To complete the table for the functions \( f(x) = 4(2^x) \) and \( g(x) = 5x + 6 \), we will calculate the values for each function at \( x = 1, 2, 3, \) and \( 4 \).
Calculations
For \( f(x) = 4(2^x) \):
- \( f(1) = 4(2^1) = 4 \times 2 = 8 \)
- \( f(2) = 4(2^2) = 4 \times 4 = 16 \)
- \( f(3) = 4(2^3) = 4 \times 8 = 32 \)
- \( f(4) = 4(2^4) = 4 \times 16 = 64 \)
For \( g(x) = 5x + 6 \):
- \( g(1) = 5(1) + 6 = 5 + 6 = 11 \)
- \( g(2) = 5(2) + 6 = 10 + 6 = 16 \)
- \( g(3) = 5(3) + 6 = 15 + 6 = 21 \)
- \( g(4) = 5(4) + 6 = 20 + 6 = 26 \)
Completed Table
| x | f(x) | g(x) | |---|------|------| | 1 | 8 | 11 | | 2 | 16 | 16 | | 3 | 32 | 21 | | 4 | 64 | 26 |
Analysis of Growth
As \( x \) gets larger, we see:
- At \( x = 1 \), \( f(1) = 8 \) and \( g(1) = 11 \) ( \( g(1) > f(1) \) )
- At \( x = 2 \), \( f(2) = 16 \) and \( g(2) = 16 \) ( \( g(2) = f(2) \) )
- At \( x = 3 \), \( f(3) = 32 \) and \( g(3) = 21 \) ( \( f(3) > g(3) \) )
- At \( x = 4 \), \( f(4) = 64 \) and \( g(4) = 26 \) ( \( f(4) > g(4) \) )
Conclusion
Although both functions grow as \( x \) increases, \( g(x) \) is greater than \( f(x) \) for \( x = 1 \) and equal for \( x = 2 \). However, \( f(x) \) exceeds \( g(x) \) starting from \( x = 3 \) and grows much faster thereafter due to its exponential growth compared to the linear growth of \( g(x) \).
Therefore, the function \( f(x) \) eventually exceeds \( g(x) \) starting from \( x = 3 \).