + 3) \) is common to both the numerator and denominator. Canceling this factor leaves a hole at \( x = -3 \).
This function has a removable discontinuity.
Step 3: Analyze the third function
Factor the numerator and denominator of \( f(x) = \frac{x^2 + 4x + 4}{x^2 + 2x - 8} \):
There are no common factors that can be canceled between the numerator and the denominator. The discontinuities at \( x = -4 \) and \( x = 2 \) are vertical asymptotes (non-removable).
This function does not have a removable discontinuity.
Step 4: Analyze the fourth function
Factor the numerator and denominator of \( f(x) = \frac{x + 7}{x^2 + 5x - 14} \):
The factor \( (x + 7) \) is common to both the numerator and denominator. Canceling this factor leaves a hole at \( x = -7 \).
This function has a removable discontinuity.
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The functions that have removable discontinuities (holes) are:
- \( f(x) = \frac{x - 1}{x^2 - 1} \)
- \( f(x) = \frac{x^2 - 9}{x^2 + 7x + 12} \)
- \( f(x) = \frac{x + 7}{x^2 + 5x - 14} \)