Answer :
To determine where the function [tex]\( f(x) = (x-3)(x+1) \)[/tex] is positive and decreasing, let's analyze the function step-by-step.
1. Roots of the function:
The roots of the function are the values of [tex]\( x \)[/tex] for which [tex]\( f(x) = 0 \)[/tex]. We can find these by setting the function equal to zero:
[tex]\[ (x-3)(x+1) = 0 \][/tex]
This gives us the roots [tex]\( x = 3 \)[/tex] and [tex]\( x = -1 \)[/tex].
2. Intervals between roots:
The roots divide the real number line into three intervals:
[tex]\[ x < -1, \quad -1 < x < 3, \quad x > 3 \][/tex]
3. Sign of the function in each interval:
- For [tex]\( x < -1 \)[/tex]:
[tex]\[ (x-3) < 0 \quad \text{and} \quad (x+1) < 0 \quad \Rightarrow \quad (x-3)(x+1) > 0 \][/tex]
Here, the product of two negative terms is positive.
- For [tex]\( -1 < x < 3 \)[/tex]:
[tex]\[ (x-3) < 0 \quad \text{and} \quad (x+1) > 0 \quad \Rightarrow \quad (x-3)(x+1) < 0 \][/tex]
Here, the product of a negative term and a positive term is negative.
- For [tex]\( x > 3 \)[/tex]:
[tex]\[ (x-3) > 0 \quad \text{and} \quad (x+1) > 0 \quad \Rightarrow \quad (x-3)(x+1) > 0 \][/tex]
Here, the product of two positive terms is positive.
4. Behavior of the function (increasing or decreasing):
To determine where the function is decreasing, we analyze the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = (x-3)(x+1) \][/tex]
Using the product rule for the derivative:
[tex]\[ f'(x) = (x-3) \cdot \frac{d}{dx}(x+1) + (x+1) \cdot \frac{d}{dx}(x-3) \][/tex]
[tex]\[ f'(x) = (x-3) \cdot 1 + (x+1) \cdot 1 \][/tex]
[tex]\[ f'(x) = x - 3 + x + 1 \][/tex]
[tex]\[ f'(x) = 2x - 2 \][/tex]
5. Critical points of the derivative:
Set the derivative equal to zero to find critical points:
[tex]\[ 2x - 2 = 0 \][/tex]
[tex]\[ x = 1 \][/tex]
6. Sign of the derivative in each interval:
- For [tex]\( x < 1 \)[/tex]
[tex]\[ 2x - 2 < 0 \quad \Rightarrow \quad f'(x) < 0 \][/tex]
The function is decreasing in the interval [tex]\( x < 1 \)[/tex].
- For [tex]\( x > 1 \)[/tex]
[tex]\[ 2x - 2 > 0 \quad \Rightarrow \quad f'(x) > 0 \][/tex]
The function is increasing in the interval [tex]\( x > 1 \)[/tex].
7. Combining the information:
We want the function to be positive and decreasing. From our analysis:
- The function is positive for [tex]\( x < -1 \)[/tex] and [tex]\( x > 3 \)[/tex].
- The function is decreasing for [tex]\( x < 1 \)[/tex].
The interval where the function is both positive and decreasing is [tex]\( x < -1 \)[/tex].
Therefore, the correct answer is:
[tex]\[ \text{all real values of } x \text{ where } x < -1 \][/tex]
1. Roots of the function:
The roots of the function are the values of [tex]\( x \)[/tex] for which [tex]\( f(x) = 0 \)[/tex]. We can find these by setting the function equal to zero:
[tex]\[ (x-3)(x+1) = 0 \][/tex]
This gives us the roots [tex]\( x = 3 \)[/tex] and [tex]\( x = -1 \)[/tex].
2. Intervals between roots:
The roots divide the real number line into three intervals:
[tex]\[ x < -1, \quad -1 < x < 3, \quad x > 3 \][/tex]
3. Sign of the function in each interval:
- For [tex]\( x < -1 \)[/tex]:
[tex]\[ (x-3) < 0 \quad \text{and} \quad (x+1) < 0 \quad \Rightarrow \quad (x-3)(x+1) > 0 \][/tex]
Here, the product of two negative terms is positive.
- For [tex]\( -1 < x < 3 \)[/tex]:
[tex]\[ (x-3) < 0 \quad \text{and} \quad (x+1) > 0 \quad \Rightarrow \quad (x-3)(x+1) < 0 \][/tex]
Here, the product of a negative term and a positive term is negative.
- For [tex]\( x > 3 \)[/tex]:
[tex]\[ (x-3) > 0 \quad \text{and} \quad (x+1) > 0 \quad \Rightarrow \quad (x-3)(x+1) > 0 \][/tex]
Here, the product of two positive terms is positive.
4. Behavior of the function (increasing or decreasing):
To determine where the function is decreasing, we analyze the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = (x-3)(x+1) \][/tex]
Using the product rule for the derivative:
[tex]\[ f'(x) = (x-3) \cdot \frac{d}{dx}(x+1) + (x+1) \cdot \frac{d}{dx}(x-3) \][/tex]
[tex]\[ f'(x) = (x-3) \cdot 1 + (x+1) \cdot 1 \][/tex]
[tex]\[ f'(x) = x - 3 + x + 1 \][/tex]
[tex]\[ f'(x) = 2x - 2 \][/tex]
5. Critical points of the derivative:
Set the derivative equal to zero to find critical points:
[tex]\[ 2x - 2 = 0 \][/tex]
[tex]\[ x = 1 \][/tex]
6. Sign of the derivative in each interval:
- For [tex]\( x < 1 \)[/tex]
[tex]\[ 2x - 2 < 0 \quad \Rightarrow \quad f'(x) < 0 \][/tex]
The function is decreasing in the interval [tex]\( x < 1 \)[/tex].
- For [tex]\( x > 1 \)[/tex]
[tex]\[ 2x - 2 > 0 \quad \Rightarrow \quad f'(x) > 0 \][/tex]
The function is increasing in the interval [tex]\( x > 1 \)[/tex].
7. Combining the information:
We want the function to be positive and decreasing. From our analysis:
- The function is positive for [tex]\( x < -1 \)[/tex] and [tex]\( x > 3 \)[/tex].
- The function is decreasing for [tex]\( x < 1 \)[/tex].
The interval where the function is both positive and decreasing is [tex]\( x < -1 \)[/tex].
Therefore, the correct answer is:
[tex]\[ \text{all real values of } x \text{ where } x < -1 \][/tex]
