Answer :
To determine which function could be [tex]\( r(x) \)[/tex], we need to analyze the provided options, taking into account the domain of the function and the behavior of the function as [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex] and [tex]\( \infty \)[/tex].
### Domain Analysis:
Given domain: [tex]\( (-\infty,-7) \cup (-7,5) \cup (5, \infty) \)[/tex]:
- The function [tex]\( r(x) \)[/tex] is undefined at [tex]\( x = -7 \)[/tex] and [tex]\( x = 5 \)[/tex].
### Behavior Analysis:
As [tex]\( x \)[/tex] approaches [tex]\(-\infty\)[/tex] and [tex]\(\infty\)[/tex], the value of the function should approach 1.
Let’s analyze each function one by one.
#### Option A: [tex]\( f(x) = \frac{x-5}{x+7} \)[/tex]
- Check the domain:
- This function is undefined at [tex]\( x = -7 \)[/tex]. But it doesn’t include [tex]\( x = 5 \)[/tex]. So, this function does not satisfy the given domain.
- Check end behavior:
- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to \frac{x}{x} = 1 \)[/tex].
Although the end behavior is correct, the domain does not match. So, this is incorrect.
#### Option B: [tex]\( f(x) = \frac{2x - 35}{x^3 + 2x - 35} \)[/tex]
- Check the domain:
- This function is a bit more complex, but it is not obvious from the numerator and denominator where the function might be undefined without factorization.
- Despite that, this form seems to not simplify straightforwardly to a situation where [tex]\( x = -7 \)[/tex] and [tex]\( x = 5 \)[/tex]. Hence, further, analysis would show it's not the expected form. And usually polynomial of higher power in denominator fraction doesn’t typically approach a constant ratio unless it's a rational form.
It's best to move on to the other options before extensively factoring.
#### Option C: [tex]\( f(x) = \frac{x^2 - 3x - 15}{x^2 + 2x - 35} \)[/tex]
- Check domain:
- Factor both the numerator and the denominator:
- [tex]\( x^2 - 3x - 15 = (x - 5)(x + 3) \)[/tex]
- [tex]\( x^2 + 2x - 35 = (x + 7)(x - 5) \)[/tex]
- The denominator will be zero at [tex]\( x = -7 \)[/tex] or [tex]\( x = 5 \)[/tex], making the function undefined at these points as required.
- Check end behavior:
- As [tex]\( x \to \infty \)[/tex], the highest power terms dominate, so [tex]\( f(x) \to \frac{x^2}{x^2} = 1 \)[/tex].
This function matches both the domain and the end behavior.
#### Option D: [tex]\( f(x) = \frac{x^2 + 2x - 35}{x^2 - 3x - 15} \)[/tex]
- Check domain:
- Factor both the numerator and the denominator:
- [tex]\( x^2 + 2x - 35 = (x + 7)(x - 5) \)[/tex]
- [tex]\( x^2 - 3x - 15 = (x - 5)(x + 3) \)[/tex]
- This function is undefined at [tex]\( x = 5 \)[/tex], but not at [tex]\( x = -7 \)[/tex].
- Despite exhibiting end behavior appropriately as [tex]\( x \to \infty, \to 1 \)[/tex].
While this satisfies behavior, not satisfying domain criteria.
### Conclusion:
The correct answer is [tex]\( f(x) = \frac{x^2 - 3x - 15}{x^2 + 2x - 35} \)[/tex], which is option C.
### Domain Analysis:
Given domain: [tex]\( (-\infty,-7) \cup (-7,5) \cup (5, \infty) \)[/tex]:
- The function [tex]\( r(x) \)[/tex] is undefined at [tex]\( x = -7 \)[/tex] and [tex]\( x = 5 \)[/tex].
### Behavior Analysis:
As [tex]\( x \)[/tex] approaches [tex]\(-\infty\)[/tex] and [tex]\(\infty\)[/tex], the value of the function should approach 1.
Let’s analyze each function one by one.
#### Option A: [tex]\( f(x) = \frac{x-5}{x+7} \)[/tex]
- Check the domain:
- This function is undefined at [tex]\( x = -7 \)[/tex]. But it doesn’t include [tex]\( x = 5 \)[/tex]. So, this function does not satisfy the given domain.
- Check end behavior:
- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to \frac{x}{x} = 1 \)[/tex].
Although the end behavior is correct, the domain does not match. So, this is incorrect.
#### Option B: [tex]\( f(x) = \frac{2x - 35}{x^3 + 2x - 35} \)[/tex]
- Check the domain:
- This function is a bit more complex, but it is not obvious from the numerator and denominator where the function might be undefined without factorization.
- Despite that, this form seems to not simplify straightforwardly to a situation where [tex]\( x = -7 \)[/tex] and [tex]\( x = 5 \)[/tex]. Hence, further, analysis would show it's not the expected form. And usually polynomial of higher power in denominator fraction doesn’t typically approach a constant ratio unless it's a rational form.
It's best to move on to the other options before extensively factoring.
#### Option C: [tex]\( f(x) = \frac{x^2 - 3x - 15}{x^2 + 2x - 35} \)[/tex]
- Check domain:
- Factor both the numerator and the denominator:
- [tex]\( x^2 - 3x - 15 = (x - 5)(x + 3) \)[/tex]
- [tex]\( x^2 + 2x - 35 = (x + 7)(x - 5) \)[/tex]
- The denominator will be zero at [tex]\( x = -7 \)[/tex] or [tex]\( x = 5 \)[/tex], making the function undefined at these points as required.
- Check end behavior:
- As [tex]\( x \to \infty \)[/tex], the highest power terms dominate, so [tex]\( f(x) \to \frac{x^2}{x^2} = 1 \)[/tex].
This function matches both the domain and the end behavior.
#### Option D: [tex]\( f(x) = \frac{x^2 + 2x - 35}{x^2 - 3x - 15} \)[/tex]
- Check domain:
- Factor both the numerator and the denominator:
- [tex]\( x^2 + 2x - 35 = (x + 7)(x - 5) \)[/tex]
- [tex]\( x^2 - 3x - 15 = (x - 5)(x + 3) \)[/tex]
- This function is undefined at [tex]\( x = 5 \)[/tex], but not at [tex]\( x = -7 \)[/tex].
- Despite exhibiting end behavior appropriately as [tex]\( x \to \infty, \to 1 \)[/tex].
While this satisfies behavior, not satisfying domain criteria.
### Conclusion:
The correct answer is [tex]\( f(x) = \frac{x^2 - 3x - 15}{x^2 + 2x - 35} \)[/tex], which is option C.
