Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
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.
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. 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. Stay informed by coming back for more detailed answers.