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To determine the intervals where the function [tex]\( f(x) = \frac{9 - x^2}{x^2 - 4} \)[/tex] is positive, we need to analyze the sign of the numerator and denominator, along with considering any critical points where the denominator might change sign.
### 1. Analyze the Function
The function [tex]\( f(x) = \frac{9 - x^2}{x^2 - 4} \)[/tex] can be rewritten as:
[tex]\[ f(x) = \frac{(3 - x)(3 + x)}{(x - 2)(x + 2)} \][/tex]
This factorization helps us identify the roots and potential sign changes of the function.
### 2. Critical Points and Asymptotes
The roots of the numerator are [tex]\( x = 3 \)[/tex] and [tex]\( x = -3 \)[/tex]. The roots of the denominator, which are also vertical asymptotes, are [tex]\( x = 2 \)[/tex] and [tex]\( x = -2 \)[/tex].
### 3. Intervals to Check
Given the critical points [tex]\(x = -3, -2, 2,\)[/tex] and [tex]\( 3\)[/tex], we need to check the following intervals:
- [tex]\( (-\infty, -3) \)[/tex]
- [tex]\( (-3, -2) \)[/tex]
- [tex]\( (-2, 2) \)[/tex]
- [tex]\( (2, 3) \)[/tex]
- [tex]\( (3, \infty) \)[/tex]
### 4. Sign Analysis on Each Interval
#### Interval [tex]\( (-\infty, -3) \)[/tex]
Choose a test point [tex]\( x = -4 \)[/tex]:
[tex]\[ f(-4) = \frac{9 - (-4)^2}{(-4)^2 - 4} = \frac{9 - 16}{16 - 4} = \frac{-7}{12} \][/tex]
[tex]\( f(-4) < 0 \)[/tex]
So, [tex]\( f(x) \)[/tex] is not positive on [tex]\((-∞, -3)\)[/tex].
#### Interval [tex]\( (-3, -2) \)[/tex]
Choose a test point [tex]\( x = -2.5 \)[/tex]:
[tex]\[ f(-2.5) = \frac{9 - (-2.5)^2}{(-2.5)^2 - 4} = \frac{9 - 6.25}{6.25 - 4} = \frac{2.75}{2.25} \][/tex]
[tex]\( f(-2.5) > 0 \)[/tex]
So, [tex]\( f(x) \)[/tex] is positive on [tex]\( (-3, -2) \)[/tex].
#### Interval [tex]\( (-2, 2) \)[/tex]
Notice that within this interval, [tex]\( x = 0 \)[/tex] is a convenient test point:
[tex]\[ f(0) = \frac{9 - 0^2}{0^2 - 4} = \frac{9}{-4} = -2.25 \][/tex]
[tex]\( f(0) < 0 \)[/tex]
So, [tex]\( f(x) \)[/tex] is not positive on [tex]\( (-2, 2) \)[/tex].
#### Interval [tex]\( (2, 3) \)[/tex]
Choose a test point [tex]\( x = 2.5 \)[/tex]:
[tex]\[ f(2.5) = \frac{9 - (2.5)^2}{(2.5)^2 - 4} = \frac{9 - 6.25}{6.25 - 4} = \frac{2.75}{2.25} \][/tex]
[tex]\( f(2.5) > 0 \)[/tex]
So, [tex]\( f(x) \)[/tex] is positive on [tex]\( (2, 3) \)[/tex].
#### Interval [tex]\( (3, \infty) \)[/tex]
Choose a test point [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = \frac{9 - 4^2}{4^2 - 4} = \frac{9 - 16}{16 - 4} = \frac{-7}{12} \][/tex]
[tex]\( f(4) < 0 \)[/tex]
So, [tex]\( f(x) is not positive on \( (3, \infty) \)[/tex].
### Conclusion
[tex]\( f(x) \)[/tex] is positive on the intervals:
- [tex]\( (-3, -2) \)[/tex]
- [tex]\( (2, 3) \)[/tex]
Therefore, the intervals for which [tex]\( f(x) \)[/tex] is positive are:
- [tex]\( (-3, -2) \)[/tex]
- [tex]\( (2, 3) \)[/tex]
### 1. Analyze the Function
The function [tex]\( f(x) = \frac{9 - x^2}{x^2 - 4} \)[/tex] can be rewritten as:
[tex]\[ f(x) = \frac{(3 - x)(3 + x)}{(x - 2)(x + 2)} \][/tex]
This factorization helps us identify the roots and potential sign changes of the function.
### 2. Critical Points and Asymptotes
The roots of the numerator are [tex]\( x = 3 \)[/tex] and [tex]\( x = -3 \)[/tex]. The roots of the denominator, which are also vertical asymptotes, are [tex]\( x = 2 \)[/tex] and [tex]\( x = -2 \)[/tex].
### 3. Intervals to Check
Given the critical points [tex]\(x = -3, -2, 2,\)[/tex] and [tex]\( 3\)[/tex], we need to check the following intervals:
- [tex]\( (-\infty, -3) \)[/tex]
- [tex]\( (-3, -2) \)[/tex]
- [tex]\( (-2, 2) \)[/tex]
- [tex]\( (2, 3) \)[/tex]
- [tex]\( (3, \infty) \)[/tex]
### 4. Sign Analysis on Each Interval
#### Interval [tex]\( (-\infty, -3) \)[/tex]
Choose a test point [tex]\( x = -4 \)[/tex]:
[tex]\[ f(-4) = \frac{9 - (-4)^2}{(-4)^2 - 4} = \frac{9 - 16}{16 - 4} = \frac{-7}{12} \][/tex]
[tex]\( f(-4) < 0 \)[/tex]
So, [tex]\( f(x) \)[/tex] is not positive on [tex]\((-∞, -3)\)[/tex].
#### Interval [tex]\( (-3, -2) \)[/tex]
Choose a test point [tex]\( x = -2.5 \)[/tex]:
[tex]\[ f(-2.5) = \frac{9 - (-2.5)^2}{(-2.5)^2 - 4} = \frac{9 - 6.25}{6.25 - 4} = \frac{2.75}{2.25} \][/tex]
[tex]\( f(-2.5) > 0 \)[/tex]
So, [tex]\( f(x) \)[/tex] is positive on [tex]\( (-3, -2) \)[/tex].
#### Interval [tex]\( (-2, 2) \)[/tex]
Notice that within this interval, [tex]\( x = 0 \)[/tex] is a convenient test point:
[tex]\[ f(0) = \frac{9 - 0^2}{0^2 - 4} = \frac{9}{-4} = -2.25 \][/tex]
[tex]\( f(0) < 0 \)[/tex]
So, [tex]\( f(x) \)[/tex] is not positive on [tex]\( (-2, 2) \)[/tex].
#### Interval [tex]\( (2, 3) \)[/tex]
Choose a test point [tex]\( x = 2.5 \)[/tex]:
[tex]\[ f(2.5) = \frac{9 - (2.5)^2}{(2.5)^2 - 4} = \frac{9 - 6.25}{6.25 - 4} = \frac{2.75}{2.25} \][/tex]
[tex]\( f(2.5) > 0 \)[/tex]
So, [tex]\( f(x) \)[/tex] is positive on [tex]\( (2, 3) \)[/tex].
#### Interval [tex]\( (3, \infty) \)[/tex]
Choose a test point [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = \frac{9 - 4^2}{4^2 - 4} = \frac{9 - 16}{16 - 4} = \frac{-7}{12} \][/tex]
[tex]\( f(4) < 0 \)[/tex]
So, [tex]\( f(x) is not positive on \( (3, \infty) \)[/tex].
### Conclusion
[tex]\( f(x) \)[/tex] is positive on the intervals:
- [tex]\( (-3, -2) \)[/tex]
- [tex]\( (2, 3) \)[/tex]
Therefore, the intervals for which [tex]\( f(x) \)[/tex] is positive are:
- [tex]\( (-3, -2) \)[/tex]
- [tex]\( (2, 3) \)[/tex]
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