Answer :
To determine where the function [tex]\( f(x) = \frac{10}{x^2 - 7x - 30} \)[/tex] is positive or negative, let's start by analyzing the expression in the denominator, [tex]\( x^2 - 7x - 30 \)[/tex].
First, we need to find the roots of the denominator, as these will help us determine the intervals where the function changes sign or is undefined.
### Step 1: Solving the Quadratic Equation
We solve the quadratic equation [tex]\( x^2 - 7x - 30 = 0 \)[/tex]:
[tex]\[ x^2 - 7x - 30 = 0 \][/tex]
To find the roots, we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In this case, [tex]\( a = 1 \)[/tex], [tex]\( b = -7 \)[/tex], and [tex]\( c = -30 \)[/tex]. Plugging these values into the quadratic formula:
[tex]\[ x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(-30)}}{2(1)} \][/tex]
[tex]\[ x = \frac{7 \pm \sqrt{49 + 120}}{2} \][/tex]
[tex]\[ x = \frac{7 \pm \sqrt{169}}{2} \][/tex]
[tex]\[ x = \frac{7 \pm 13}{2} \][/tex]
So, we get two roots:
[tex]\[ x = \frac{7 + 13}{2} = \frac{20}{2} = 10 \][/tex]
[tex]\[ x = \frac{7 - 13}{2} = \frac{-6}{2} = -3 \][/tex]
Thus, the roots are [tex]\( x = -3 \)[/tex] and [tex]\( x = 10 \)[/tex].
### Step 2: Analyzing the Intervals
Now we analyze the sign of [tex]\( f(x) \)[/tex] in the following intervals created by these roots:
1. [tex]\( (-\infty, -3) \)[/tex]
2. [tex]\( (-3, 10) \)[/tex]
3. [tex]\( (10, \infty) \)[/tex]
Notice that at [tex]\( x = -3 \)[/tex] and [tex]\( x = 10 \)[/tex], the function [tex]\( f(x) \)[/tex] is undefined due to division by zero.
### Step 3: Testing the Intervals
To determine the sign of [tex]\( f(x) \)[/tex] within each interval, we take a test point from each interval and evaluate the denominator [tex]\( x^2 - 7x - 30 \)[/tex]:
- In the interval [tex]\( (-\infty, -3) \)[/tex], we choose a point like [tex]\( x = -4 \)[/tex]:
[tex]\[ (-4)^2 - 7(-4) - 30 = 16 + 28 - 30 = 14 \][/tex]
Since the denominator is positive, [tex]\( f(x) = \frac{10}{14} \)[/tex] is positive.
- In the interval [tex]\( (-3, 10) \)[/tex], we choose a point like [tex]\( x = 0 \)[/tex]:
[tex]\[ 0^2 - 7(0) - 30 = -30 \][/tex]
Since the denominator is negative, [tex]\( f(x) = \frac{10}{-30} \)[/tex] is negative.
- In the interval [tex]\( (10, \infty) \)[/tex], we choose a point like [tex]\( x = 11 \)[/tex]:
[tex]\[ 11^2 - 7(11) - 30 = 121 - 77 - 30 = 14 \][/tex]
Since the denominator is positive, [tex]\( f(x) = \frac{10}{14} \)[/tex] is positive.
### Conclusion
From this analysis, we see:
- [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].
Thus, the correct statements among the given options are:
- [tex]\( f(x) \)[/tex] is positive for all [tex]\( x < -3 \)[/tex].
- [tex]\( f(x) \)[/tex] is positive for all [tex]\( x > 10 \)[/tex].
So the answers should reflect that:
- [tex]\( f(x) \)[/tex] is positive for all [tex]\( x < -3 \)[/tex]
- [tex]\( f(x) \)[/tex] is positive for all [tex]\( x > 10 \)[/tex]
First, we need to find the roots of the denominator, as these will help us determine the intervals where the function changes sign or is undefined.
### Step 1: Solving the Quadratic Equation
We solve the quadratic equation [tex]\( x^2 - 7x - 30 = 0 \)[/tex]:
[tex]\[ x^2 - 7x - 30 = 0 \][/tex]
To find the roots, we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In this case, [tex]\( a = 1 \)[/tex], [tex]\( b = -7 \)[/tex], and [tex]\( c = -30 \)[/tex]. Plugging these values into the quadratic formula:
[tex]\[ x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(-30)}}{2(1)} \][/tex]
[tex]\[ x = \frac{7 \pm \sqrt{49 + 120}}{2} \][/tex]
[tex]\[ x = \frac{7 \pm \sqrt{169}}{2} \][/tex]
[tex]\[ x = \frac{7 \pm 13}{2} \][/tex]
So, we get two roots:
[tex]\[ x = \frac{7 + 13}{2} = \frac{20}{2} = 10 \][/tex]
[tex]\[ x = \frac{7 - 13}{2} = \frac{-6}{2} = -3 \][/tex]
Thus, the roots are [tex]\( x = -3 \)[/tex] and [tex]\( x = 10 \)[/tex].
### Step 2: Analyzing the Intervals
Now we analyze the sign of [tex]\( f(x) \)[/tex] in the following intervals created by these roots:
1. [tex]\( (-\infty, -3) \)[/tex]
2. [tex]\( (-3, 10) \)[/tex]
3. [tex]\( (10, \infty) \)[/tex]
Notice that at [tex]\( x = -3 \)[/tex] and [tex]\( x = 10 \)[/tex], the function [tex]\( f(x) \)[/tex] is undefined due to division by zero.
### Step 3: Testing the Intervals
To determine the sign of [tex]\( f(x) \)[/tex] within each interval, we take a test point from each interval and evaluate the denominator [tex]\( x^2 - 7x - 30 \)[/tex]:
- In the interval [tex]\( (-\infty, -3) \)[/tex], we choose a point like [tex]\( x = -4 \)[/tex]:
[tex]\[ (-4)^2 - 7(-4) - 30 = 16 + 28 - 30 = 14 \][/tex]
Since the denominator is positive, [tex]\( f(x) = \frac{10}{14} \)[/tex] is positive.
- In the interval [tex]\( (-3, 10) \)[/tex], we choose a point like [tex]\( x = 0 \)[/tex]:
[tex]\[ 0^2 - 7(0) - 30 = -30 \][/tex]
Since the denominator is negative, [tex]\( f(x) = \frac{10}{-30} \)[/tex] is negative.
- In the interval [tex]\( (10, \infty) \)[/tex], we choose a point like [tex]\( x = 11 \)[/tex]:
[tex]\[ 11^2 - 7(11) - 30 = 121 - 77 - 30 = 14 \][/tex]
Since the denominator is positive, [tex]\( f(x) = \frac{10}{14} \)[/tex] is positive.
### Conclusion
From this analysis, we see:
- [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].
Thus, the correct statements among the given options are:
- [tex]\( f(x) \)[/tex] is positive for all [tex]\( x < -3 \)[/tex].
- [tex]\( f(x) \)[/tex] is positive for all [tex]\( x > 10 \)[/tex].
So the answers should reflect that:
- [tex]\( f(x) \)[/tex] is positive for all [tex]\( x < -3 \)[/tex]
- [tex]\( f(x) \)[/tex] is positive for all [tex]\( x > 10 \)[/tex]
