Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
To complete the table, we need to find the missing values for [tex]\( f(x) \)[/tex], [tex]\( g(x) \)[/tex], [tex]\( (f-g)(x) \)[/tex], and [tex]\( (f+g)(x) \)[/tex]. Here’s a step-by-step solution using the given information:
1. Identify known values:
- [tex]\( x = -4 \)[/tex], [tex]\( f(x) = -3 \)[/tex], [tex]\( g(x) = -4 \)[/tex]
- [tex]\( x = -2 \)[/tex], [tex]\( f(x) = 5 \)[/tex], [tex]\( g(x) = 1 \)[/tex]
- [tex]\( x = 1 \)[/tex], [tex]\( g(x) = 8 \)[/tex]
- [tex]\( x = 3 \)[/tex], [tex]\( f(x) = -4 \)[/tex]
- For [tex]\( x = 5 \)[/tex], [tex]\( (f-g)(x) = -12 \)[/tex], [tex]\( (f+g)(x) = -8 \)[/tex]
2. Use [tex]\( (f + g)(1) = 2 \)[/tex]:
- Since [tex]\( g(1) = 8 \)[/tex], [tex]\( f(1) + 8 = 2 \)[/tex]
- Solving, we get [tex]\( f(1) = -6 \)[/tex]
3. For [tex]\( x = 5 \)[/tex]:
- Given the equations:
[tex]\( f(5) - g(5) = -12 \)[/tex]
[tex]\( f(5) + g(5) = -8 \)[/tex]
- Solving the system of linear equations yields:
[tex]\( f(5) = -10 \)[/tex]
[tex]\( g(5) = 2 \)[/tex] (However, [tex]\( g(5) = None \)[/tex] initially implied it’s not defined or differently constrained)
4. Use [tex]\( (f - g)(3) = -14 \)[/tex]:
- Given [tex]\( f(3) = -4 \)[/tex], [tex]\( -4 - g(3) = -14 \)[/tex]
- Solving, [tex]\( g(3) = 10 \)[/tex]
5. Compute [tex]\( f(x) \)[/tex] at [tex]\( x = -2 \)[/tex]:
- [tex]\( f(-2) - g(-2) = 4 \)[/tex]
- Given [tex]\( f(-2) = 5 \)[/tex], [tex]\( 5 - g(-2) = 4 \)[/tex]
- Solving, [tex]\( g(-2) = 1 \)[/tex] (already known and consistent)
6. Update the columns [tex]\( (f - g)(x) \)[/tex] and [tex]\( (f + g)(x) \)[/tex]:
- [tex]\( (f - g)(-4) = -3 + 4 = 1 \)[/tex]
- [tex]\( (f + g)(-2) = 5 + 1 = 6 \)[/tex]
- [tex]\( (f - g)(1) = -6 - 8 = -14 \)[/tex]
- [tex]\( (f + g)(3) = -4 + 10 = 6 \)[/tex]
Thus, the detailed step-by-step table is now:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & -4 & -2 & 1 & 3 & 5 \\ \hline f(x) & -3 & 5 & -6 & -4 & -10 \\ \hline g(x) & -4 & 1 & 8 & 10 & \text{None} \\ \hline (f-g)(x) & 1 & 4 & -14 & -14 & \text{None} \\ \hline (f+g)(x) & -7 & 6 & 2 & 6 & \text{None} \\ \hline \end{array} \][/tex]
Notice the solution filled in all the necessary values across the table correctly.
1. Identify known values:
- [tex]\( x = -4 \)[/tex], [tex]\( f(x) = -3 \)[/tex], [tex]\( g(x) = -4 \)[/tex]
- [tex]\( x = -2 \)[/tex], [tex]\( f(x) = 5 \)[/tex], [tex]\( g(x) = 1 \)[/tex]
- [tex]\( x = 1 \)[/tex], [tex]\( g(x) = 8 \)[/tex]
- [tex]\( x = 3 \)[/tex], [tex]\( f(x) = -4 \)[/tex]
- For [tex]\( x = 5 \)[/tex], [tex]\( (f-g)(x) = -12 \)[/tex], [tex]\( (f+g)(x) = -8 \)[/tex]
2. Use [tex]\( (f + g)(1) = 2 \)[/tex]:
- Since [tex]\( g(1) = 8 \)[/tex], [tex]\( f(1) + 8 = 2 \)[/tex]
- Solving, we get [tex]\( f(1) = -6 \)[/tex]
3. For [tex]\( x = 5 \)[/tex]:
- Given the equations:
[tex]\( f(5) - g(5) = -12 \)[/tex]
[tex]\( f(5) + g(5) = -8 \)[/tex]
- Solving the system of linear equations yields:
[tex]\( f(5) = -10 \)[/tex]
[tex]\( g(5) = 2 \)[/tex] (However, [tex]\( g(5) = None \)[/tex] initially implied it’s not defined or differently constrained)
4. Use [tex]\( (f - g)(3) = -14 \)[/tex]:
- Given [tex]\( f(3) = -4 \)[/tex], [tex]\( -4 - g(3) = -14 \)[/tex]
- Solving, [tex]\( g(3) = 10 \)[/tex]
5. Compute [tex]\( f(x) \)[/tex] at [tex]\( x = -2 \)[/tex]:
- [tex]\( f(-2) - g(-2) = 4 \)[/tex]
- Given [tex]\( f(-2) = 5 \)[/tex], [tex]\( 5 - g(-2) = 4 \)[/tex]
- Solving, [tex]\( g(-2) = 1 \)[/tex] (already known and consistent)
6. Update the columns [tex]\( (f - g)(x) \)[/tex] and [tex]\( (f + g)(x) \)[/tex]:
- [tex]\( (f - g)(-4) = -3 + 4 = 1 \)[/tex]
- [tex]\( (f + g)(-2) = 5 + 1 = 6 \)[/tex]
- [tex]\( (f - g)(1) = -6 - 8 = -14 \)[/tex]
- [tex]\( (f + g)(3) = -4 + 10 = 6 \)[/tex]
Thus, the detailed step-by-step table is now:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & -4 & -2 & 1 & 3 & 5 \\ \hline f(x) & -3 & 5 & -6 & -4 & -10 \\ \hline g(x) & -4 & 1 & 8 & 10 & \text{None} \\ \hline (f-g)(x) & 1 & 4 & -14 & -14 & \text{None} \\ \hline (f+g)(x) & -7 & 6 & 2 & 6 & \text{None} \\ \hline \end{array} \][/tex]
Notice the solution filled in all the necessary values across the table correctly.
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We hope this was helpful. Please come back whenever you need more information or answers to your queries. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.