Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
Let's analyze the function [tex]\( f(x) = -(x + 6)(x + 2) \)[/tex] step-by-step to determine which statements about this function are true.
### 1. Finding the First Derivative
To understand the intervals where the function is increasing or decreasing, we first need to find the critical points by computing the first derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f(x) = -(x + 6)(x + 2) \][/tex]
Using the product rule and chain rule:
[tex]\[ f'(x) = -[(x + 6)'(x + 2) + (x + 6)(x + 2)'] \][/tex]
[tex]\[ f'(x) = -[(1)(x + 2) + (x + 6)(1)] \][/tex]
[tex]\[ f'(x) = -(x + 2 + x + 6) \][/tex]
[tex]\[ f'(x) = -(2x + 8) \][/tex]
[tex]\[ f'(x) = -2(x + 4) \][/tex]
### 2. Finding the Critical Points
Next, we set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ -2(x + 4) = 0 \][/tex]
[tex]\[ x + 4 = 0 \][/tex]
[tex]\[ x = -4 \][/tex]
So, the critical point is at [tex]\( x = -4 \)[/tex].
### 3. Analyzing the Intervals Around the Critical Points
We need to determine the sign of [tex]\( f'(x) \)[/tex] in the intervals around the critical point to identify where the function is increasing or decreasing. The critical point divides the real number line into two intervals: [tex]\( x < -4 \)[/tex] and [tex]\( x > -4 \)[/tex].
#### Interval 1: [tex]\( x < -4 \)[/tex]
Choose a test point in this interval, say [tex]\( x = -5 \)[/tex]:
[tex]\[ f'(-5) = -2(-5 + 4) = -2(-1) = 2 \][/tex]
Since [tex]\( f'(-5) > 0 \)[/tex], [tex]\( f(x) \)[/tex] is increasing in the interval [tex]\( x < -4 \)[/tex].
#### Interval 2: [tex]\( x > -4 \)[/tex]
Choose a test point in this interval, say [tex]\( x = -3 \)[/tex]:
[tex]\[ f'(-3) = -2(-3 + 4) = -2(1) = -2 \][/tex]
Since [tex]\( f'(-3) < 0 \)[/tex], [tex]\( f(x) \)[/tex] is decreasing in the interval [tex]\( x > -4 \)[/tex].
### 4. Conclusion
Based on the analysis above, we conclude:
- The function [tex]\( f(x) \)[/tex] is increasing for all real values of [tex]\( x \)[/tex] where [tex]\( x < -4 \)[/tex].
- The function [tex]\( f(x) \)[/tex] is decreasing for all real values of [tex]\( x \)[/tex] where [tex]\( x > -4 \)[/tex].
Thus, the correct statement about the function is:
The function is increasing for all real values of [tex]\( x \)[/tex] where [tex]\( x < -4 \)[/tex].
### 1. Finding the First Derivative
To understand the intervals where the function is increasing or decreasing, we first need to find the critical points by computing the first derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f(x) = -(x + 6)(x + 2) \][/tex]
Using the product rule and chain rule:
[tex]\[ f'(x) = -[(x + 6)'(x + 2) + (x + 6)(x + 2)'] \][/tex]
[tex]\[ f'(x) = -[(1)(x + 2) + (x + 6)(1)] \][/tex]
[tex]\[ f'(x) = -(x + 2 + x + 6) \][/tex]
[tex]\[ f'(x) = -(2x + 8) \][/tex]
[tex]\[ f'(x) = -2(x + 4) \][/tex]
### 2. Finding the Critical Points
Next, we set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ -2(x + 4) = 0 \][/tex]
[tex]\[ x + 4 = 0 \][/tex]
[tex]\[ x = -4 \][/tex]
So, the critical point is at [tex]\( x = -4 \)[/tex].
### 3. Analyzing the Intervals Around the Critical Points
We need to determine the sign of [tex]\( f'(x) \)[/tex] in the intervals around the critical point to identify where the function is increasing or decreasing. The critical point divides the real number line into two intervals: [tex]\( x < -4 \)[/tex] and [tex]\( x > -4 \)[/tex].
#### Interval 1: [tex]\( x < -4 \)[/tex]
Choose a test point in this interval, say [tex]\( x = -5 \)[/tex]:
[tex]\[ f'(-5) = -2(-5 + 4) = -2(-1) = 2 \][/tex]
Since [tex]\( f'(-5) > 0 \)[/tex], [tex]\( f(x) \)[/tex] is increasing in the interval [tex]\( x < -4 \)[/tex].
#### Interval 2: [tex]\( x > -4 \)[/tex]
Choose a test point in this interval, say [tex]\( x = -3 \)[/tex]:
[tex]\[ f'(-3) = -2(-3 + 4) = -2(1) = -2 \][/tex]
Since [tex]\( f'(-3) < 0 \)[/tex], [tex]\( f(x) \)[/tex] is decreasing in the interval [tex]\( x > -4 \)[/tex].
### 4. Conclusion
Based on the analysis above, we conclude:
- The function [tex]\( f(x) \)[/tex] is increasing for all real values of [tex]\( x \)[/tex] where [tex]\( x < -4 \)[/tex].
- The function [tex]\( f(x) \)[/tex] is decreasing for all real values of [tex]\( x \)[/tex] where [tex]\( x > -4 \)[/tex].
Thus, the correct statement about the function is:
The function is increasing for all real values of [tex]\( x \)[/tex] where [tex]\( x < -4 \)[/tex].
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.