Question
Given f of x is equal to the quantity x plus 4 end quantity divided by the quantity x squared minus 3x minus 28 end quantity, which of the following is true?
f(x) is positive for all x > –4
f(x) is negative for all x > –4
f(x) is positive for all x < 7
f(x) is negative for all x < 7
f(x) is positive for all x > –4
f(x) is negative for all x > –4
f(x) is positive for all x < 7
f(x) is negative for all x < 7
Answers
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Answered by
GPT-4o mini
To analyze the function \( f(x) = \frac{x + 4}{x^2 - 3x - 28} \), we need to consider its behavior in relation to the numerator and denominator.
### Step 1: Finding Critical Points
First, we identify the points where \( f(x) \) can change sign by examining the numerator and denominator:
- **Numerator**: \( x + 4 = 0 \) implies \( x = -4 \)
- **Denominator**: \( x^2 - 3x - 28 = 0 \)
To find the roots of the denominator, we can factor:
\[
x^2 - 3x - 28 = (x - 7)(x + 4) = 0
\]
This gives roots at \( x = 7 \) and \( x = -4 \).
### Step 2: Analyzing the Function
The function \( f(x) \) is undefined at the points where the denominator is zero, which occur at \( x = -4 \) and \( x = 7 \). We will evaluate the sign of \( f(x) \) in the intervals defined by these critical points: \( (-\infty, -4) \), \( (-4, 7) \), and \( (7, \infty) \).
**Interval 1: \( (-\infty, -4) \)**
Choose a test point, such as \( x = -5 \):
\[
f(-5) = \frac{-5 + 4}{(-5)^2 - 3(-5) - 28} = \frac{-1}{25 + 15 - 28} = \frac{-1}{12} < 0
\]
**Interval 2: \( (-4, 7) \)**
Choose a test point, such as \( x = 0 \):
\[
f(0) = \frac{0 + 4}{0^2 - 3(0) - 28} = \frac{4}{-28} = \frac{-1}{7} < 0
\]
**Interval 3: \( (7, \infty) \)**
Choose a test point, such as \( x = 8 \):
\[
f(8) = \frac{8 + 4}{8^2 - 3(8) - 28} = \frac{12}{64 - 24 - 28} = \frac{12}{12} = 1 > 0
\]
### Step 3: Conclusions
From the analysis:
- In the interval \( (-\infty, -4) \), \( f(x) < 0 \)
- In the interval \( (-4, 7) \), \( f(x) < 0 \)
- In the interval \( (7, \infty) \), \( f(x) > 0 \)
### Answering the Question:
Now, regarding the options provided:
1. \( f(x) \) is positive for all \( x > -4 \) (False)
2. \( f(x) \) is negative for all \( x > -4 \) (True in the interval \( (-4, 7) \), but false for \( x > 7 \))
3. \( f(x) \) is positive for all \( x < 7 \) (False, since it is negative until \( x = 7 \))
4. \( f(x) \) is negative for all \( x < 7 \) (This is true for \( x < 7 \))
The correct answer is:
**\( f(x) \) is negative for all \( x < 7 \).**
### Step 1: Finding Critical Points
First, we identify the points where \( f(x) \) can change sign by examining the numerator and denominator:
- **Numerator**: \( x + 4 = 0 \) implies \( x = -4 \)
- **Denominator**: \( x^2 - 3x - 28 = 0 \)
To find the roots of the denominator, we can factor:
\[
x^2 - 3x - 28 = (x - 7)(x + 4) = 0
\]
This gives roots at \( x = 7 \) and \( x = -4 \).
### Step 2: Analyzing the Function
The function \( f(x) \) is undefined at the points where the denominator is zero, which occur at \( x = -4 \) and \( x = 7 \). We will evaluate the sign of \( f(x) \) in the intervals defined by these critical points: \( (-\infty, -4) \), \( (-4, 7) \), and \( (7, \infty) \).
**Interval 1: \( (-\infty, -4) \)**
Choose a test point, such as \( x = -5 \):
\[
f(-5) = \frac{-5 + 4}{(-5)^2 - 3(-5) - 28} = \frac{-1}{25 + 15 - 28} = \frac{-1}{12} < 0
\]
**Interval 2: \( (-4, 7) \)**
Choose a test point, such as \( x = 0 \):
\[
f(0) = \frac{0 + 4}{0^2 - 3(0) - 28} = \frac{4}{-28} = \frac{-1}{7} < 0
\]
**Interval 3: \( (7, \infty) \)**
Choose a test point, such as \( x = 8 \):
\[
f(8) = \frac{8 + 4}{8^2 - 3(8) - 28} = \frac{12}{64 - 24 - 28} = \frac{12}{12} = 1 > 0
\]
### Step 3: Conclusions
From the analysis:
- In the interval \( (-\infty, -4) \), \( f(x) < 0 \)
- In the interval \( (-4, 7) \), \( f(x) < 0 \)
- In the interval \( (7, \infty) \), \( f(x) > 0 \)
### Answering the Question:
Now, regarding the options provided:
1. \( f(x) \) is positive for all \( x > -4 \) (False)
2. \( f(x) \) is negative for all \( x > -4 \) (True in the interval \( (-4, 7) \), but false for \( x > 7 \))
3. \( f(x) \) is positive for all \( x < 7 \) (False, since it is negative until \( x = 7 \))
4. \( f(x) \) is negative for all \( x < 7 \) (This is true for \( x < 7 \))
The correct answer is:
**\( f(x) \) is negative for all \( x < 7 \).**
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