To analyze the function [tex]\( f(x) = \frac{10}{x^2 - 7x - 30} \)[/tex] and determine for which intervals it is positive or negative, we first need to find the roots of the denominator [tex]\( x^2 - 7x - 30 \)[/tex]. These roots are the points where the function is undefined (vertical asymptotes).
### Step 1: Find the roots of the denominator
The denominator can be factored as follows:
[tex]\[ x^2 - 7x - 30 = (x - 10)(x + 3) \][/tex]
This gives us the roots:
[tex]\[ x = 10 \quad \text{and} \quad x = -3 \][/tex]
### Step 2: Determine the intervals for analysis
The roots divide the x-axis into three intervals:
1. [tex]\( x < -3 \)[/tex]
2. [tex]\( -3 < x < 10 \)[/tex]
3. [tex]\( x > 10 \)[/tex]
### Step 3: Analyze the sign of [tex]\( f(x) \)[/tex] in each interval
We will test a value from each of these intervals to determine whether [tex]\( f(x) \)[/tex] is positive or negative in that interval.
#### Interval [tex]\( x < -3 \)[/tex]
Choose [tex]\( x = -4 \)[/tex]:
[tex]\[ f(-4) = \frac{10}{(-4 - 10)(-4 + 3)} = \frac{10}{(-14)(-1)} = \frac{10}{14} = \frac{5}{7} > 0 \][/tex]
So, [tex]\( f(x) \)[/tex] is positive for [tex]\( x < -3 \)[/tex].
#### Interval [tex]\( -3 < x < 10 \)[/tex]
Choose [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{10}{(0 - 10)(0 + 3)} = \frac{10}{(-10)(3)} = \frac{10}{-30} = -\frac{1}{3} < 0 \][/tex]
So, [tex]\( f(x) \)[/tex] is negative for [tex]\( -3 < x < 10 \)[/tex].
#### Interval [tex]\( x > 10 \)[/tex]
Choose [tex]\( x = 11 \)[/tex]:
[tex]\[ f(11) = \frac{10}{(11 - 10)(11 + 3)} = \frac{10}{(1)(14)} = \frac{10}{14} = \frac{5}{7} > 0 \][/tex]
So, [tex]\( f(x) \)[/tex] is positive for [tex]\( x > 10 \)[/tex].
### Conclusion
Based on the analysis, we have:
- [tex]\( f(x) \)[/tex] is positive for [tex]\( x < -3 \)[/tex]
- [tex]\( f(x) \)[/tex] is negative for [tex]\( -3 < x < 10 \)[/tex]
- [tex]\( f(x) \)[/tex] is positive for [tex]\( x > 10 \)[/tex]
Among the given options, the true statements are:
- [tex]\( f(x) \)[/tex] is positive for all [tex]\( x > 10 \)[/tex]
- [tex]\( f(x) \)[/tex] is negative for all [tex]\( -3 < x < 10 \)[/tex] (if included)
Therefore, the correct response is:
[tex]\[ \boxed{f(x) \text{ is positive for all } x > 10.} \][/tex]
